Category: SEMESTER-3

CALCULAS 1ST ORDER DERIVATIVE Part II

DIFFERENTIATION SOLUTION OF SNDEY SEM-3

অবকলন SEMESTER-3 Unit-3 CALCULAS 1st order derivative

1ST ORDER DERIVATIVE Part I দেখতে এখানে CLICK করো।

Fill in the Blanks

1. ^d/_dx cos(logx) = _______
Ⓐ – sin(logx) Ⓑ ¹/_x cos(logx)
Ⓒ –¹/_x sin(logx) Ⓓ ¹/_x sin(logx)

Solution: ^d/_dx cos(logx)
= – sin(logx). ^d/_dx(logx)
= – sin(logx). ¹/_x = –¹/_x sin(logx)
Ans: Ⓒ –¹/_x sin(logx)

2. ^d/_dxlog(sin x) = _______
Ⓐ tan x Ⓑ cot x Ⓒ – tan x Ⓓ – cot x

Solution: ^d/_dxlog(sin x)
= ¹/_sinx. ^d/_dx(sin x)
= ¹/_sinx.cosx = cotx
Ans: Ⓑ cot x

3. ^d/_dx(cot x°) = _______
Ⓐ -cosec³ x° Ⓑ cosec³ x°
Ⓒ –¹⁸⁰/_πcosec² x°
Ⓓ –^π/₁₈₀cosec² x°

Solution: ^d/_dx(cot x°)
=^d/_dx(cot ^πx/₁₈₀)
= – cosec² ^πx/₁₈₀. ^π/₁₈₀
= –^π/₁₈₀cosec² x°
Ans: Ⓓ –^π/₁₈₀cosec² x°

4. y = 10^{10^x} হলে, ^dy/_dx = _______
Ⓐ 10^{10^x}(log_e10)² Ⓑ 10^{10^x}log_e10
Ⓒ 10^{10^x}.10^xlog_e10 Ⓓ 10^{10^x}.10^x(log_e10)²

Solution: y = 10^{10^x}
∴^dy/_dx = 10^{10^x}.log_e10.^d/_dx(10^x)
= 10^{10^x}.log_e10.10^x.log_e10
= 10^{10^x}.10^x(log_e10)²
Ans: Ⓓ 10^{10^x}.10^x(log_e10)²

\(5.\ f(x) = \sqrt{x^2 + 1}\) হলে \(f'(-1) =\) _______

Ⓐ ¹/_√2 Ⓑ –¹/_√2
Ⓒ 1 Ⓓ -1

\(Solution:\ f(x) = \sqrt{x^2 + 1}\\∴ f'(x) =\frac{1}{2\sqrt{x^2 + 1}}.\frac{d}{dx}(x^2 + 1)\\=\frac{1}{2\sqrt{x^2 + 1}}.2x\\=\frac{x}{\sqrt{x^2 + 1}}\\∴ f'(-1) =\frac{-1}{\sqrt{(-1)^2 + 1}}=\frac{-1}{√2}\\Ans:\ Ⓑ\ \frac{-1}{√2}\)

6. y = ¹/_{3x – 1} হলে, x = 0 বিন্দুতে ^dy/_dx = _______
Ⓐ -3 Ⓑ ¹/₃
Ⓒ –¹/₉Ⓓ 3

Solution: y = ¹/_{3x – 1} = (3x – 1)^-1
∴ ^dy/_dx =-1(3x – 1)^-2.^d/_dx(3x – 1)
=-1(3x – 1)^-2.3
x = 0 বিন্দুতে,
^dy/_dx = -1(3.0 – 1)^-2.3
= -1(-1)^-2.3 = -3
Ans: Ⓐ -3

7. f(x) = log(tan^x/₂) হলে f'(x) = _______
Ⓐ sin x Ⓑ ¹ / _{2sin^x/₂cos ^x/₂}
Ⓒ -cosecx Ⓓ cosecx

Solution: f(x) = log(tan^x/₂)
∴ f'(x) = ¹ / _tan^x/₂ .sec² ^x/₂ .¹/₂
= ¹ / _{2sin^x/₂cos ^x/₂}
= ¹/_sinx = cosecx
Ans: Ⓓ cosecx

8. logx-এর সাপেক্ষে x² -এর অবকল সহগ = _______
Ⓐ 2x²Ⓑ 2
Ⓒ 2x Ⓓ 2x³

Solution: ^d/_d(logx)(x²)
= ^d/_dx(x²).¹/_{^d/_dx(logx)}
= 2x.¹/_{¹/_x} = 2x²
Ans: Ⓐ 2x²

9. যদি y = cosec³x হয়, তবে ^dy/_dx = _______
Ⓐ – 3cosecxcot x
Ⓑ – 3cosec³xcot³ x
Ⓒ – cosec³xcot x
Ⓓ – 3cosec³xcot x

Solution: y = cosec³x
∴ ^dy/_dx = 3cosec²x.^d/_dx(cosecx)
= 3cosec²x.(- cosecxcotx)
= – 3cosec³xcot x
Ans: Ⓓ – 3cosec³xcot x

10. যদি y = m^{ax2 + bx + c} হয়, তবে ^dy/_dx = _______
Ⓐ (2ax + b)m^{ax2 + bx + c}.log_em
Ⓑ (2ax + b)m^{ax2 + bx + c}
Ⓒ m^{ax2 + bx + c}.log_em
Ⓓ (2ax + b).log_em

Solution: y = m^{ax2 + bx + c}
∴ ^dy/_dx =m^{ax2 + bx + c}.log_em.^d/_dx(ax² + bx + c)
= m^{ax2 + bx + c}.log_em.(2ax + b)
Ans: Ⓐ (2ax + b)m^{ax2 + bx + c}.log_em

11. y = log(ax + b)³ হলে ^dy/_dx = _______
Ⓐ ³/_{ax + b} Ⓑ ^3x/_{ax + b}
Ⓒ ^3a/_{ax + b}Ⓓ এদের কোনোটিই নয়

Solution: y = log(ax + b)³ = 3log(ax + b)
∴ ^dy/_dx
= 3.¹/_{ax + b}.(a.1) = ^3a/_{ax + b}
Ans: Ⓒ ^3a/_{ax + b}

\(12.\ y = \sqrt{tan^{-1}x}\) হলে \(\frac{dy}{dx} =\ Ⓐ\ \frac{1}{2(1 + x^2)\sqrt{tan^{-1}x}}\\Ⓑ\ \frac{2}{(1 + x^2)\sqrt{tan^{-1}x}}\\Ⓒ\ -\frac{1}{2(1 + x^2)\sqrt{tan^{-1}x}}\\Ⓓ\ \frac{1}{\sqrt{tan^{-1}x}}\)

\(Solution:\ y = \sqrt{tan^{-1}x}\\∴ \frac{dy}{dx} =\frac{1}{2\sqrt{tan^{-1}x}}.\frac{d}{dx}(tan^{-1}x)\\=\frac{1}{2\sqrt{tan^{-1}x}}.\frac{1}{1 + x^2}\\=\frac{1}{2(1 + x^2)\sqrt{tan^{-1}x}}\\Ans:\ Ⓐ\ \frac{1}{2(1 + x^2)\sqrt{tan^{-1}x}}\)

13. যদি y = f(x)ⁿ হয়, তবে ^dy/_dx = ______
Ⓐ n{f(x)}^{n – 1}
Ⓑ n{f(x)}^{n – 1}.f'(x)
Ⓒ n{f(x)}ⁿ .f'(x)
Ⓓ nf'(x)

Solution: y = f(x)ⁿ
∴ ^dy/_dx = nf(x)^{n – 1}.^d/_dx{f(x)}
= n{f(x)}^{n – 1}.f'(x)
Ans: Ⓑ n{f(x)}^{n – 1}.f'(x)

14. যদি y = e^φ(x) হয়, তবে ^dy/_dx = ______
Ⓐ e^φ(x). φ'(x)
Ⓑ e^φ‘^(x) Ⓒ e^φ(x)
Ⓓ এদের কোনোটিই নয়

Solution: y = e^φ(x)
∴ ^dy/_dx
= e^φ(x).^d/_dx{φ(x)} = e^φ(x). φ'(x)
Ans: Ⓐ e^φ(x). φ'(x)

15. যদি \(y = log{\sqrt{f(x)}}\) হয়, তবে \(\frac{dy}{dx}=\\Ⓐ\ \frac{1}{2}f'(x)\quad Ⓑ\ \frac{1}{2f'(x)}\\Ⓒ\ \frac{f'(x)}{2f(x)}\quad Ⓓ\ \frac{2f'(x)}{f(x)}\\Solution:\ y = log{\sqrt{f(x)}}\\∴ \frac{dy}{dx}= \frac{1}{\sqrt{f(x)}}.\frac{d}{dx}(f(x))\\= \frac{1}{\sqrt{f(x)}}.\frac{1}{2\sqrt{f(x)}}.f'(x)\\= \frac{f'(x)}{2f(x)}\\Ans:\ Ⓒ\ \frac{f'(x)}{2f(x)}\)

SEMESTER-3
সূচিপত্র

👉 UNIT-1 সম্বন্ধ ও অপেক্ষক

1. সম্বন্ধ
2. অপেক্ষক
3. বিপরীত ত্রিকোণমিতিক অপেক্ষকসমূহ

👉 UNIT-2 বীজগণিত

1. ম্যাট্রিক্সের প্রকারভেদ ও ম্যাট্রিক্স বীজগণিত
- ম্যাট্রিক্স PART 1
- ম্যাট্রিক্স PART 2
2. নির্ণায়ক
- মাইনর ও সহ-উৎপাদক
3. একটি ম্যাট্রিক্সের অ্যাডজয়েন্ট ও বিপরীত ম্যাট্রিক্স এবং সরল সহসমীকরণের সমাধান

👉 UNIT-3 কলনবিদ্যা

1. সন্ততা এবং অন্তরকলনযোগ্যতা
- সন্ততা
- অন্তরকলনযোগ্যতা
2. অবকলন বা অন্তরকলন
- অবকলন – I
- অবকলন PART – II
3. দ্বিতীয় ক্রমের অন্তরকলজ
4. অন্তরকলজের ব্যাখ্যা
5. স্পর্শক ও অভিলম্ব
6. বর্ধিষ্ণু ও ক্ষয়িষ্ণু অপেক্ষক
. চরম ও অবম মান

👉 UNIT-4 সম্ভাবনা

1. সম্ভাবনা
- সম্ভাবনা
- Bayes’ উপপাদ্য
2. সমসম্ভব চলক ও তার বিভাজন
3. দ্বিপদ বিভাজন

👉 Semester III -এর প্রশ্নপত্রের সম্পূর্ণ সমাধান

Semester III 2025 এর প্রশ্নপত্র ও তার সম্পূর্ণ সমাধান

16. যদি y = 3^u(x) হয়, তবে ^dy/_dx = ______
Ⓐ3^u(x).log3
Ⓑ 3^u(x).u'(x)
Ⓒ log3.u'(x)
Ⓓ 3^u(x).log3.u'(x)

Solution: y = 3^u(x)
∴ ^dy/_dx
=3^u(x)log3.^d/_dx{u(x)}
= 3^u(x).log3.u'(x)
Ans: Ⓓ 3^u(x).log3.u'(x)

17. যদি y = sin{f(x)} হয়, তবে ^dy/_dx = ______
Ⓐ cos{f(x)}.f'(x)
Ⓑ sec{f(x)}.f'(x)
Ⓒ sin^-1{f(x)}f'(x)
Ⓓএদের কোনোটিই নয়

Solution: y = sin{f(x)}
∴ ^dy/_dx
= cos f(x).^d/_dx{f(x)}
= cos{f(x)}.f'(x)
Ans: Ⓐ cos{f(x)}.f'(x)

18. যদি y = sec{φ(x)} হয়, তবে ^dy/_dx = ______
Ⓐ secφ(x).tan^-1φ(x).φ'(x)
Ⓑ sec^-1φ(x).tanφ(x).φ'(x)
Ⓒ sec^-1φ(x).ctan^-1φ(x).φ'(x)
Ⓓ secφ(x)tanφ(x).φ'(x)

Solution: y = sec{φ(x)}
∴ ^dy/_dx
= secφ(x)tanφ(x).^d/_dx{φ(x)}
= secφ(x)tanφ(x).φ'(x)
Ans: Ⓓ secφ(x)tan ¹φ(x).φ'(x)

\(19.\ \frac{d}{dx}(\sqrt{logx})^3 = \\Ⓐ\ \frac{3\sqrt{logx}}{2x}\quad Ⓑ\ \frac{2\sqrt{logx}}{3x}\\Ⓒ\ \frac{\sqrt{logx}}{2x}\quad Ⓓ\ \frac{\sqrt{logx}}{3x}\\Solution:\ \frac{d}{dx}(\sqrt{logx})^3\\=3(\sqrt{logx})^2.\frac{1}{2\sqrt{logx}}.\frac{1}{x}\\= \frac{3\sqrt{logx}}{2x}\\Ans:\ Ⓐ\ \frac{3\sqrt{logx}}{2x}\)

20. ^d/_dx(e^x⁹) = ______
Ⓐ 9x⁸.e^x⁹ Ⓑ x⁸.e^x⁹
Ⓒ 9e^x⁸ Ⓓ 8e^x⁸

Solution: ^d/_dx(e^x⁹)
= e^x⁹.^d/_dx(x⁹)
= e^x⁹.9x⁸ = 9x⁸.e^x⁹
Ans: Ⓐ 9x⁸.e^x⁹

21. ^d/_dxlog(cot x) = ______
Ⓐ secx cosecx
Ⓑ – secx cosecx
Ⓒ sinxcosx
Ⓓ – sinxcosx

Solution: ^d/_dxlog(cot x)
= ¹/_{cot x}.(-cosec² x)
= – secx cosecx
Ans: Ⓑ – secx cosecx

22. ^d/_dxlog(logx) (x > 1) = ______
Ⓐ ¹/_logx Ⓑ ¹/_x²
Ⓒ xlogx Ⓓ ¹/_xlogx

Solution: ^d/_dxlog(logx)
= ¹/_logx.^d/_dx(logx)
= ¹/_logx.¹/_x = ¹/_xlogx
Ans: Ⓓ ¹/_xlogx

23. ^d/_dx(e^{e^x}) = ______
Ⓐ e^{e^x} Ⓑ (e^x)² Ⓒ e^x.e^{e^x}
Ⓓএদের কোনোটিই নয়

Solution: ^d/_dx(e^{e^x})
= e^{e^x}.^d/_dx(e^x)
= e^{e^x}.e^x = e^x.e^{e^x}
Ans: Ⓒ e^x.e^{e^x}

24. x-এর সাপেক্ষে log(cosecx + cot x) এর অন্তরকলজ kcosecx হলে k = ______
Ⓐ 1 Ⓑ -1 Ⓒ 2 Ⓓ –¹/₂

Solution: ^d/_dx[log(cosecx + cot x)]
=¹/_{cosecx + cot x}.(- cosecxcot x – cosec²x)
= ^{– cosecx(cot x + cosecx)}/_{cosecx + cot x}
= -cosecx
Ans: Ⓑ -1

25. ^d/_dx(cosec⁶x) = ______
Ⓐ – cosec⁶xcotx
Ⓑ – 6cosec⁶xcotx
Ⓒ – 6cosecxcotx
Ⓓ- 6cosecxcot⁶x

Solution: ^d/_dx(cosec⁶x)
= 6cosec⁵x.(- cosecx.cotx)
= – 6cosec⁶xcotx
Ans: Ⓑ – 6cosec⁶xcotx

26. ^d/_dx(cot^-1x)⁴ = ^k/_{1 + x²} (cot^-1x)³ হলে k = ______
Ⓐ 2 Ⓑ -2 Ⓒ 4 Ⓓ -4

Solution: ^d/_dx(cot^-1x)⁴ = ^k/_{1 + x²}(cot^-1x)³
⇒ 4(cot^-1x)³.(- ¹/_{1 + x²}) = ^k/_{1 + x²}(cot^-1x)³
⇒ ^-4/_{1 + x²} = ^k/_{1 + x²}
∴ k = -4
Ans: Ⓓ -4

\(27. log⁡\left[ e^x\left( \frac{x – 2}{x + 2} \right)^\frac{3}{4} \right]\) অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ ______\(Ⓐ\ \frac{x^2- 1}{x^2 – 2}\quad Ⓑ\ \frac{x^2 – 1}{x^4 – 2}\\Ⓒ\ \frac{x^2 – 1}{x^2 – 4}\quad Ⓓ\ \frac{x^2 + 1}{x^2 + 4}\)

\(Solution: \frac{d}{dx}\left( log⁡\left[ e^x\left( \frac{x – 2}{x + 2} \right)^\frac{3}{4} \right] \right)\\=\frac{d}{dx}\left[ loge^x+\frac{3}{4}\left\{ log(x-2)-log(x+2) \right\} \right]\\=\frac{d}{dx}\left[x+\frac{3}{4}\left\{ log(x-2)-log(x+2) \right\} \right]\\=1+\frac{3}{4}\left[ \frac{1}{x-2}.1-\frac{1}{x+2}.1 \right]\\=1+\frac{3}{4}.\frac{x+2-x+2}{(x-2)(x+2)}\\=1+\frac{3}{4}.\frac{4}{x^2 – 4}\\=1+\frac{3}{x^2 – 4}\\=\frac{x^2 – 1}{x^2 – 4}\\Ans:\ Ⓒ\ \frac{x^2 – 1}{x^2 – 4}\)

28. tan^-1(^{sin x}/_{1 + cos x}) অপেক্ষকটির x -এর সাপেক্ষে অন্তরকলজ ______
Ⓐ ¹/₂ Ⓑ 1 Ⓒ 0 Ⓓ 2

Solution: ^d/_dx[tan^-1(^{sin x}/_{1 + cos x})]
= ^d/_dx[tan^-1(^{2sin ^x/₂ cos ^x/₂}/_{2cos² ^x/₂})]
=^d/_dx[tan^-1(tan ^x/₂)]
= ^d/_dx(^x/₂) = ¹/₂
Ans: Ⓐ ¹/₂

29. নীচের ______ অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ ^{2xlogxcosec2x – logtan x}/_x(logx)²
Ⓐ log_x sinx
Ⓑ log_sinx x
Ⓒ log_x tan x
Ⓓ log_{sin x}(sec(x) + 10^x²

\(Solution: \frac{d}{dx}(log_x⁡tanx)\\=\frac{d}{dx}\left( \frac{logtanx}{logx} \right)\\=\frac{\frac{1}{tanx}.sec^2⁡x.logx-logtanx.\frac{1}{x}}{(logx)^2}\\=\frac{\frac{xlogx}{sinxcosx}– logtanx}{x(logx)^2}\\=\frac{\frac{2xlogx}{sin2x}- logtanx}{x(logx)^2}\\=\frac{2xlogxcosec2x – logtanx}{x(logx)^2}\)

Ans: Ⓒ log_x tan x

30. ^d/_dx(x^x)^x = ______
Ⓐ x²(x + 2xlogx)
Ⓑ x^{x² + 1}(x + 2logx)
Ⓒ x^2x(x + 2xlogx)
Ⓓ x^x²(x + 2xlogx)

Solution: ^d/_dx(x^x)^x
= ^d/_dx(e^{xlogx^x})
= ^d/_dx(e^{x² logx})
=e^{x² logx}.^d/_dx(x² logx)
= (x^x)^x(2x.logx + x².¹/_x)
= x^x²(x + 2xlogx)
Ans: Ⓓ x^x²(x + 2xlogx)

31. y = tanx^sinx হলে, [^dy/_dx]_x_{= ^π/₄} = ______
Ⓐ 2 Ⓑ √2 Ⓒ √3 Ⓓ 1

Solution: y = tanx^sinx
∴ logy = logtanx^sinx = sinxlogtanx
∴ 1/y. ^dy/_dx = cosxlogtanx + sinx.¹/_tanxsec²x
⇒ ^dy/_dx = tanx^sinx [cosxlogtanx + secx]
[^dy/_dx]_{x = ^π/₄}
= (tan ^π/₄)^{sin ^π/₄} [cos ^π/₄ logtan ^π/₄ + sec ^π/₄]
= (1)^√2[¹/_√2 log1 + √2]
=1[0 + √2] = √2
Ans: Ⓑ √2

32. যদি xy + 1 = cos(xy) হয়, তবে x = ^π/₂, y = 0 বিন্দুতে ^dy/_dx = ______
Ⓐ 0 Ⓑ 1 Ⓒ 2 Ⓓ 3

Solution: xy + 1 = cos(xy)
∴ 1.y + x.^dy/_dx + 0 = -sin(xy)( 1.y + x.^dy/_dx)
x = ^π/₂, y = 0 বিন্দুতে,
1.0 + ^π/₂.^dy/_dx = -sin(^π/₂.0)( 1.0 + ^π/₂.^dy/_dx)
⇒ ^π/₂.^dy/_dx = -sin 0.^π/₂.^dy/_dx
⇒ ^π/₂.^dy/_dx = 0
∴ ^dy/_dx = 0
Ans: Ⓐ 0

33. x = a(cos t + logtan^t/₂) , y = a sint হলে ^dy/_dx = ______
Ⓐ sint Ⓑ cost
Ⓒ tant Ⓓ cot t

Solution: x = a(cos t + logtan^t/₂)

\(∴ \frac{dx}{dt}= a\left( -sint+ \frac{1}{tan\frac{t}{2}}.sec^2\frac{t}{2}.\frac{1}{2} \right)\\= a\left( -sint+ \frac{1}{2 sin⁡\frac{t}{2}.cos\frac{t}{2}} \right)\)

= a(-sint + ¹/_sin⁡t)
= a.^{1 – sin^2⁡⁡t}/_sin⁡t = ^{acos^2⁡⁡t}/_sin⁡t
y = a sint
∴ ^dy/_dt = acost

\(∴ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\= \frac{acost}{\frac{acos^2⁡t}{sin⁡t}}= tant\)

Ans: Ⓒ tant

34. যদি tan y = ^2t/_{1 – t²}, sin x = ^2t/_{1 + t²} হয়, তবে ^dy/_dx = ______
Ⓐ 0 Ⓑ 1 Ⓒ -1 Ⓓ 2

Solution: tan y = ^2t/_{1 – t²}
⇒ y = tan^-1 ^2t/_{1 – t²} = 2tan^-1 t
sin x = ^2t/_{1 + t²}
⇒ x = sin^-1 ^2t/_{1 + t²} = 2tan^-1 t
∴ y = x
∴ ^dy/_dx = 1
Ans: Ⓑ 1

35. যদি x = a(θ + sin θ) , y = a(1 + cos θ) হয়, তবে [^dy/_dx]_θ_{= ^π/₂} ______
Ⓐ 0 Ⓑ -1 Ⓒ -2 Ⓓ 6

Solution: x = a(θ + sin θ)
∴ ^dx/_dθ = a(1 + cos θ),
y = a(1 + cos θ)
∴ ^dy/_dθ = – asin θ
∴ ^dy/_dx = ^dy/dθ / _dx/dθ
= ^{-asin θ} / _{a(1 + cos θ)}
= – ^{2sin ^θ/₂cos ^θ/₂} / _{2cos ^θ/₂}
=- tan ^θ/₂
∴ [^dy/_dx]_{θ = ^π/₂}
= – tan ^π/₄ = – 1
Ans: Ⓑ -1

36. যদি \(y= sin\sqrt{x^2+a^2}\) হয়, তবে \(\frac{dy}{dx} =\) ______

\(Ⓐ\ \frac{xsin\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\quad Ⓑ\ \frac{cos\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\\Ⓒ\ \frac{xcos\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\quad Ⓓ\ \frac{cos\sqrt{x^2+a^2}}{x^2- a^2}\)

\(Solution:\ y = sin\sqrt{x^2+a^2}\\∴ \frac{dy}{dx}= cos\sqrt{x^2+a^2}.\frac{1}{2\sqrt{x^2+a^2}}.2x=\frac{xcos\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\\Ans:\ Ⓒ\ \frac{xcos\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\)

\(37.\ y = \frac{x^3\sqrt{x^2 – 12}}{\sqrt[3]{20 – x}}\) হলে, \(\left[ \frac{dy}{dx} \right]_{x=4}=\)______

Ⓐ 100 Ⓑ 110
Ⓒ 120 Ⓓ 140

\(Solution: y = \frac{x^3\sqrt{x^2 – 12}}{\sqrt[3]{20 – x}}\\∴logy = log \frac{x^3\sqrt{x^2 – 12}}{\sqrt[3]{20 – x}} = 3logx + \frac{1}{2}log(x^2 – 12) – \frac{1}{3}log(20 – 3x)\\∴ \frac{1}{y}.\frac{dy}{dx}= \frac{3}{x}+ \frac{1}{2}.\frac{1}{(x^2 – 12)}.2x – \frac{1}{3}.\frac{1}{20 -3x}.(-3)\\⇒ \frac{dy}{dx} = \frac{x^3\sqrt{x^2 – 12}}{\sqrt[3]{20 – x}}\left[ \frac{3}{x} + \frac{x}{x^2 – 12} + \frac{1}{20 -3x} \right]\\∴ \left[ \frac{dy}{dx} \right]_{x=4}\\= \frac{4^3\sqrt{4^2 – 12}}{\sqrt[3]{20 – 4}}\left[ \frac{3}{4} + \frac{4}{16 – 12} + \frac{1}{20 -3.4} \right]\\= \frac{64.2}{2}[\frac{3}{4} + 1 + \frac{1}{8}]\)

= 48 + 64 + 8 = 120
Ans: Ⓒ 120

38. tan² y = ^{1 + cos 2x}/_{1 – cos 2x} হলে, ^dy/_dx = ______
Ⓐ 0 Ⓑ ±1
Ⓒ ±2 Ⓓ ±4

Solution: tan² y = ^{1 + cos 2x}/_{1 – cos 2x}
⇒ tan² y = cot²x
⇒ tan y = ± cotx = tan(^π/₂ ± x)
∴ y = ^π/₂ ± x
∴ ^dy/_dx = ±1
Ans: Ⓑ ±1

39. y = ^{1 + sin} ^θ ^{+ cos} ^θ/_{1 + sin} _θ _{– cos} _θ হলে ^dy/_dθ + ¹/_{1 – cos} _θ = ______
Ⓐ 0 Ⓑ 1 Ⓒ -1 Ⓓ 2

Solution: y = ^{1 + sin} ^θ ^{+ cos} ^θ/_{1 + sin} _θ _{– cos} _θ
⇒ y = ^{1 + cos} ^θ ^{+ sin} ^θ/_{1 – cos} _{θ +} _sin _θ

\(⇒\frac{2cos^2 \frac{θ}{2} + 2sin \frac{θ}{2}.cos\frac{θ}{2}}{2sin^2\frac{θ}{2} + 2sin\frac{θ}{2}.cos\frac{θ}{2}}\\=\frac{2cos\frac{θ}{2}(cos\frac{θ}{2}+ sin\frac{θ}{2})}{2sin\frac{θ}{2}(sin\frac{θ}{2} + cos\frac{θ}{2})}\\=cot\frac{θ}{2}\\∴ \frac{dy}{dθ}=-cosec^2\frac{θ}{2}.\frac{1}{2} \\=\frac{1}{2sin^2\frac{θ}{2}}\\=\frac{1}{1 -cos⁡2.\frac{θ}{2}}\\=\frac{1}{1 – cosθ}\)

∴ ^dy/_dθ – ¹/_{1 – cosθ} = 0
Ans: Ⓐ 0

40. sin y = xsin(a + y) হলে, যদি ^dy/_dx = ^sina/_{1 – kxcosa + x²} হয়, তবে k = ______
Ⓐ 1 Ⓑ 2 Ⓒ -2 Ⓓ -1

Solution: sin y = xsin(a + y)
⇒ sin y = xsina cosy + xcosa siny
⇒sin y(1 – xcosa) = xsinacosy
⇒ tan y = ^xsina/_{1 – xcosa}
⇒ y = tan^-1 ^xsina/_{1 – xcosa}

\(∴ \frac{dy}{dx}=\frac{1}{1 + \left( \frac{xsina}{1 – xcosa} \right)^2}.\frac{sina(1 – xcosa) – xsina(-1.cosa)}{(1 – xcosa)^2}\\⇒ \frac{dy}{dx}=\frac{(1 – xcosa)^2}{(1 – xcosa)^2 + (xsina)^2}.\frac{sina -xsinacosa + xsinacosa}{(1 – xcosa)^2}\\⇒ \frac{dy}{dx}=\frac{sina}{1- 2xcosa + x^2 cos^2 a + x^2 sin^2 a}\\⇒ \frac{dy}{dx}=\frac{sina}{1 – 2xcosa + x^2}\\∴ k = 2\\Ans:\ Ⓑ\ 2\)

\(41.\ \frac{d}{dx}\left[\frac{sin x}{sin(x – \frac{π}{4})} \right]= kcosec^2 \left( x – \frac{π}{4} \right)\)

হলে k-এর মান ______
Ⓐ ¹/₂ Ⓑ –¹/_√2
Ⓒ ¹/₃ Ⓓ –¹/_√3

\(Solution:\ \frac{d}{dx}\left[\frac{sin x}{sin(x – \frac{π}{4})} \right]\\=\frac{cos x.sin(x – \frac{π}{4}) -sinx.cos(x – \frac{π}{4}).1}{sin^2 (x – \frac{π}{4})}\\= \frac{sin(x – \frac{π}{4} – x)}{sin^2 (x – \frac{π}{4})}\\= \frac{-sin \frac{π}{4}}{sin^2 (x – \frac{π}{4})}\)

= –¹/_√2cosec² (x – ^π/₄)
∴ k = – ¹/_√2
Ans: Ⓑ –¹/_√2

42. নীচের ______ সম্পর্কটির ক্ষেত্রে ^dy/_dx = ^y/_x হবে।
Ⓐ x^y = y^x Ⓑ x^y.y^x = 1
Ⓒ (cos x)^y = (sin y)^x
Ⓓ x³y⁴ = (x + y)⁷

Solution: x³y⁴ = (x + y)⁷
∴ log x³y⁴ = (x + y)⁷
⇒ 3logx + 4logy = 7log(x + y)
∴ ³/_x + ⁴/_y.^dy/_dx = ⁷/_{x + y}(1 + ^dy/_dx)
⇒ ^dy/_dx(⁴/_y – ⁷/_{x + y}) = ⁷/_{x + y} – ³/_x
⇒^dy/_dx.^{4x + 4y – 7y}/_{y(x + y)} = ^{7x – 3x – 3y}/_{x(x + y)}
⇒ ^dy/_dx.^{4x – 3y}/_y = ^{4x – 3y}/_x
∴ ^dy/_dx = ^y/_x
Ans: Ⓓ x³y⁴ = (x + y)⁷

43. ^d/_dx[(sin x)^{cos x} + (cos x)^{sin x}] = ______
Ⓐ (sin x)^{sin x}[sin x tan x – cos xlog cos x] + (sin x)^{cos x}[tan xlog tan x – sin x tan x ]
Ⓑ (cos x)^{sin x}[cos x cotx + sin xlog cos x] + (sin x)^{sin x}[cot xlog tan x + sin x cot x ]
Ⓒ (sin x)^{cos x}[cos x cotx – sin xlog sin x] + (cos x)^{sin x}[cos xlog cos x – sin x tan x]
Ⓓ এদের কোনোটিই নয়

Solution: ^d/_dx[(sin x)^{cos x} + (cos x)^{sin x}]
= ^d/_dx(sin x)^{cos x} + ^d/_dx(cos x)^{sin x}
= ^d/_dx[e^{cosxlog(sin x)}] + ^d/_dx[e^{sin xlog(cosx)}]
=e^{cosxlog(sin x)}.[ cosx.¹/_sinx.cosx + (-sinx).log(sinx)] + e^{sin xlog(cosx)}.[cosx.log(cosx) + sinx.¹/_cosx.(-sinx)]
= (sin x)^{cos x}[cos x cotx – sin xlog sin x] + (cos x)^{sin x}[cos xlog cos x – sin x tan x]
Ans: Ⓒ (sin x)^{cos x}[cos x cotx – sin xlog sin x] + (cos x)^{sin x}[cos xlog cos x – sin x tan x]

\(44.\ y = sin[\left[ 2tan^{-1}\sqrt{\frac{1-x}{1 + x}} \right]⁡\) হলে, \(\frac{dy}{dx}= ——\\Ⓐ\ \frac{x}{\sqrt{1 – x^2}}\quad Ⓑ\ -\frac{x}{\sqrt{1 – x^2}}\\Ⓒ\ -\frac{1}{\sqrt{1 – x^2}}\quad Ⓓ\ \frac{1}{\sqrt{1 – x^2}}\)

Solution: ধরি, x = cos θ

\(∴ y = sin\left[ 2tan^{-1}\sqrt{\frac{1-x}{1 + x}} \right]\\=sin\left[ 2tan^{-1}\sqrt{\frac{1-cos θ}{1 + cos θ}} \right]\\=sin\left[ 2tan^{-1}\sqrt{\frac{2sin^2 \frac{θ}{2}}{2cos^2 \frac{θ}{2}}} \right]\\=sin\left[ 2tan^{-1}\sqrt{tan \frac{θ}{2}} \right]\\=sin2.\frac{θ}{2}\\=sinθ= \sqrt{1-x^2}\\∴ \frac{dy}{dx}=\frac{1}{2\sqrt{1-x^2}}.(-2x)= -\frac{x}{\sqrt{1-x^2}}\\Ans:\ Ⓑ\ -\frac{x}{\sqrt{1 – x^2}}\)

45. x = ^{t – e^-t²}/_2t² হলে t ^dx/_dt + 2x – ¹/_2t = ______
Ⓐ e^-t Ⓑ e^{^-t²}
Ⓒ e^-2t Ⓓ e^-2t²

Solution: x = ^{t – e^-t²}/_2t²
⇒ x.2t² = t – e^-t²
⇒ ^dx/_dt.2t² + 4tx = 1 – e^-t²(-2t)
⇒t.^dx/_dt + 2x = ¹/_2t + e^-t²
⇒ t.^dx/_dt + 2x – ¹/_2t = e^-t²
Ans: Ⓑ e^-t²

Column Matching

1. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।স্তম্ভ Aস্তম্ভ B

স্তম্ভ A	স্তম্ভ B
[i] ^d/_dx(sin^{-1 2x}/_{1 + x}²) =	[a] ¹_{/(1 + x)√x}
[ii] ^d/_dx(sin^-1(e^x2) =	[b] ²/_{1 + x2}
[iii] ^d/_dx(cos^-1 ^{1 – x}/_{1 + x}) =	[c] ^2x/_{1 + x⁴}
[iv] ^d/_dx(tan^-1(x²) =	[d] ^2xex2/_{sqt(1 – e(2×2))}

Ⓐ [i] – [b], [ii] – [d], [iii] – [a], [iv] – [c]
Ⓑ [i] – [b], [ii] – [a], [iii] – [d], [iv] – [c]
Ⓒ [i] – [c], [ii] – [a], [iii] – [d], [iv] – [b]
Ⓓ [i] – [d], [ii] – [a], [iii] – [b], [iv] – [c]

Solution: [i] ^d/_dx(sin^{-1 2x}/_{1 + x}²)
= ^d/_dx(tan^-12x)
= ²/_{1 + x2} → [b]

\([ii]\ \frac{d}{dx}\left( sin^{-1}⁡\left( {e^{x^2}} \right) \right)\\= \frac{1}{\sqrt{1-(e^{x^2})^2}}.e^{x^2}.2x\\=\frac{2xe^{x^2}} {\sqrt{1-e^{2x^2}}}\) → [d]

[iii] ^d/_dx(cos^-1 ^{1 – x}/_{1 + x})
= ^d/_dx(2tan^-1 √x)
= 2.¹/_{1 + (√x)²}.¹/_2√x = ¹_{/(1 + x)√x} → [a]
[iv] ^d/_dx(tan^-1(x²)
= ¹/_{1 + (x²)²}.2x
= ^2x/_{1 + x⁴} → [a]
Ans: Ⓐ [i] – [b], [ii] – [d], [iii] – [a], [iv] – [c]

2. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

স্তম্ভ A	স্তম্ভ B
[i] ^d/_dx(sin 4xcos x) =	[a] 8(2sin8x – sin4x)
[ii] ^d/_dx(4sin2xsin 6x) =	[b] – 2sin4x – sin2x
[iii] ^d/_dx(sinxsin2xsin3x) =	[c] ¹/₂(5cos5x + 3cos3x)
[iv] ^d/_dx(cosxcos 3x) =	[d] ¹/₂(cos2x + 2cos4x – 3cos6x)

Solution: [i] ^d/_dx(sin 4xcos x)
=^d/_dx(¹/₂.2sin 4xcos x)
= ¹/₂×^d/_dx(sin 5x + sin 3x)
= ¹/₂(5cos5x + 3cos3x) → [c]
[ii] ^d/_dx(4sin2xsin 6x)
= 2×^d/_dx(2sin2xsin6x)
= 2×^d/_dx(cos4x – cos8x)
=2(-4sin4x + 8sin8x)
= 8(2sin8x – sin4x)→ [a]

[iii] ^d/_dx(sinxsin2xsin3x)
= ^d/_dx[¹/₂.(2sinxsin2x)sin3x]
= ¹/₂×^d/_dx[(cosx – cos3x)sin3x]
=¹/₄[(2cosx sin3x) – (2cos3xsin3x)]
= ¹/₄[sin 4x + sin 2x – sin6x]
=¹/₄(4cos4x + 2cos2x – 6cos6x)
= ¹/₂(cos2x + 2cos4x – 3cos6x) → [d]
[iv] ^d/_dx(cosxcos 3x)
=^d/_dx(¹/₂.2cos3xcos x)
=¹/₂×^d/_dx(-4cos4x – 2cos2x)
= ¹/₂(-4sin4x-2sin2x)
= -2sin4x – sin2x→ [b]
Ans: Ⓒ [i] – [c], [ii] – [a], [iii] – [d], [iv] – [b]

3. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

স্তম্ভ A	স্তম্ভ B
[i] ^d/_dx(e^ax cos bx) =	[a] sin^{p – 1}x cos^{q – 1}x×(pcos²x – qsin²x)
[ii] ^d/_dx (sin^px cos^q x) =	[b] e^ax(acos bx – bsin bx)
[iii] ^d/_dx(sin^px/cos^q x) =	[c] ¹/₂ e^xcos x(sin 2x + 3cos 2x – 1)
[iv] ^d/_dx(e^xsinxcos²x) =	[d] sin^{p – 1}x/cos^{q + 1}x (pcos²x + qsin²x)

Ⓐ [i] – [b], [ii] – [], [iii] – [a], [iv] – [c]
Ⓑ [i] – [b], [ii] – [a], [iii] – [d], [iv] – [c]
Ⓒ [i] – [c], [ii] – [a], [iii] – [d], [iv] – [b]
Ⓓ [i] – [d], [ii] – [a], [iii] – [b], [iv] – [c]

Solution: [i] ^d/_dx(e^ax cos bx)
= e^ax.a.cos bx + e^ax (-sinbx.b)
= e^ax(acos bx – bsin bx) → [b]
[ii] ^d/_dx(sin^px cos^q x)
= (psin^{p – 1}x.cosx)cos^qx + sin^px.(qcos^{q – 1}x.-sinx)
=sin^{p – 1}x.cos^{q – 1}x.pcos²x + sin^{p – 1}x.cos^{q – 1}x.qsin²x
= sin^{p – 1}xcos^{q – 1}x(pcos²x – qsin²x) → [a]

\([iii]\ \frac{d}{dx}\left( \frac{sin^p⁡x}{cos^q⁡x} \right)\\= \frac{(psin^{p-1}x.cosx)cos^qx – sin^p⁡x(q cos^{q-1}⁡x.(-sinx))}{(cos^q⁡x)^2}\\= \frac{psin^{p-1}x cos^{q+1}x + q sin^{p+1}x cos^{q-1}⁡x}{cos^{2q}⁡x}\\ = \frac{sin^{p-1}x cos^{q-1}x (pcos^2 x + q sin^2x)}{cos^2q⁡x}\\= \frac{sin^{p-1}⁡x}{cos^{q+1})x}(pcos^2 x+qsin^2⁡x)\) → [d]

[iv] ^d/_dx(e^xsinxcos²x)
= e^xsinx. ^d/_dx(cos²x) + e^x.^d/_dx(sinx)cos²x + ^d/_dx(e^x)sinxcos²x
=e^xsinx.2cosx.(-sinx) + e^x.cosx.cos²x + e^xsinxcos²x
= ¹/₂e^xcos x(-4sin²x + 2cos²x + 2sinxcosx)
= ¹/₂e^xcos x[-2(1 – cos2x) + (1 + cos2x) + sin2x]
== ¹/₂e^xcos x[-2 + 2cos2x + 1 + cos2x + sin2x]
= ¹/₂e^xcos x(sin 2x + 3cos 2x – 1) → [c]
Ans: Ⓑ [i] – [b], [ii] – [a], [iii] – [d], [iv] – [c]

4. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।স্তম্ভ Aস্তম্ভ B

স্তম্ভ A	স্তম্ভ B
[i] logx-এর সাপেক্ষে x² -এর অবকল সহগ	[a] xsec²x
[ii] logx-এর সাপেক্ষে tanx-এর অবকল সহগ	[b] secx
[iii] cosecx-এর সাপেক্ষে cotx -এর অবকল সহগ	[c] 2x²
[iv] tan^-1x-এর সাপেক্ষে log₁₀x-এর অবকল সহগ	[d] ^{1 + x²}/_x log₁₀e

Ⓐ [i] – [c], [ii] – [a], [iii] – [d], [iv] – [b]
Ⓑ [i] – [c], [ii] – [b], [iii] – [a], [iv] – [d]
Ⓒ [i] – [a], [ii] – [c], [iii] – [b], [iv] – [d]
Ⓓ [i] – [c], [ii] – [a], [iii] – [b], [iv] – [d]

Solution: [i] ^d/_d(logx)(x²)
= ^d/_dx(x²).¹/_{^d/_dx(logx)}
= 2x.¹/_x = 2x² → [c]
[ii] ^d/_d(logx)(tanx
= ^d/_dx(tanx).¹/_{^d/_dx(logx)}
= sec²x.¹/_{¹/_x} = xsec²x → [a]
[ii] ^d/_d(cosecx)(cotx)
= ^d/_dx(cotx).¹/_{^d/_dx(cosecx)}
= -cosec²x.¹/_cosecxcotx = secx → [b]

[ii] ^d/_d(tan^-1x)(log₁₀x)
= ^d/_dx(log_x10).¹/_{^d/_dx(tan^-1x)}
=^d/_dx(log_ex. log₁₀e).¹/_{¹/_{1 + x²}}
= log₁₀e.¹/_x(1 + x²)
= ^{1 + x²}/_x log₁₀e → [d]
Ans: Ⓓ [i] – [c], [ii] – [a], [iii] – [b], [iv] – [d]

5. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

স্তম্ভ A	স্তম্ভ B
[i] x² + y² = 4 হলে, ^dy/_dx =	=[a] y³
[ii] y = ^x/_{√(1 + x²)} হলে x^{3 dy}/_dx =	[b] ^y²/_{x(y – x)}
[iii] y= e^{^y/_x} হলে ^dy/_dx =	[c] ^y/_{x(1 + y)}
[iv] ye^y হলে ^dy/_dx =	[d] – ^x/_y

Ⓐ [i] – [d], [ii] – [a], [iii] – [b], [iv] – [c]
Ⓑ [i] – [d], [ii] – [a], [iii] – [c], [iv] – [b]
Ⓒ [i] – [a], [ii] – [d], [iii] – [b], [iv] – [c]
Ⓓ [i] – [d], [ii] – [b], [iii] – [c], [iv] – [a]

Solution: [i] x² + y² = 4
∴ 2x + 2y^dy/_dx = 0
⇒ 2y^dy/_dx = – ^x/_y → [d]

\([ii]\ y\ = \frac{x}{\sqrt{1 + x^2}}\\∴ \frac{dy}{dx} = \frac{1.\sqrt{1 + x^2}-x.\frac{1}{2\sqrt{1 + x^2}}.2x}{(\sqrt{1 + x^2})^2}\\=\frac{1 + x^2- x^2}{(1 + x^2)\sqrt{1 + x^2}}=\frac{1}{(\sqrt{1 + x^2})^3}\\∴ x^3\frac{dy}{dx}=\frac{x^3}{(\sqrt{1 + x^2})^3}\\⇒ x^3\frac{dy}{dx} = \left( \frac{x}{\sqrt{1 + x^2}} \right)^3=y^3 → [a]\)

[iii] y= e^{^y/_x}

\(∴ \frac{dy}{dx} = e^\frac{y}{x}.\frac{\frac{dy}{dx}.x – y.1}{x^2}\\⇒ \frac{dy}{dx} = y.\frac{x\frac{dy}{dx} – y}{x^2}\)

⇒ x^2dy/_dx = xy ^dy/_dx – y²
⇒ x(x – y)^dy/_dx = – y²
⇒^dy/_dx = ^{– y²}/_{x(x – y)}
⇒ ^dy/_dx = ^y²/_{x(y – x)}→ [b]

[iv] ye^y = x
∴ ^dy/_dx. e^y + y. e^y. ^dy/_dx = 1
⇒ e^y .^dy/_dx(1 + y) = 1
⇒ ^x/_y.^dy/_dx(1 + y) = 1
⇒^dy/_dx = → [c]
Ans: Ⓐ [i] – [d], [ii] – [a], [iii] – [b], [iv] – [c]

Rearrangement of Sentences/Events

1. cos y = xcos(a + y) থেকে ^dy/_dx নির্ণয়ের ধাপগুলি নীচে দেওয়া হল –

[i] x = ^{cos y}/_{cos(a + y)}
[ii] ^dx/_dy = ^{sin(a + y – y)}/_{cos²(a + y)
[}iii] ^dx/_dy = ^{– sin ycos(a + y) + cos ysin(a + y)}/_{cos²(a + y)}[iv] ^dy/_dx = ^{cos²(a + y)}/_{sin a} ধাপগুলির সঠিক ক্রমটি হবে-
Ⓐ [i] – [ii] – [iii] – [iv]
Ⓑ [i] – [iv] – [iii] – [ii]
Ⓒ [i] – [ii] – [iv] – [iii]
Ⓓ [i] – [iii] – [ii] – [iv]

Solution: cos y = xcos(a + y)
⇒ x = ^{cos y}/_{cos(a + y)} → [i]
∴ ^dx/_dy = ^{– sin ycos(a + y) + cos ysin(a + y)}/_{cos²(a + y)}→ [iii]
⇒ ^dx/_dy = ^{sin(a + y – y)}/_{cos²(a + y)} → [iii]
⇒ ^dx/_dy = ^{sin a}/_{cos²(a + y)}
∴ ^dy/_dx = ^{cos²(a + y)}/_{sin a}→ [iv]
Ans: Ⓓ [i] – [iii] – [ii] – [iv]

2. y^z = e^{y – x} থেকে ^dy/_dx নির্ণয়ের ধাপগুলি নীচে দেওয়া হল –

    [i] xlogy = y – x
  [ii] ^dy/_dx = ^ylogey/_{y – x}
    [iii] ^d/_dx(xlog y) = ^d/_dx(y – x)
   [iv] (1 – ^x/_y) ^dy/_dx = 1 + logy
    [v] ^dy/_dx = ^{log ey}/_{log y/_logey} = ^(logey)²/_logy
    [vi] ^dy/_dx = ^logey/_{1 – x/_y}
    [vii] logy + ^x/_y ^dy/_dx = ^dy/_dx – 1
      ধাপগুলির সঠিক ক্রমটি হবে –
Ⓐ [i] – [ii] – [iii] – [iv] – [v] – [vi] [vii]
Ⓑ [i] – [iii] – [vii] – [iv] – [ii] – [vi] – [v]
Ⓒ [i] – [iii] – [iv] – [vii] – [vi] – [ii] – [v]
Ⓓ [i] – [iii] – [vi] – [ii] – [iv] – [vii] – [v]

Solution: y^x = e^{y – x}
বা, logy^x = loge^{y – x}
বা, xlogy = (y – x)loge = (y – x) → [i]
    ^d/_dx(xlog y) = ^d/_dx(y – x) → [iii]
∴ 1.logy + x. ¹/_y ^dy/_dx = ^dy/_dx – 1
⇒ logy + ^x/_y ^dy/_dx = ^dy/_dx – 1 → [vii]
⇒(1 – ^x/_y)^dy/_dx= 1 + logy → [iv]
⇒ ^{y – x}/_y^dy/_dx = log_ee + logy = logey
⇒ ^dy/_dx = ^ylogey/_{y – x} → [ii]
⇒^dy/_dx = ^logey/_{1 – x/_y} → [vi]    Ans: Ⓑ [i] – [iii] – [vii] – [iv] – [ii] – [vi] – [v]

3. f(x) = (^{a + x}/_{b + x})^x থেকে f'(0) -এর মান নির্ণয় করার ধাপগুলি নীচে দেওয়া হল –
[i] f'(x) = ^du/_dx + ^dv/_dx নির্ণয় করতে হবে।
[ii] u = (^{a + x}/_{b + x})^x এবং v = cos x ধরতে হবে।
[iii] f'(x) -এ x = 0 বসাতে হবে।
[iv] ^du/_dx এবং ^dv/_dx নির্ণয় করতে হবে।

ধাপগুলির সঠিক ক্রমটি হবে-
Ⓐ [i] – [ii] – [iv] – [iii]
Ⓑ [ii] – [iv] – [i] – [iii]
Ⓒ [ii] – [i] – [iv] – [iii]
Ⓓ [ii] – [i] – [iii] – [iv]

Solution: f(x) =(^{a + x}/_{b + x})^x + cosx
[ii] u = (^{a + x}/_{b + x})^x এবং v = cos x
[i] f'(x) = ^du/_dx + ^dv/_dxনির্ণয় করতে হবে।
[iv] ^du/_dx এবং ^dv/_dx নির্ণয় করতে হবে।
u = (^{a + x}/_{b + x})^x
⇒ logu = log(^{a + x}/_{b + x})^x
           = x[log(a + x) – log(b + x)]
∴ ¹/_u ^du/_dx = 1.[ log(a + x) – log(b + x)] + x.[¹/_{a + x} – ¹/_{b + x}]
⇒ ^du/_dx = (^{a + x}/_{b + x})^x[log(^{a + x}/_{b + x}) + ^{x(b – a)}/_{(a + x)(b + x)}]
     v = cos x
∴ ^dv/_dx = -sinx
[iii] f'(x) -এ x = 0 বসাতে হবে।
∴ f'(x) = (^{a + x}/_{b + x})^x[log(^{a + x}/_{b + x}) + ^{x(b – a)}/_{(a + x)(b + x)}] – sinx
∴ f'(0) = (^{a + 0}/_{b + 0})⁰ [log(^{a + 0}/_{b + 0}) + ^{0(b – a)}/_{(a + 0(b + 0)}] – sin 0
            = 1.log(^a/_b) + 0 – 0 = log(^a/_b)
Ans: Ⓒ [ii] – [i] – [iv] – [iii]

Relationship between Statements ____

প্রতিটি প্রশ্নে বিবৃতি A এবং বিবৃতি B দেওয়া আছে। নীচের কোন বিকল্পটি বিবৃতি A এবং বিবৃতি B-এর মধ্যে সঠিক সম্পর্ক নির্দেশ করে?
Ⓐ বিবৃতি A এবং বিবৃতি B পরস্পরবিরোধী
Ⓑ বিবৃতি B হল বিবৃতি A এর কারণ
Ⓒ বিবৃতি A হল সত্য কিন্তু বিবৃতি B মিথ্যা
Ⓓ বিবৃতি A এবং বিবৃত্তি B পরস্পর নির্ভরশীল নয়

1. বিবৃতি-A: \(y=\begin{vmatrix}f(x)\ g(x)\ h(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}\) হলে \(\frac{dy}{dx} = \begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}\) (যেখানে l, m, n, a, b, c ধ্রুবক)

বিবৃতি-B: \(y=\begin{vmatrix}f(x)\ g(x)\ h(x)\\F(x)\ G(x)\ H(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}\) হলে \(\frac{dy}{dx} = \begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\F(x)\ G(x)\ H(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}\)

Solution: বিবৃতি-A:

\(y=\begin{vmatrix}f(x)\ g(x)\ h(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}\\∴\frac{dy}{dx}=\begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}+\begin{vmatrix}f(x)\ g(x)\ h(x)\\0\quad 0\quad 0\\a\quad b\quad c\end{vmatrix}+\begin{vmatrix}f(x)\ g(x)\ h(x)\\l\quad m\quad n\\0\quad 0\quad 0\end{vmatrix}\\=\begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}+0+0\\=\begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\l\quad m\quad n\\a\quad b\quad c\end{vmatrix}\)→ বিবৃতিটি সত্য

বিবৃতি-B:

\(y=\begin{vmatrix}f(x)\ g(x)\ h(x)\\F(x)\ G(x)\ H(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}\\\frac{dy}{dx} = \begin{vmatrix}f^{‘}(x)\ g^{‘}(x)\ h^{‘}(x)\\F(x)\ G(x)\ H(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}+\begin{vmatrix}f(x)\ g(x)\ h(x)\\F^{‘}(x)\ G^{‘}(x)\ H^{‘}(x)\\ϕ(x)\ ψ(x)\ τ(x)\end{vmatrix}+\begin{vmatrix}f(x)\ g(x)\ h(x)\\F(x)\ G(x)\ H(x)\\ϕ^{‘}(x)\ ψ^{‘}(x)\ τ^{‘}(x)\end{vmatrix}\)→ বিবৃতিটি মিথ্যা

Ans: Ⓒ বিবৃতি A হল সত্য কিন্তু বিবৃতি B মিথ্যা

2. বিবৃতি-A: tan y = ^{tan x + secx – 1}/_{tan x – secx + 1} হলে ^dy/_dx = ¹/₂
বিবৃতি-B: y = tan^-1(^{a cos x – b sin x}/_{b cos x + a sin x}) হলে ^dy/_dx = – 1

Solution: বিবৃতি-A: ^{tan x + secx – 1}/_{tan x – secx + 1}
= ^{tan⁡x+ secx – (sec²x – tan²⁡x}/_{tan x – secx + 1}
= ^{(tan⁡x+ secx)(1 –tan⁡x+ secx)}/_{tan x – secx + 1}
⇒ tanx + secx
⇒ ^{1 + sin x}/_cosx
= ^{(cos ^x/₂ + sin ^{^x/₂})²}/_{cos^²_^{^x}/₂} _{– sin^²⁡ _^{^x}/₂}
= ^{(cos ^x/₂ + sin ^{^x/₂})²}/_{(cos_^{^x}/₂} _{– sin⁡ _^{^x}/₂)(cos_^{^x}/₂} _{+ sin⁡ _^{^x}/₂)}
=^{cos ^x/₂ + sin ^{^x/₂}}/_{cos_^{^x}/₂} _{– sin⁡ _^{^x}/₂}
= ^{1 + tan ^{^x/₂}}/_{1 – tan_^{^x}/₂}
⇒ ^{tan ^π/₄ + tan ^{^x/₂}}/_{1 – tan^π_/₄}._{tan⁡ _^{^x}/₂}
⇒ tan(^π/₄ + ^x/₂)
∴ tan y = tan(^π/₄ + ^x/₂)
⇒ y = ^π/₄ + ^x/₂
∴ ^dy/_dx = ¹/₂ → বিবৃতিটি সত্য

বিবৃতি-B: y = tan^-1(^{a cos x – b sin x}/_{b cos x + a sin x})
= tan^-1(^{^a/_b – tan x}/_{1 + ^a/_b tan x})
=tan^-1(^{^a/_b}) – tan^-1(tan x)
= ^{^a/_b} – x
∴ ^dy/_dx = 0 – 1 = -1 → বিবৃতিটি সত্য
Ans: Ⓓ বিবৃতি A এবং বিবৃত্তি B পরস্পর নির্ভরশীল নয়

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True and False ___

1. বিবৃতি-I: যদি \(y = cos^{-1}{f(x)}\) হয়, তবে \(\frac{dy}{dx}=-\frac{f^{‘}(x)}{\sqrt{1 – [f(x)]^2}}\)

বিবৃতি-II: যদি \(y = tan^{-1}{ϕ(x)}\) হয়, তবে \(\frac{dy}{dx}=\frac{ϕ^{‘}(x)}{\sqrt{1 + [ϕ(x)]^2}}\)

Solution: বিবৃতি-I: y = cos^-1{f(x)}

\(∴\frac{dy}{dx}=-\frac{f^{‘}(x)}{\sqrt{1 – [f(x)]^2}}\)

→ বিবৃতিটি সত্য
বিবৃতি-II: y = tan^-1{ϕ(x)}

\(∴\frac{dy}{dx}=\frac{ϕ^{‘}(x)}{1 + [ϕ(x)]^2}\)

→ বিবৃতিটি মিথ্যা
Ans: Ⓑ বিবৃতি I সত্য

2. বিবৃতি-I: ^d/_dx(2x² + 3)⁴ = 16x(2x² + 3)³

বিবৃতি-II: \(\frac{d}{dx}\sqrt{x^2+a^2}= \frac{x}{\sqrt{x^2+a^2}}\)

Ⓐ বিবৃতি I ও II সত্য
Ⓑ বিবৃতি । সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓒ বিবৃতি I মিথ্যা কিন্তু বিবৃতি II সত্য
Ⓓ বিবৃতি I ও II মিথ্যা

Solution: বিবৃতি-I: ^d/_dx(2x² + 3)⁴
= 4(2x² + 3)³.2.2x
= 16x(2x² + 3)³ → বিবৃতিটি সত্য
বিবৃতি-II:

\(\frac{d}{dx}\sqrt{x^2+a^2}\\=\frac{1}{2\sqrt{x^2+a^2}}.2x\\= \frac{x}{\sqrt{x^2+a^2}}\) → বিবৃতিটি সত্য

Ans: Ⓐ বিবৃতি I ও II সত্য

3. বিবৃতি-I: x-এর সাপেক্ষে log_ax এর অন্তরকলজ ¹/_x log_ae
বিবৃতি-II: x-এর সাপেক্ষে log_xe এর অন্তরকলজ – ¹/_{x(log x)²}
Ⓐ উভয় বিবৃতি সত্য
Ⓑ উভয় বিবৃতি মিথ্যা
Ⓒ বিবৃতি I সত্য এবং II মিথ্যা
Ⓓ বিবৃতি I মিথ্যা এবং II সত্য

Solution: বিবৃতি-I: ^d/_dx(log_ax)
= ^d/_dx(log_ex. log_ae)
= ¹/_x log_ae→ বিবৃতিটি সত্য
বিবৃতি-II: ^d/_dx(log_xe)
= ^d/_dx(¹/_{log x})
= ¹/_{(log x)²}.¹/_x=¹/_{x(log x)²}→ বিবৃতিটি সত্য
Ans: Ⓐ উভয় বিবৃতি সত্য

4. বিবৃতি-I: ^d/_dx (x^x) = x^x(1 + logx)
বিবৃতি-II: ^d/_dx(2e^x – 3^x + 10^x ln(10) = (2e^x – 3m^xlog_e^m + 10^xlog_e¹⁰)
বিবৃতি-III: ^d/_dx (2^{x + 2} – e^{x + 1} + log₁₀x = 2^{x + 2}log_e² – e^x + ^log10e/_x

Ⓐ বিবৃতি I, II, III সত্য
Ⓑ বিবৃতি I, II, III মিথ্যা
Ⓒ বিবৃতি I, II সত্য কিন্তু III সত্য নয়
Ⓓ বিবৃতি II, III সত্য কিন্তু । সত্য নয়

Solution: বিবৃতি-I: ^d/_dx(x^x)
= ^d/_dx(e^xlogx)
=e^xlogx.( x.¹/_x +1.logx)
= x^x(1 + logx + 1) → বিবৃতিটি সত্য
বিবৃতি-II: ^d/_dx(2e^x – 3m^x + 10^x)
= 2e^x – 3m^x.log_em+ 10^x. log_e10 → বিবৃতিটি সত্য
বিবৃতি-III: ^d/_dx(2^{x + 2} – e^{x + 1} + log₁₀x)
= 2^{x + 2}log_e2 – e^{x + 1} + ¹/_x .log₁₀e → বিবৃতিটি মিথ্যা
Ans: Ⓒ বিবৃতি I, II সত্য কিন্তু III সত্য নয়

5. বিবৃতি-I: ^d/_dx(10^x .x¹⁰) = 10^x.x⁹ (10 + xlog10)

বিবৃতি-II:\(\frac{d}{dx}\left(x\frac{e^x + e^{3x}}{e^x + e^{-x}} \right) = e^{3x}(1 + 2x)\)

Solution: বিবৃতি-I: ^d/_dx(10^x .x¹⁰)
= 10^x.log_e10. x¹⁰ + 10^x .10x⁹
= 10^x.x⁹ (xlog10 +10) → বিবৃতিটি সত্য

বিবৃতি-II:

\(\frac{d}{dx}\left(x\frac{e^x + e^{3x}}{e^x + e^{-x}} \right)\\=\frac{d}{dx}\left(x\frac{e^{2x}(e^{-x} + e^x)}{e^x + e^{-x}} \right)\)

= ^d/_dx(xe^2x)
=1. e^2x + x.e^2x.2
= e^2x(1 + 2x) → বিবৃতিটি মিথ্যা
Ans: Ⓑ বিবৃতি । সত্য এবং II মিথ্যা

6. বিবৃতি-I: ^d/_dxlog(ax)^x = log(ax) + 1

বিবৃতি-II: \(\frac{d}{dx}logsin(\sqrt{x^2+1})= \frac{x}{\sqrt{x^2+1}}cot\sqrt{x^2+1}\)

Ⓐ বিবৃতি । সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓑ বিবৃতি I, II উভয়ই সত্য
Ⓒ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য
Ⓓ বিবৃতি I, II উভয়ই মিথ্যা

Solution: বিবৃতি-I: ^d/_dxlog(ax)^x
= ^d/_dx[x(loga + logx)]
= 1.(loga + logx) + x(0 + ¹/_x)
= log(ax) + 1→ বিবৃতিটি সত্য
বিবৃতি-II:

\(\frac{d}{dx}logsin(\sqrt{x^2+1})\\= \frac{1}{sin\sqrt{x^2+1}}.cos\sqrt{x^2+1}.\frac{1}{2\sqrt{x^2+1}}.2x\\=\frac{x}{\sqrt{x^2+1}}cot\sqrt{x^2+1}\) → বিবৃতিটি সত্য

Ans: Ⓑ বিবৃতি I, II উভয়ই সত্য

7. বিবৃতি-I: \(\frac{d}{dx}\left( 2xtan^{-1}x – log(1 + x2) \right) = \frac{2tan^{-1}⁡x}{\sqrt{1 + x^2}}\)

বিবৃতি-II:\(\frac{d}{dx}\left( \frac{2x+1}{x^3}-logtan x + x sin^{-1}(x^2 )\right) = -\frac{4}{x^3}-\frac{3}{x^4}-secxcosecx + sin^{-1}⁡(x^2 ) +\frac{2x^2}{\sqrt{1 – x^4}}\)

Solution: বিবৃতি-I:

\(\frac{d}{dx}\left( 2xtan^{-1}x – log(1 + x2) \right)\\=2.1.tan^{-1}x + 2x.\frac{1}{1 + x^2}- \frac{1}{1 + x^2}.2x = 2.tan^{-1}x\)→ বিবৃতিটি মিথ্যা

বিবৃতি-II:

\(\frac{d}{dx}\left( \frac{2x+1}{x^3}-logtan x + x sin^{-1}(x^2 )\right)\\=\frac{d}{dx}\left(\frac{2}{x^2} +\frac{1}{x^3} -logtan x + x sin^{-1}⁡(x^2 )\right)\\=-\frac{4}{x^3}-\frac{3}{x^4} -\frac{1}{tanx}.sec^2⁡x+1.sin^{-1}(x^2)+x.\frac{1}{\sqrt{1 – x^4}}).2x\\=-\frac{4}{x^3}-\frac{3}{x^4}-secxcosecx +sin^{-1}⁡(x^2 ) +\frac{2x^2}{\sqrt{1 – x^4}}\)→ বিবৃতিটি সত্য

8. বিবৃতি-I: \(\sqrt{1 + sin x}+\sqrt{\frac{1 – x^2}{1 + x^2}}\) অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ \(\frac{1}{2}\left( cos\frac{x}{2} – sin\frac{x}{2} \right))- \frac{2x}{(1 + x^2)\sqrt{1 – x^4}}\)

বিবৃতি-II: \(\frac{1}{\sqrt{x + a}-\sqrt{x + b}}+ x^2sinx^3\) অপেক্ষকটির x3 -এর সাপেক্ষে অন্তরকলজ \(\frac{1}{2(a – b)}\left[ \frac{1}{\sqrt{x + a}}+\frac{1}{\sqrt{x + b}} \right]+ 2xsinx^3 + 3x^4cos x^3\)

Solution: বিবৃতি- I:

\(\frac{d}{dx}\left[ \sqrt{1 + sin x}+\sqrt{\frac{1 – x^2}{1 + x^2}} \right]\\\frac{d}{dx}\left[cos⁡\frac{x}{2}+sin⁡\frac{x}{2} \right]+\frac{d}{dx}\left[ \left( \frac{1 – x^2}{1 + x^2} \right)^{\frac{1}{2}}\right]\\=-sin\frac{x}{2}.\frac{1}{2}+cos\frac{x}{2}.\frac{1}{2}+\frac{1}{2}.\left( \frac{1 – x^2}{1 + x^2} \right)^{-\frac{1}{2}}.\frac{-2x(1 + x^2 )-(1- x^2 ).2x}{(1 + x^2)^2}\\=-\frac{1}{2}\left(sin⁡\frac{x}{2}-cos\frac{x}{2}\right)+\frac{1}{2}\left( \frac{1 + x^2}{1 – x^2} \right)^{\frac{1}{2}}.\frac{-4x}{(1 + x^2 )^2}\\=\frac{1}{2}\left(cos⁡\frac{x}{2}-sin\frac{x}{2}\right)-\frac{1}{2}.\frac{\sqrt{1 + x^2}}{\sqrt{1 – x^2}}.\frac{4x}{(1 + x^2)^2}\\=\frac{1}{2}\left(cos⁡\frac{x}{2}-sin\frac{x}{2}\right)-\frac{1}{\sqrt{1 – x^2}}.\frac{2x}{\sqrt{1 + x^2}(1 + x^2)}\\=\frac{1}{2}\left(cos⁡\frac{x}{2}-sin\frac{x}{2}\right)-\frac{2x}{(1 + x^2)\sqrt{1 – x^4}}\) → বিবৃতিটি সত্য

বিবৃতি-II:

\(\frac{d}{dx}\left[ \frac{1}{\sqrt{x + a} -\sqrt{x + b}}+x^2 sinx^3 \right]\\=\frac{d}{dx}\left[ \frac{\sqrt{x + a})+\sqrt{x + b}}{(\sqrt{x + a} -\sqrt{x + b})(\sqrt{x + a}+\sqrt{x + b})} \right]+\frac{d}{dx}[x^2 sinx^3]\\=\frac{d}{dx}\left[ \frac{\sqrt{x + a}+\sqrt{x + b}}{a-b} \right]+2xsinx^3+x^2.cosx^3.3x^2\\=\frac{1}{(a – b)}\left[\frac{1}{2\sqrt{x + a}}.1+\frac{1}{2\sqrt{x + b}}.1 \right]+ 2xsinx^3 + 3x^4cos x^3\\=\frac{1}{2(a – b)}\left[\frac{1}{\sqrt{x + a}}+\frac{1}{\sqrt{x + b}}\right]+ 2xsinx^3 + 3x^4cos x^3\)→ বিবৃতিটি সত্য

Ans: Ⓑ বিবৃতি I, II উভয়ই সত্য

9. বিবৃতি-Ⅰ: e^xsecx অপেক্ষকটির x -এর সাপেক্ষে অন্তরকলজ e^xsecx(1 + tan x)।
বিবৃতি-II: (cos x)^logx অপেক্ষকটির x এর সাপেক্ষে অন্তরকলজ (log x)^cosx[^cosx/_xlogx – sin x – log(logx)]|

Ⓐ বিবৃতি I, II উভয়ই সত্য
Ⓑ বিবৃতি I, II উভয়ই মিথ্যা
Ⓒ বিবৃতি। সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓓ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য

Solution: বিবৃতি- I: ^d/_dx(e^xsecx)
= e^xsecx + e^xsecx.tanx
= e^xsecx(1 + tan x) → বিবৃতিটি সত্য
বিবৃতি- II: ^d/_dx [(cos x)^logx]
= ^d/_dx[e^{log⁡{(cos x)^logx}}]
= ^d/_dx[e^{logx⁡{log⁡(cosx)}}]
== e^{logx⁡{log⁡(cosx)}} .^d/_dx[logx⁡{log⁡(cosx) }]
= (cos x)^logx.(¹/_x.log⁡(cosx) + logx.¹/_cosx.(-sinx))
= (cos x)^logx(¹/_x log⁡(cosx)- logxtanx)→ বিবৃতিটি মিথ্যা
Ans: Ⓒ বিবৃতি। সত্য কিন্তু বিবৃতি II মিথ্যা

10. বিবৃতি-I: tan^-1x এর সাপেক্ষে ^{tan^-1x}/_{1 + tan^-1x} এর অবকল সহগ ¹/_{(1 – tan^-1x)²}

বিবৃতি-II: \(tan^{-1}x\) এর সাপেক্ষে \(tan^{-1}⁡\frac{\sqrt{1 + x^2}- 1}{x}\) এর অবকল সহগ 1

Ⓐ বিবৃতি। সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓑ বিবৃতি I, II উভয়ই সত্য
Ⓒ বিবৃতি I, II উভয়ই মিথ্যা
Ⓓ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য

Solution: বিবৃতি-I: ধরি, y = tan^-1x এবং

\(z = \frac{tan^{-1}⁡x}{1+tan^{-1}⁡x}=\frac{y}{1+y}\\∴ \frac{dz}{dy}=\frac{1(1 + y)- y.(0 + 1)}{(1 + y)^2} \\= \frac{1}{(1 + y)^2} =\frac{1}{(1 + tan^{-1}⁡x)^2}\)→ বিবৃতিটি মিথ্যা

বিবৃতি-II: ধরি, y = tan^-1x
∴ x = tany

\(z = tan^{-1}\frac{\sqrt{1 + x^2}– 1}{x}\\=tan^{-1}\frac{secy – 1}{tany}\\= tan^{-1}⁡\frac{1 – cosy}{siny}\\=tan^{-1}⁡\frac{2sin^2\frac{y}{2}}{2sin \frac{y}{2}.cos \frac{y}{2}}\)

11. বিবৃতি-I: sin^-1 ^2x/_{1 + x²} -এর সাপেক্ষে cos^-1 ^{1 – x²}/_{1 + x²} এর অবকল সহগ ¹/₂

বিবৃতি-II: \(sin^{-1⁡}x\) এর সাপেক্ষে \(x^{sin^{-1⁡}⁡x}\) এর অবকল সহগ \(x^{sin^{-1⁡}⁡x}\left[ \frac{\sqrt{1- x^2}}{x}sin^{-1⁡}x+ logx \right]\)

Ⓐ বিবৃতি । সত্য কিন্তু বিবৃতি II মিথ্যা
Ⓑ বিবৃতি I, II উভয়ই সত্য
Ⓒ বিবৃতি I, II উভয়ই মিথ্যা
Ⓓ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য

Solution: বিবৃতি-I: ধরি, y = cos^-1 ^{1 – x²}/_{1 + x²} = 2tan^-1x
এবং z = sin^-1 ^2x/_{1 + x²} = 2tan^-1x
∴ y = z
∴ ^dy/_dz = 1 → বিবৃতিটি মিথ্যা
বিবৃতি-II: ধরি, y = x^{sin^-1x} = e^{logx^{sin^-1x}}= e^{^{sin^-1x}logx}

\(∴ \frac{dy}{dx}= e^{sin^{-1}⁡x logx}\left( \frac{1}{\sqrt{1 -x^2}}logx+sin^{-1⁡}x.\frac{1}{x} \right)\\=x^{sin^{-1}⁡x}\left( \frac{logx}{\sqrt{1 -x^2}}+\frac{sin^{-1⁡}x}{x} \right)\)এবং \(z = sin^{-1}⁡x \\∴ \frac{dz}{dx}= \frac{1}{\sqrt{1 – x^2}}\\∴ \frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}} \\=\frac{x^{sin^{-1}⁡x}\left( \frac{logx}{\sqrt{1 -x^2}}+\frac{sin^{-1⁡}x}{x} \right)}{\frac{1}{\sqrt{1 – x^2}}}\\=x^{sin^{-1}⁡x}\left[logx+ \sqrt{1- x^2}.\frac{sin^{-1}⁡x}{x}\right]\\=x^{sin^{-1}⁡x}\left[ \frac{\sqrt{1- x^2}}{x}sin^{-1}x +logx \right]\)→বিবৃতিটি সত্য

Ans: Ⓓ বিবৃতি । মিথ্যা কিন্তু বিবৃতি II সত্য

12 বিবৃতি-I: যদি \(y = e^{\sqrt{cos x}}\) হয়, তবে \(\frac{dy}{dx} = \frac{sin xe^{\sqrt{cos x}}}{2\sqrt{cos x}}\\\)বিবৃতি-II: যদি \(y = \frac{1}{\sqrt{log(secx)}}\) হয়, তবে \(\frac{dy}{dx} = – \frac{1}{2}tan x[log(secx)]^{-\frac{3}{2}}\)

Solution: বিবৃতি-I: y = e^{√cos x}
∴ ^dy/_dx = e^{√cos x}.¹/_{2√cos x}.(-sin x)
= – ^{sinxe^{√cos x}}/_2√cosx → বিবৃতিটি মিথ্যা

বিবৃতি-II: \(y = \frac{1}{\sqrt{log(secx)}}=log(secx)^{-\frac{1}{2}}\)

∴ ^dy/_dx = –¹/₂[log(secx)] ^–³/₂.¹/_secx.secx.tanx
= –¹/₂tanx[log(secx)] ^–³/₂ → বিবৃতিটি সত্য
Ans: Ⓑ বিবৃতি II সত্য

Case Based ___

1. y = f(t) এবং x = Φ(t) হলে ^dy/_dx = ^f’(t)/_Φ’(t) যেখানে Φ’(t) ≠ 0

Solution: x = acos θ
∴ ^dx/_dθ = – asin θ
y = bsin θ
∴ ^dy/_dθ = bcos θ
∴ ^dy/_dx = ^dy/dθ/_dx/dθ = ^bcosθ/_-asinθ = – ^b/_acot θ
Ans: Ⓒ – ^b/_acot θ

[ii] x = acos ^θ/₂; y = bsin ^θ/₂ হলে ^dy/_dx =
Ⓐ ^b/_atan ^θ/₂ Ⓑ –^b/_atan ^θ/₂
Ⓒ – ^b/_acot ^θ/₂ Ⓓ এদের কোনোটিই নয়

Solution: x = acos ^θ/₂
∴ ^dx/_dθ = – asin ^θ/₂. ¹/₂
y = bsin ^θ/₂
∴ ^dy/_dθ = bcos ^θ/₂. ¹/₂
∴ ^dy/_dx = ^dy/dθ/_dx/dθ = ^bcos ^{^θ/2}/_{-asin ^θ/2} = – ^b/_acot θ
Ans: Ⓒ – ^b/_acot ^θ/₂

Solution: x = acos³ θ
∴ ^dx/_dθ = a.3cos² θ.(- sin θ) = – 3acos² θ.sin θ
y = asin³ θ
∴ ^dy/_dθ = a.3sin² θ cos θ = 3asin² θ cos θ
∴ ^dy/_dx = ^dy/dθ/_dx/dθ = ^3asin² ^θ ^cos ^θ/_– _3acos² _θ.sin _θ = -tan θ
Ans: Ⓒ -tan θ

2. মনে করো, y = f(x) অপেক্ষকের বিপরীত অপেক্ষক x = g(y) এবং f'(x) = ¹/_{1 + x²}

Solution: y = f(x)
এবং x = g(y)
⇒ x = g{f(x)}
∴ 1 = g'{f(x)}. f'(x)
⇒ 1 = g'{f(x)}. ¹/_{1 + x²}
⇒g'{f(x)} = 1 + x²
⇒ g'(y) = 1 + [g(y)]²
∴ g'(x) = 1 + [g(x)]²
Ans: Ⓐ 1 + [g(x)]²

Solution: f(x) = tanx হলে f'(x) = ¹/_{1 + x2} হয়।
Ans: Ⓐ tanx

3. মনে করো, f(x) = (^{a + x}/_{1 + x})^{a + 1 + 2x}
[i] f'(x) = kf(x) হলে k =
Ⓐ ^{(1 – a)(1 + a + x)}/_{(1 + x)(a + x)} + 2log(^{a + x}/_{1 + x})
Ⓑ ^{(1 – a)(1 + a + 2x)}/_{(1 + x)(a + x)} + log(^{a + x}/_{1 + x})
Ⓒ ^{(1 – a)(1 + a + 2x)}/_{(1 + x)(a + x)} + 2log(^{a + x}/_{1 + x})
Ⓓ এদের কোনোটিই নয়

Solution: f(x) = (^{a + x}/_{1 + x})^{a + 1 + 2x}
⇒ logf(x) = log(^{a + x}/_{1 + x})^{a + 1 + 2x}
⇒logf(x) = (a + 1 + 2x)[log(a + x) – log(1 + x)]
⇒ ¹/_f(x)f'(x) = 2[log(a + x) – log(1 + x)] + (a + 1 + 2x)[¹/_{a + x} – ¹/_{1 + x}]
⇒ ¹/_f(x)f'(x) = 2log(^{a + x}/_{1 + x}) + (a + 1 + 2x)[^{1 + x – a – x}/_{(a + x)(1 + x)}]
⇒¹/_f(x)f'(x) = 2log(^{a + x}/_{1 + x}) + ^{(a + 1 + 2x)(1 – a)}/_{(a + x)(1 + x)}
⇒ f'(x) = f(x)[2log(^{a + x}/_{1 + x}) + ^{(a + 1 + 2x)(1 – a)}/_{(a + x)(1 + x)}]
∴ k = 2log(^{a + x}/_{1 + x}) + ^{(a + 1 + 2x)(1 – a)}/_{(a + x)(1 + x)}
Ans: Ⓒ ^{(1 – a)(1 + a + 2x)}/_{(1 + x)(a + x)} + 2log(^{a + x}/_{1 + x})

[ii] f'(0) =
Ⓐ a^{a + 1} [2log a + ^{1 – a2}/_a]
Ⓑ a^{a + 1} [log(a) + ^{1 – a2}/_a]
Ⓒ a^{a + 1} [2 loga – ^{1 – a2}/_a]
Ⓓ এদের কোনোটিই নয়

Solution: f'(x) = f(x)[2log(^{a + x}/_{1 + x}) + ^{(a + 1 + 2x)(1 – a)}/_{(a + x)(1 + x)}]
∴ f'(0) = (^{a + 0}/_{1 + 0})^{a + 1 + 2.0}[2log(^{a + 0}/_{1 + 0}) + ^{(a + 1 + 2.0)(1 – a)}/_{(a + 0)(1 + 0)}]
= a^{a + 1}[2log a + ^{(a + 1)(1 – a)}/_a]
= a^{a + 1}[2log a + ^{1 – a2}/_a]
Ans: Ⓐ a^{a + 1} [2log a + ^{1 – a2}/_a]

4. মনে করো, (1 + x)ⁿ = C₀ + C₁x + C₂x² + . . . +C_nxⁿ
[i] C₀ + 2C₁ + 3C₂ + . . . + (n +1)C_n =
Ⓐ (n + 2)2^{n + 1}
Ⓑ (n + 2)2ⁿ
Ⓒ (n + 1)2^{n + 1}
Ⓓ (n + 2)2^{n – 1}

Solution: (1 + x)ⁿ = C₀ + C₁x + C₂x² + . . . +C_nxⁿ
∴ x(1 + x)ⁿ = xC₀ + C₁x² + C₂x³ + . . . + C_nx^{n + 1} x -এর সাপেক্ষে অবকলন করে পাই,
1.(1 + x)ⁿ + x.n(1 + x)^{n – 1} = 1.C₀ + 2xC₁x + 3xC₂x² + . . . + (n + 1)C_nxⁿ x = 1 বসিয়ে পাই,
(1 + 1)ⁿ + 1.n(1+1)^{n – 1} = C₀ + 2.1.C₁ + 3.1.C₂.1² + . . . + (n + 1)C_n.1ⁿ⇒ 2ⁿ + n.2^{n – 1} = C₀ + 2C₁ + 3C₂ + . . . + (n + 1)C_n
⇒ 2.2^{n – 1} + n.2^{n – 1} = C₀ + 2C₁ + 3C₂ + . . . + (n + 1)C_n
⇒ (n + 2)2^{n – 1} = C₀ + 2C₁ + 3C₂ + . . . + (n + 1)C_nAns: Ⓓ (n + 2)2^{n – 1}

Solution: (1 + x)ⁿ = C₀ + C₁x + C₂x² + . . . + C_nxⁿ x -এর সাপেক্ষে অবকলন করে পাই,
n(1 + x)^{n – 1} = C₁ + 2C₂x + 3C₃x² + . . . + nC_nx^{n -1}x = -1 বসিয়ে পাই,
n(1 – 1)^{n – 1} = C₁ + 2C₂(-1) + 3C₃(-1)² + . . . + nC_n(-1)^{n -1}
⇒ 0 = C₁ – 2C₂ + 3C₃ + . . . + (-1)^{n -1}nC_nAns: Ⓑ 0

2025 SEMESTER III MATHEMATICS QUESTION PAPER WITH SOLUTION →

← অবকলন DIFFERENTIATION-PART I

October 18, 2026

DIFFERENTIATION SOLUTION OF SNDEY SEM-3
SOLUTION OF DIFFERENTIATION SNDEY SEMESTER-3
SOLUTION OF DIFFERENTIATION SNDEY SEMESTER-3
অবকলন বা অন্তরকলন
Differentiation
Unit 3
Chapter 2
Part I
Part I
Part IIএর সমাধান দেখতে এখানে CLICK করো।
বহুবিকল্পভিত্তিক প্রশ্নাবলি (MCQ)                                                প্রতিটি প্রশ্নের মান 1
Conventional Type
1. y = e^{sin x} হলে ^dy/_dx =
Ⓐ – e^{sin x} cos x Ⓑ e^{sin x} sin x
Ⓒ e^{sin x} cos x Ⓓ e^{cos x}
Solution: y = e^{sin x}
∴ ^dy/_dx = e^{sin x}.^d/_dx(sin x) = e^{sin x}cos x
Ans: Ⓒ e^{sin x} cos x
2. y = sin³x হলে ^dy/_dx =
Ⓐ sin³x cos x Ⓑ 3sin²x cos x
Ⓒ – 3sin²x cos x Ⓓ 3sin²x cos x
Solution: y = sin³x
∴ ^dy/_dx = 3sin²x.^d/_dx(sin x) = 3sin²x cos x
Ans: Ⓑ 3sin²x cos x
3. f(x) = 2^3x² হলে f'(x) =
Ⓐ 2^3x².6x         Ⓑ 2^3x².log 2
Ⓒ 2^3x².3xlog 2   Ⓓ 2^3x².6xlog 2
Solution: f(x) = 2^3x²
∴ f'(x) = 2^3x².log 2.^d/_dx(3x²)
= 2^3x².log 2.6x = 2^3x².6xlog 2
Ans: Ⓓ 2^3x².6xlog 2
4. যদি ^dy/_dxlog(x² – 5) = ^φ(x)/_{x² – 5} হয় তবে φ(x)-এর মান হবে –
Ⓐ 2x Ⓑ x²
Ⓒ x Ⓓ 2x – 5
Solution: ^dy/_dxlog(x² – 5) = ^φ(x)/_{x² – 5}
⇒ ¹/_{x² – 5}.2x = ^φ(x)/_{x² – 5}
∴ φ(x) = 2x
Ans: Ⓐ 2x
5. যদি ^dy/_dx(tan^-1 x)² = k tan^-1 x.¹/_{1 + x²} হয় তবে k-এর মান হবে –
Ⓐ 1 Ⓑ -2
Ⓒ 2Ⓓ – 1
Solution: ^dy/_dx(tan^-1 x)²
= 2tan^-1 x.^d/_dx(tan^-1 x)
= 2tan^-1 x.¹/_{1 + x²}
∴ k = 2
Ans: Ⓒ 2
6. x² + y² = a² হলে ^dy/_dx-এর মান হবে –
Ⓐ ^x/_y Ⓑ – ^x/_y
Ⓒ ^y/_x Ⓓ – ^y/_x
Solution: x² + y² = a²
∴ 2x + 2y.^dy/_dx = 0
⇒ ^dy/_dx = – ^x/_y
Ans: Ⓑ – ^x/_y
7. ^d/_dx (5^f(x))-এর মান হবে –
Ⓐ 5^f(x).f’(x)
Ⓑ 5^f(x).log_e 5
Ⓒ 5fx.f'(x)loge 5
Ⓓ 5^f(x).f’(x)log_e 5
Solution: ^d/_dx (5^f(x))
= 5^f(x).log_e 5.^d/_dx[f(x)]
= 5^f(x).log_e 5.f’(x)
Ans: Ⓓ 5^f(x).f’(x)log_e 5
8. ^d/_dx (2x³ – 5)¹⁰ = (2x³ – 5)⁹f(x) হয়, তবে f(x) হবে-
Ⓐ 60x² Ⓑ 30x²
Ⓒ 20x² Ⓓ 45x²
Solution: ^d/_dx (2x³ – 5)¹⁰ = (2x³ – 5)⁹f(x)
⇒ 10(2x³ – 5)⁹ ×(2.3x² – 0) = (2x³ – 5)⁹f(x)
⇒ 60x² = f(x)
Ans: Ⓐ 60x²
9. ^d/_dx (x^x)-এর মান হবে –
Ⓐ x. x^{x – 1} Ⓑ x^xlog x
Ⓒ x^x(¹/_x + log x) Ⓓ x^x(1 + log x)
Solution: ^d/_dx (x^x)
= ^d/_dx (e^{xlog x})
= e^{xlog x}×^d/_dx (xlog x)
=e^{xlog x}×(1.log x + x.¹/_x)
= x^x(log x + 1)
Ans: Ⓓ x^x(1 + log x)
10. যদি y = log₁₀ x হয় তবে ^dy/_dx হবে –
Ⓐ ¹/_xlog₁₀ e Ⓑ ¹/_xlog_e 10
Ⓒ 1xlog10 x Ⓓ ¹/_10x
Solution: y = log₁₀ x
= log₁₀ e×log_e x
= log₁₀ e×¹/_x
Ans: Ⓐ ¹/_xlog₁₀ e
11. ^d/_dx(10^mx)-এর মান হবে –
Ⓐ 10^mxlog_e 10    Ⓑ 10^mxmlog_e 10
Ⓒ m10^mx     Ⓓ এদের কোনোটিই নয়।

Solution: ^d/_dx(10^mx)
= 10^mx.m.log_e 10
Ans: Ⓑ 10^mxmlog_e 10
12. x = a cos θ, y = a sin θ হলে ^dy/_dx এর মান হবে –
Ⓐ tan θ Ⓑ – tan θ
Ⓒ -cot θ Ⓓ cot θ
Solution:  x = a cos θ
∴ ^dx/_dθ = -asin θ
এবং y = asin θ
∴ ^dy/_dθ = acos θ
∴ ^dy/_dθ = ^dy/dθ/_dx/dθ
     = ^{acos θ}/_{-asin θ} = -cot θ
Ans: Ⓒ -cot θ
13. x = 3 বিন্দুতে ^d/_dx{|x – 1| + |x – 5|}-এর মান নিম্নের কোনটি হবে?
Ⓐ -2       Ⓑ 0
Ⓒ 2          Ⓓ 4
```
Solution: |x - 1| + |x - 5|
        = {-(x-1)-(x-5)  যখন  x<1
          {(x-1)-(x-5)   যখন  1≤x<5
          {(x-1) +(x-5)  যখন  x≥ 5)
        = {6-2x  যখন  x<1
          {4     যখন  1≤x<5
          {2x-6  যখন  x≥ 5
  ^d/_dx{|x - 1| + |x - 5|}
= ^d/_dx(4)... x = 3 বিন্দুতে
= 0
Ans: Ⓑ 0
```
SEMESTER-3
সূচিপত্র
👉 UNIT-1 সম্বন্ধ ও অপেক্ষক
- 1. সম্বন্ধ
- 2. অপেক্ষক
- 3. বিপরীত ত্রিকোণমিতিক অপেক্ষকসমূহ
👉 UNIT-2 বীজগণিত
- 1. ম্যাট্রিক্সের প্রকারভেদ ও ম্যাট্রিক্স বীজগণিত
  - ম্যাট্রিক্স PART 1
  - ম্যাট্রিক্স PART 2
- 2. নির্ণায়ক
  - মাইনর ও সহ-উৎপাদক
- 3. একটি ম্যাট্রিক্সের অ্যাডজয়েন্ট ও বিপরীত ম্যাট্রিক্স এবং সরল সহসমীকরণের সমাধান
👉 UNIT-3 কলনবিদ্যা
- 1. সন্ততা এবং অন্তরকলনযোগ্যতা
  - সন্ততা
  - অন্তরকলনযোগ্যতা
- 2. অবকলন বা অন্তরকলন
  - অবকলন – I
  - অবকলন PART – II
- 3. দ্বিতীয় ক্রমের অন্তরকলজ
- 4. অন্তরকলজের ব্যাখ্যা
- 5. স্পর্শক ও অভিলম্ব
- 6. বর্ধিষ্ণু ও ক্ষয়িষ্ণু অপেক্ষক
- . চরম ও অবম মান
👉 UNIT-4 সম্ভাবনা
- 1. সম্ভাবনা
  - সম্ভাবনা
  - Bayes’ উপপাদ্য
- 2. সমসম্ভব চলক ও তার বিভাজন
- 3. দ্বিপদ বিভাজন
👉 Semester III -এর প্রশ্নপত্রের সম্পূর্ণ সমাধান
- Semester III 2025 এর প্রশ্নপত্র ও তার সম্পূর্ণ সমাধান
14. u ও v যদি x-এর অন্তরকলনযোগ্য অপেক্ষক হয়, তবে ^d/_dx(tan^-1 ^u/_v) =
\(Ⓐ\ \frac{v \frac{du}{dx}}{u^2 + v^2}\quad\\Ⓑ\ \frac{u \frac{dv}{dx}}{u^2 + v^2}\\Ⓒ\ \frac{v \frac{du}{dx}-u\frac{dv}{dx}}{u^2 + v^2}\quad\\ Ⓓ\ \frac{v \frac{du}{dx}+u\frac{dv}{dx}}{u^2 + v^2}\)
\(Solution:\ \frac{d}{dx}(tan^{-1} \frac{u}{v})\\= \frac{1}{1 +\left( \frac{u}{v} \right)^2}.\frac{d}{dx}.\left( \frac{u}{v} \right)\\=\frac{v^2}{u^2 + v^2}.\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\\=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{u^2 + v^2}\\Ans:\ Ⓒ\ \frac{v\frac{du}{dx}-u\frac{dv}{dx}}{u^2 + v^2}\)
15. যদি ^d/_dx(tan^-1 x) = ¹/_{1 + x²} হয়, তবে ^d/_dx(cot^-1 x) =
Ⓐ -(1 + x²)    Ⓑ – ¹/_{1 + x²}
Ⓒ ²/_{1 + x²}         Ⓓ ¹/_{1 – x²}
Solution: ^d/_dx(cot^-1 x)
= ^d/_dx(^π/₂ – tan^-1 x)
= 0 – ¹/_{1 + x²} = – ¹/_{1 + x²}
Ans: Ⓑ – ¹/_{1 + x²}
16. 2tan^-1 ^x/_a -এর x-এর সাপেক্ষে অন্তরকলজ করে পাই, ^k/_{(x² + a²)} হলে k =
Ⓐ 1            Ⓑ 2
Ⓒ 3          Ⓓ 4
\(Solution: \frac{d}{dx}(2tan^{-1}\frac{x}{a}) = \frac{ka}{x^2 + a^2}\\⇒ 2.\frac{1}{1 + \left( \frac{x}{a} \right)^2}.\frac{d}{dx}\left( \frac{x}{a} \right) = \frac{ka}{x^2 + a^2}\\⇒ \frac{2a^2}{a^2 + x^2}.\frac{1}{a} = \frac{ka}{x^2 + a^2}\\ ⇒ k = 2\\Ans:\ Ⓑ 2\)
17. log(cot^-1 x) -এর x-এর সাপেক্ষে অন্তরকলজ করে পাই, ¹/_{k(1 + x²)} । এক্ষেত্রে k =
Ⓐ x                   Ⓑ √x
Ⓒ tan^-1x        Ⓓ – cot^-1 x
\(Solution: \frac{d}{dx}[log(cot^{-1}x)]= \frac{1}{k(1 + x^2)}\\⇒ \frac{1}{cot^{-1}x}.\frac{d}{dx}(cot^{-1} x)] = \frac{1}{k(1 + x^2)}\\⇒ \frac{1}{cot^{-1}x}.\left(-\frac{1}{1 + x^2} \right) = \frac{1}{k(1 + x^2)}\\⇒ \frac{-1}{cot^{-1}x}= \frac{1}{k}\)
⇒ – cot^-1 x = k
Ans: Ⓓ – cot^-1 x
18. ^d/_dx sec(tan^-1x) =
\(Ⓐ\ \frac{x}{\sqrt{1 + x^2}}\quad Ⓑ\ -\frac{x}{\sqrt{1 + x^2}}\\Ⓒ\ \frac{1}{\sqrt{1 + x^2}}\quad Ⓓ\ \frac{x}{\sqrt{1 – x^2}}\)
Solution: ^d/_dx sec(tan^-1x)
\(= \frac{d}{dx}sec(sec^{-1}\sqrt{1 + x^2})\\=\frac{d}{dx}(\sqrt{1 + x^2})\\=\frac{1}{2\sqrt{1 + x^2}}×2x=\frac{x}{\sqrt{1 + x^2}}\\Ans:\ Ⓐ\ \frac{x}{\sqrt{1 + x^2}}\)
19. ^d/_dx (2sec^-1 2x – 3sin^-1 x + 3cos^-1 x² ) =
\(Ⓐ\ \frac{2}{x\sqrt{4x^2 – 1}}-\frac{3}{\sqrt{1 – x^2}}-\frac{2x}{\sqrt{1 – x^4}}\\Ⓑ\ \frac{2}{x\sqrt{4x^2 – 1}}+\frac{3}{\sqrt{1 – x^2}}-\frac{2x}{\sqrt{1 – x^4}}\\Ⓒ\ \frac{2}{x\sqrt{4x^2 – 1}}+\frac{3}{\sqrt{1 – x^2}}+\frac{2x}{\sqrt{1 – x^4}}\)Ⓓ এদের কোনোটিই নয়
Solution: ^d/_dx (2sec^-1 2x – 3sin^-1 x + 3cos^-1 x² )
\(=2.\frac{1}{2x\sqrt{(2x)^2 – 1}}×2-3.\frac{1}{\sqrt{1 – x^2}}-\frac{1}{\sqrt{1 – (x^2 )^2}}×2x\\=\frac{2}{x\sqrt{4x^2 – 1}}-\frac{3}{\sqrt{1 – x^2}}-\frac{2x}{\sqrt{1 – x^4}}\\Ans:\ Ⓐ\ \frac{2}{x\sqrt{4x^2 – 1}}-\frac{3}{\sqrt{1 – x^2}}-\frac{2x}{\sqrt{1 – x^4}}\)
20. ^d/_dx(2cosec^-1 3x + 3cosec2x) হল
Ⓐ একটি ঋণাত্মক পূর্ণসংখ্যা
Ⓑ একটি ধনাত্মক পূর্ণসংখ্যা
Ⓒ 0
Ⓓ একটি x-এর অপেক্ষক
Solution: ^d/_dx[2cosec^-1 3x + 3cosec2x]
\(=-2.\frac{1}{3x\sqrt{(3x)^2 – 1}}.3+3(-cosec2x.cot2x).2\\=-\frac{2}{x\sqrt{9x^2 – 1}}-6cosec2xcot 2x\)Ans: Ⓓ একটি x-এর অপেক্ষক
\(21. \frac{d}{dx}(cos^{-1}x+cos^{-1}\sqrt{1-x^2})\) হল
Ⓐ একটি ঋণাত্মক পূর্ণসংখ্যা
Ⓑ একটি ধনাত্মক পূর্ণসংখ্যা
Ⓒ 0
Ⓓ একটি x-এর অপেক্ষক
\(Solution:\ \frac{d}{dx}(cos^{-1}x+cos^{-1}\sqrt{1-x^2})\\=\frac{d}{dx} (cos^{-1}x+sin^{-1}x)\\=\frac{d}{dx} (\frac{π}{2})=0\\Ans:\ Ⓒ\ 0\)
22. ^d/_dx (¹/_{px + q}) =
Ⓐ ^p/_{(px + q)²}                 Ⓑ ^q/_{(px + q)²}
Ⓒ – ^p/_{(px + q)²}         Ⓓ – ^q/_{(px + q)²}
Solution: ^d/_dx (¹/_{px + q})
=^d/_dx(px + q)^-1
= -1.(px + q)^-2.p.1
= – ^p/_{(px + q)²}
Ans: Ⓒ – ^p/_{(px + q)²}
23. ^d/_dx(e^2x)⁴ =
Ⓐ e^2x         Ⓑ 8e^8x        Ⓒ ^{e^8x}/₈
Ⓓ এদের কোনোটিই নয়

Solution: ^d/_dx(e^2x)⁴
= 4(e^2x)³ ^d/_dx(e^2x )
=4(e^2x)³.2e^2x
=8(e^2x)⁴ = 8e^8x
Ans: Ⓑ 8e^8x
24. d/dx(10^10x)=
Ⓐ 10^{10x + 1}.log 10x
Ⓑ 10^{10x + 1}
Ⓒ 10^{10x + 1}.log 10
Ⓓ 10^x.log 10
Solution: ^d/_dx(10^10x)
= 10^10x.10.log_e⁡10
= 10^{10x + 1}.log_e⁡10
Ans: Ⓒ 10^{10x + 1}.log ⁡10
25. ^d/_dx(2^2x² ^{+ 5x}) =
Ⓐ (4x + 5).2^2x² ^{+ 5x}
Ⓑ 2^2x² ^{+ 5x}.log 2
Ⓒ (4x+5) log⁡2
Ⓓ (4x+5)2^2x² ^{+ 5x}.log 2
Solution: ^d/_dx(2^2x² ^{+ 5x})
= 2^2x² ^{+ 5x}.log 2.^d/_dx(2x² + 5x)
= 2^2x² ^{+ 5x}.log 2.(4x+5)
Ans: Ⓓ (4x+5)2^2x² ^{+ 5x}.log 2
\(26.\ \frac{d}{dx}log\left( \sqrt{x^2+a^2} \right) =\\Ⓐ\ \frac{2x}{x^2 + a^2}\\Ⓑ\ \frac{x}{2(x^2 + a^2)}\\Ⓒ\ \frac{x}{x^2 + a^2}\)Ⓓ এদের কোনোটিই নয়
\(Solution:\ \frac{d}{dx}log\left( \sqrt{x^2+a^2} \right)\\=\frac{1}{x^2+a^2}.\frac{d}{dx}(x^2+a^2)\\=\frac{2x}{x^2 + a^2}\\Ans:\ Ⓐ \frac{2x}{x^2 + a^2}\)
27. ^d/_dxlog[log(logx)] =
Ⓐ ^x/_{log(log x)}
Ⓑ ^x/_{log x.log(log x)}
Ⓒ ^x/_{xlog x}
Ⓓ এদের কোনোটিই নয়
Solution: ^d/_dxlog[log(logx)]
= ¹/_log(logx)^d/_dx[log(logx)]
=¹/_log(logx)×¹/_{log x}×^d/_dx(logx)
= ¹/_log(logx)×¹/_{log x}×¹/_x
Ans: Ⓓ এদের কোনোটিই নয়
28. ^d/_dx(10^{log(cos x)}) =
Ⓐ tan x . 10^{log(cos x)}.log 10
Ⓑ – tan x . 10^{log(cos x)}.log 10
Ⓒ tan x . 10^{log(sin x)}.log 10
Ⓓ – tan x . 10^{log(sin x)}.log 10
Solution: ^d/_dx(10^{log(cos x)})
= 10^{log(cos x)} .log_e10.^d/_dx(log(cos x))
=10^{log(cos x)} .log_e10.¹/_{cos x}.(-sin x)
=- tan x . 10^{log(cos x)}.log 10
Ans: Ⓑ – tan x . 10^{log(cos x)}.log 10
29. ^d/_dx sin(cos x²) =
Ⓐ 2x sin x².cos(cos x²)
Ⓑ -2x sin x².cos(cos x²)
Ⓒ x sin x².cos(cos x)
Ⓓ এদের কোনোটিই নয়
Solution: ^d/_dx sin(cos x²)
= cos(cos x²).^d/_dx(cos x²)
=cos(cos x²).(-sin x²).2x
=-2x sin x².cos(cos x²)
Ans: Ⓑ -2x sin x².cos(cos x²)
S N DEY SEMESTER-3 অবকলন বা অন্তরকলন (Differentiation)
30. ^d/_dx(sin x^o) =
Ⓐ cos x^o                Ⓑ – cos x^o
Ⓒ ^π/₁₈₀cos x^o     Ⓓ ^π/₂ cos x^o
Solution: ^d/_dx(sin x^o)
=^d/_dx(sin ^πx/₁₈₀)
= cos ^πx/₁₈₀×^π/₁₈₀.1 = ^π/₁₈₀cos x^o
Ans: Ⓒ ^π/₁₈₀ cos x^o
31. x-এর সাপেক্ষে \(log(x+\sqrt{x^2+ a^2})\) -এর অন্তরকলজ \(\frac{k}{\sqrt{x^2+ a^2}}\)
হলে k =
Ⓐ 1             Ⓑ -1
Ⓒ 2            Ⓓ -2
\(Solution:\ \frac{d}{dx}\left[ log(x+\sqrt{x^2+ a^2}) \right]\\=\frac{1}{x+\sqrt{x^2+ a^2}}\left( 1+\frac{2x}{2\sqrt{x^2+ a^2}}\right)\\=\frac{1}{x+\sqrt{x^2+ a^2}}×\frac{\sqrt{x^2+ a^2}+x}{2\sqrt{x^2+ a^2}}\\=\frac{1}{\sqrt{x^2+ a^2}}\)
∴ k = 1
Ans: Ⓐ 1
32. x-এর সাপেক্ষে \(log(\sqrt{x-2}+ \sqrt{x – 3})\) -এর অন্তরকলজ \(\frac{1}{\sqrt{(x-a)(x-b)}}\)
হলে a + b =
Ⓐ 0        Ⓑ 2
Ⓒ 3        Ⓓ 5
\(Solution:\ \frac{d}{dx}\left[ log(\sqrt{x-2}+ \sqrt{x – 3}) \right]\\=\frac{1}{\sqrt{x-2}+ \sqrt{x – 3}}\left( \frac{1}{2\sqrt{x-2}}+\frac{1}{2\sqrt{x-3}} \right)\\=\frac{1}{\sqrt{x-2}+ \sqrt{x – 3}}×\frac{\sqrt{x – 3} + \sqrt{x – 2}}{2\sqrt{x – 2}\sqrt{x – 3}}\\=\frac{1}{\sqrt{(x-2)(x-3)}}\\\quad \frac{1}{\sqrt{(x-a)(x-b)}}\)
   এর সঙ্গে তুলনা করে পাই,
x = 2, y = 3
∴ x + y = 2 + 3 = 5
Ans: Ⓓ 5
33. x-এর সাপেক্ষে log tan(^π/₄ + ^x/₂) -এর অন্তরকলজ
Ⓐ sin x        Ⓑ cos x
Ⓒ tab x        Ⓓ sec x
Solution: ^d/_dx[log tan(^π/₄ + ^x/₂)]
= ¹/_{tan(^π/4 + ^x/2)} .sec²⁡(^π/₄ + ^x/₂).¹/₂
= ¹/_{2 sin⁡(^π/4 + ^x/2)cos⁡(^π/4 + ^x/2)}
==¹/_{sin2⁡(^π/4 + ^x/2)}
= ₁/_{sin⁡(^π/2 + ^x/2)}
= ¹/_cos⁡x = sec⁡x
Ans: Ⓓ sec x
34. ^d/_dx(secⁿ x) = naⁿb হলে a² – b² =
Ⓐ 1           Ⓑ 0           Ⓒ -1
Ⓓ এদের কোনোটিই নয়
Solution: ^d/_dx(secⁿ x) = naⁿb
⇒ n sec^{n – 1} x.sec xtan x = naⁿb
⇒ secⁿ x.tan x = aⁿb
∴ a = sec x, b = tan x
∴ a² – b² = sec² x – tan² x = 1
Ans: Ⓐ 1
35.যদি \(y=tan^{-1} \frac{4x}{1 + 5x^2}+tan^{-1}\frac{2 + 3x}{3 – 2x}\) হয়,তবে \(\frac{dy}{dx}=\ Ⓐ\ \frac{1}{1 + 25x^2}\\Ⓑ\ \frac{5}{1 + 5x^2}\\Ⓒ\ \frac{5}{1 + 25x^2}\\Ⓓ\ \frac{1}{1 + 5x^2}\)
\(Solution:\ y=tan^{-1} \frac{4x}{1 + 5x^2}+tan^{-1}\frac{2 + 3x}{3 – 2x}\\⇒y= tan^{-1}\frac{5x – x}{1 + 5x.x}+ tan^{-1}\frac{\frac{2}{3} + x}{1 -\frac{2}{3}x}\)
⇒y = tan^-15x – tan^-1x + tan^-1 ²/₃ + tan^-1 x
⇒y= tan^-15x + tan^-1 ²/₃
\(∴\frac{dy}{dx}=\frac{1}{1+(5x)^2}×5+0=\frac{5}{1 + 25x^2}\\Ans:\ Ⓒ\ \frac{5}{1 + 25x^2}\)
36. ^d/_dx(sin^-1x)^m =
\(Ⓐ\ \frac{m}{\sqrt{1 – x^2}}(sin^{-1}x)^{m-1}\\Ⓑ\ \frac{m}{\sqrt{1 + x^2}}(sin^{-1}x)^{m-1}\\Ⓒ\ \frac{m}{\sqrt{1 + x^2}}(sin^{-1}x)^m\\Ⓓ\ \frac{m}{\sqrt{1 – x^2}}(sin^{-1}x)^m\\Solution:\ \frac{d}{dx}(sin^{-1}x)^m\\=m(sin^{-1}x)^{m-1}×\frac{1}{\sqrt{1 – x^2}}\\=\frac{m}{\sqrt{1 – x^2}}(sin^{-1}x)^{m-1}\\Ans:\ Ⓐ\ \frac{m}{\sqrt{1 – x^2}}(sin^{-1}x)^{m-1}\)
\(37.\ \frac{d}{dx}\left( cos^{-1}\frac{x}{a} \right)^3=\\Ⓐ\ \frac{2}{\sqrt{a^2- x^2}}\left( cos^{-1}\frac{x}{a} \right)^2\\Ⓑ\ \ \frac{3}{\sqrt{a^2- x^2}}\left( cos^{-1}\frac{x}{a} \right)^3\\Ⓒ\ -\ \frac{3}{\sqrt{a^2- x^2}}\left( cos^{-1}\frac{x}{a} \right)^2\\Ⓓ\ -\frac{2}{\sqrt{a^2- x^2}}\left( cos^{-1}\frac{x}{a} \right)^3\)
\(Solution:.\ \frac{d}{dx}\left( cos^{-1}\frac{x}{a} \right)^3\\=3\left( cos^{-1}\frac{x}{a} \right)^2×\frac{d}{dx}\left(cos^{-1}\frac{x}{a} \right)\\=3\left( cos^{-1}\frac{x}{a} \right)^2×{\frac{-1}{\sqrt{1-\left( \sqrt{\frac{x}{a}} \right)^2}}}×\frac{1}{a}\\=3\left( cos^{-1}\frac{x}{a} \right)^2×\frac{-a}{a^2- x^2}×\frac{1}{a}\\= -\frac{3}{\sqrt{a^2- x^2}}\left( cos^{-1}\frac{x}{a} \right)^2\\Ans:\ Ⓒ\ -\frac{3}{\sqrt{a^2- x^2}}\left( cos^{-1}\frac{x}{a} \right)^2\)
38. ^d/_dx (tan^-1sec x) = ^a/_{1 + b²} হলে ^a/_b =
Ⓐ sin x         Ⓑ cos x
Ⓒ sec x     Ⓓ tan x

Solution: ^d/_dx (tan^-1(sec x))
= ¹/_{1 + sec² x}.secx.tanx
= ^secx.tanx/_{1 + sec² x}
প্রশ্নানুযায়ী,
     ^secx.tanx/_{1 + sec² x} = ^a/_{1 + b²}
∴ a = sec x.tan x এবং b = sec² x
∴ ^a/_b = ^secx.tanx/_{sec² x} = sin x
Ans: Ⓐ sin x
\(39.\ \frac{d}{dx}\left(\sqrt{x^2+1}-log\left[\frac{1}{x}+\sqrt{1+\frac{1}{x^2}} \right]]\right)=\\ Ⓐ\ \frac{\sqrt{x^2+1}}{x}\quad Ⓑ\ \frac{x}{\sqrt{x^2+1}}\\Ⓒ\ \frac{\sqrt{x^2-1}}{x^2}\quad Ⓓ\ \frac{x^2 – 1}{x^2 + 1}\)
\(Solution:\ \frac{d}{dx}\left(\sqrt{x^2+1}-log\left[\frac{1}{x}+\sqrt{1+\frac{1}{x^2}} \right]\right)\\=\frac{d}{dx}\left( \sqrt{x^2+1}-log\frac{1 + \sqrt{x^2+1}}{x} \right)\\=\frac{d}{dx}[\sqrt{x^2+1}-log(1 + \sqrt{x^2+1})+logx]\\=\frac{1}{2\sqrt{x^2+1}}×2x-\frac{1}{1 + \sqrt{x^2+1}}×\frac{1}{2\sqrt{x^2+1}}×2x+\frac{1}{x}\\=\frac{x}{\sqrt{x^2+1}}-\frac{x}{\sqrt{x^2+1} (1 + \sqrt{x^2+1})}+\frac{1}{x}\\=\frac{x(1 + \sqrt{x^2+1}- 1)}{\sqrt{x^2+1}(1 + \sqrt{x^2+1})}+\frac{1}{x}\\=\frac{x\sqrt{x^2+1}}{\sqrt{x^2+1}(1 + \sqrt{x^2+1})}+\frac{1}{x}\\=\frac{x}{1 + \sqrt{x^2+1}}+\frac{1}{x} \\=\frac{x^2 + 1 + \sqrt{x^2+1}}{x(1 + \sqrt{x^2+1})}\\ =\frac{\sqrt{x^2+1}.(\sqrt{x^2+1}+ 1)}{x(1 + \sqrt{x^2+1})}\\=\frac{\sqrt{x^2+1}}{x}\\Ans:\ Ⓐ\ \frac{\sqrt{x^2+1}}{x}\)
\(40.\ (0, \frac{π}{2})\) অন্তরালে \(cot^{-1} \sqrt{\frac{1 – sin x}{1 + sin x}}\) অপেক্ষকটির x এর সাপেক্ষে অবকলন হল \(Ⓐ\ 0\quad Ⓑ\ \frac{1}{2}\\Ⓒ\ cosec^{-1}(sec x)\\Ⓓ\ \frac{sin x}{1 + cot x}\)
\(Solution:\ \frac{1 – sin x}{1 – sin x}\\=\frac{\left( cos\frac{⁡x}{2} – sin⁡\frac{⁡x}{2} \right)^2}{\left( cos\frac{⁡x}{2} + sin⁡\frac{⁡x}{2} \right)^2}\\=\left( \frac{cos⁡\frac{⁡x}{2} – sin⁡\frac{⁡x}{2}}{cos⁡\frac{⁡x}{2} + sin⁡\frac{⁡x}{2}} \right)^2\\= \left( \frac{1 – tan\frac{⁡x}{2}}{1+tan\frac{⁡x}{2}} \right)^2\\= \left( \frac{tan⁡\frac{π}{4} – tan\frac{⁡x}{2}}{1+tan⁡\frac{π}{4}.tan\frac{⁡x}{2}} \right)^2\\=tan^2⁡(\frac{π}{4}-\frac{x}{2})\\∴ cot^{-1}\sqrt{\frac{1 – sin x}{1 – sin x}}\)
= cot^-1tan(^π/₄ − ^x/₂)
= cot^-1cot[^π/₂− (^π/₄ − ^x/₂)]
=cot^-1cot[^π/₂ + ^x/₂]
= ^π/₂ + ^x/₂
\(∴ \frac{d}{dx}\left[ cot^{-1}\sqrt{\frac{1 – sin x}{1 – sin x}} \right]\)
= ^d/_dx(^π/₂ + ^x/₂)
= 0 + ¹/₂ = ¹/₂
Ans: Ⓑ ¹/₂
41. x² + xy + y² = 100 হলে, ^dy/_dx =
Ⓐ – ^{2x + y}/_{x + 2y} Ⓑ – ^{x + y}/_{x + 2y}
Ⓒ ^{2x + y}/_{x + 2y} Ⓓ এদের কোনোটিই নয়
Solution: x² + xy + y² = 100
∴ 2x + (1.y + x.^dy/_dx) + 2y.^dy/_dx = 0
⇒ (x + 2y) ^dy/_dx = -(2x + y)
⇒ ^dy/_dx = – ^{2x + y}/_{x + 2y}
Solution: Ⓐ – ^{2x + y}/_{x + 2y}
42. x³ + y³ = 3axy হলে, ^dy/_dx =
\(Ⓐ\ \frac{x^2 + ay}{ax – y^2}\quad Ⓑ\ \frac{x^2 – ay}{ax – y^2}\\Ⓒ\ \frac{x^2 – ay}{ax – y^2}\quad Ⓓ \frac{x^2 + ay}{ax + y^2}\)
Solution: x³ + y³ = 3axy
∴ 3x² + 3y².^dy/_dx = 3a(1.y + x.^dy/_dx)
⇒ (y² – ax) ^dy/_dx = ay – x²
\(⇒\frac{dy}{dx}=\frac{ay – x^2}{y^2 – ax}\\⇒\frac{dy}{dx}=\frac{x^2 – ay}{ax – y^2}\\Ans:\ Ⓑ\ \frac{x^2 – ay}{ax – y^2}\)
\(43. x^{\frac{2}{3}}+y^{\frac{2}{3}}=1\) হলে, \(\frac{dy}{dx} =Ⓐ\ -\left( \frac{y}{x} \right)^3\quad Ⓑ\ -\left( \frac{y}{x} \right)^{\frac{1}{3}}\\Ⓒ\ -\left( \frac{x}{y} \right)^{\frac{1}{3}}\quad Ⓓ \left( \frac{y}{x} \right)^{\frac{1}{3}}\\Solution:\ x^{\frac{2}{3}}+y^{\frac{2}{3}}=1\\∴\frac{2}{3} x^{\frac{-1}{3}}+\frac{2}{3} y^{\frac{-1}{3}}\frac{dy}{dx}= 0\\⇒\frac{dy}{dx}= -\frac{x^\frac{-1}{3}}{y^\frac{-1}{3}} = -\left( \frac{x}{y} \right)^{-\frac{1}{3}}=-\left( \frac{y}{x} \right)^{\frac{1}{3}}\\Ans:\ Ⓑ\ -\left( \frac{y}{x} \right)^{\frac{1}{3}}\)
\(44.\ \frac{x^2}{a^2} +\frac{y^2}{b^2}= 1\) হলে, \(\frac{dy}{dx} =Ⓐ\ -\frac{xb^2}{ya^2}\quad Ⓑ\ \frac{xb^2}{a^2}\\Ⓒ\ \frac{xa^2}{yb^2}\quad Ⓓ -\frac{xa^2}{yb^2}\\Solution:\ \frac{x^2}{a^2} +\frac{y^2}{b^2}= 1\\∴\ \frac{1}{a^2}.2x+\frac{1}{b^2}.2y.\frac{dy}{dx}= 0\\⇒\frac{dy}{dx}= -\frac{xb^2}{ya^2}\\Ans:\ Ⓐ\ -\frac{xb^2}{ya^2}\)
45. x = ylog xy হলে, ^dy/_dx =
Ⓐ ^{x – y}/_{x + y} Ⓑ ^{y(x – y)}/_{x + y}
Ⓒ ^{x + y}/_{x – y} Ⓓ ^{y(x – y)}/_{x(x + y)}
Solution: x = ylog xy = ylog x + ylogy
∴ 1 = ^dy/_dx .log x + y.¹/_x + ^dy/_dx .log y + y.¹/_y .^dy/_dx
⇒ 1 – ^y/_x = (log x + log y + 1)^dy/_dx⇒ ^{x – y}/_x = (log xy + 1)^dy/_dx
⇒^{x – y}/_x = (^x/_y + 1) )^dy/_dx
⇒ ^dy/_dx = ^{y(x – y)}/_{x(x + y)}
Ans: Ⓓ ^{y(x – y)}/_{x(x + y)}
46. ax² + 2hxy + by² + 2gx + 2fy + c = 0 হলে, ^dy/_dx =
Ⓐ ^{ax + by + g}/_{bx + ay + f}
Ⓑ – ^{ax + hy + g}/_{hx + by + f}
Ⓒ ^{ax + hy + f}/_{hx + by + g}
Ⓓ এদের কোনোটিই নয়
Solution: ax² + 2hxy + by² + 2gx + 2fy + c = 0
∴ 2ax + 2by. ^dy/_dx + 2h(1.y + x. ^dy/_dx) + 2g + 2f. ^dy/_dx + 0 = 0
⇒ ^dy/_dx.(2by + 2hx + 2f) = – 2ax – 2hy – 2g
\(⇒\frac{dy}{dx}=\frac{- 2ax – 2hy – 2g}{2by + 2hx + 2f}\\⇒\frac{dy}{dx}=-\frac{2ax + 2hy + 2g}{2by + 2hx + 2f}\\Ans:\ Ⓑ\ -\frac{2ax + 2hy + 2g}{2by + 2hx + 2f}\)
47. (x² + y²)² = xy হলে ^dy/_dx =
\(Ⓐ\ \frac{y – 4x(x^2 + y^2)}{4y(x^2 + y^2 )- x}\\Ⓑ\ \frac{y – x(x^2 + y^2)}{y(x^2 + y^2 )- x}\\ Ⓒ\ \frac{y – 4x(x^2 + y^2)}{y(x^2 + y^2 )- x}\\Ⓓ\ \frac{y – x(x^2 + y^2)}{4y(x^2 + y^2 )- x}\)
Solution: (x² + y²)² = xy
∴ 2(x² + y²)(2x + 2y.^dy/_dx) = 1.y + x. .^dy/_dx
⇒ ^dy/_dx.[x – 4y(x² + y²)] ^dy/_dx = 4x(x² + y²) – y
\(⇒\frac{dy}{dx}=\frac{4x(x^2 + y^2)- y}{x – 4y(x^2 + y^2 )}\\⇒\frac{dy}{dx}=\frac{y – 4x(x^2 + y^2)}{4y(x^2 + y^2 )- x}\\Ans:\ Ⓐ\ \frac{y – 4x(x^2 + y^2)}{4y(x^2 + y^2 )- x}\)
48. x = at², y = 2at হলে ^dy/_dx =
Ⓐ t                 Ⓑ ¹/_t
Ⓒ – ¹/_t          Ⓓ¹/_t3
Solution: x = at²
∴ ^dx/_dt = 2at
এবং y = 2at
∴ ^dy/_dt = 2a
∴ ^dy/_dt
= ^dy/dt/_dx/dt
= ^2a/_2at = ¹/_t
Ans: Ⓑ ¹/_t
49. যদি x = t logt, y = ^logt/_t হয়, তবে t = 1 বিন্দুতে ^dy/_dx =
Ⓐ 1           Ⓑ -1
Ⓒ 5          Ⓓ -5
Solution: x = t logt
∴ ^dx/_dt = 1.logt + t.¹/_t = logt + 1
এবং y = ^logt/_t
\(∴ \frac{dy}{dt}=\frac{\frac{1}{t}.t – logt.1}{t^2}=\frac{1 – logt}{t^2}\\∴ \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dz}{dt}}=\frac{1 – logt}{t^2(1 + logt)}\)t = 1 বিন্দুতে \(\frac{dy}{dx}=\frac{1 – log⁡1}{1^2 (1 + log⁡1)}=\frac{1 – 0}{1(1 + 0)}=1\)
Ans: Ⓐ 1
50. যদি \(y = \sqrt{\frac{x^2 + 1}{x^2 – 1}}\) হয়, তবে \(\frac{dy}{dx}=\ Ⓐ\ \frac{2x}{(x^2- 1)\sqrt{x^4- 1}}\\Ⓑ\ \frac{2x}{(x^2- 1)\sqrt{x^4- 1}}\\Ⓒ\ \frac{x}{(x^2- 1)\sqrt{x^4- 1}}\\Ⓓ\ -\frac{2x}{(x^2- 1)\sqrt{x^4- 1}}\)
\(Solution:\ y = \sqrt{\frac{x^2 + 1}{x^2 – 1}}\\⇒ log y =log\sqrt{\frac{x^2 + 1}{x^2 – 1}}\\⇒log y = \frac{1}{2}[log (x^2 + 1) – log (x^2 – 1)]\\∴ \frac{1}{y}.\frac{dy}{dx}=\left[ \frac{2x}{x^2 + 1}- \frac{2x}{x^2 – 1} \right]\\⇒\frac{dy}{dx}=\frac{1}{2}.y.2x\left[ \frac{1}{x^2 + 1}- \frac{1}{x^2 – 1} \right]\\⇒\frac{dy}{dx}=x \sqrt{\frac{x^2 + 1}{x^2 – 1}}\left[ \frac{x^2-1-x^2-1)}{(x^2 + 1)(x^2 – 1} \right]\\⇒\frac{dy}{dx}=x \sqrt{\frac{x^2 + 1}{x^2 – 1}}\left[ \frac{-2}{(x^2 + 1)(x^2 – 1)} \right]\\⇒\frac{dy}{dx}=-\frac{2x}{(x^2- 1)\sqrt{x^4- 1}}\\Ans:\ Ⓓ\ -\frac{2x}{(x^2- 1)\sqrt{x^4- 1}}\)
51. যদি y = (1 – x)(1 – 2x)(1 – 3x²) হয়, তবে ^dy/_dx =
Ⓐ (1 – x)(1 – 2x)(1 – 3x)[¹/_{1 – x} + ²/_{1 – 2x} + ^6x/_{1 – 3x²}]
Ⓑ – (1 – x)(1 – 2x)(1 – 3x²)[¹/_{1 – x} + ²/_{1 – 2x} – ^6x/_{1 – 3x²}]
Ⓒ – (1 – x)(1 – 2x)(1 – 3x²)[¹/_{1 – x} + ²/_{1 – 2x} + ^6x/_{1 – 3x²}]
Ⓓ এদের কোনোটিই নয়
Solution: y = (1 – x)(1 – 2x)(1 – 3x²)
⇒ log y = log[(1 – x)(1 – 2x)(1 – 3x²)]
⇒ log y = log(1 – x) + log(1 – 2x) + log(1 – 3x²)
∴ ¹/_y.^dy/_dx = ^-1/_{1 – x} + ^-2/_{1 – 2x} + ^-6x/_{1 – 3x²}
⇒ ^dy/_dx = y[^-1/_{1 – x} + ^-2/_{1 – 2x} + ^-6x/_{1 – 3x²}]
⇒ ^dy/_dx = -(1 – x)(1 – 2x)(1 – 3x²)[¹/_{1 – x} + ²/_{1 – 2x} + ^6x/_{1 – 3x²}]
Ans: Ⓑ – (1 – x)(1 – 2x)(1 – 3x²)[¹/_{1 – x} + ²/_{1 – 2x} + ^6x/_{1 – 3x²}]
52. যদি \(y = \sqrt{\frac{(x – 3)(x^2 + 4)}{3x^2 + 4x + 5}}\) হয় তবে \(\frac{dy}{dx}=\ Ⓐ\ \frac{1}{2}\sqrt{\frac{(x – 3)(x^2 + 4)}{3x^2 + 4x + 5}}\\Ⓑ\ \frac{1}{2}\left[ \frac{1}{x-3}+\frac{2x}{x^2+4}-\frac{6x+4}{3x^2 + 4x + 5} \right]\\Ⓒ\ \frac{1}{2}\sqrt{\frac{(x – 3)(x^2 + 4)}{3x^2 + 4x + 5}}\left[ \frac{1}{x-3}+\frac{2x}{x^2+4}-\frac{6x+4}{3x^2 + 4x + 5} \right]\) Ⓓ এদের কোনোটিই নয়
\(Solution:\ y = \sqrt{\frac{(x – 3)(x^2 + 4)}{3x^2 + 4x + 5}}\\⇒ log y =log\sqrt{\frac{(x – 3)(x^2 + 4)}{3x^2 + 4x + 5}}\\⇒log y=\frac{1}{2}[log (x – 3) + log (x^2 + 4) – log (3x^2 + 4x + 5)]\\∴ \frac{1}{y}.\frac{dy}{dx}=\frac{1}{2}\left[ \frac{1}{x – 3}+ \frac{2x}{x^2 + 4}-\frac{6x + 4}{3x^2 + 4x + 5} \right]\\⇒\frac{dy}{dx}=\frac{1}{2}.y\left[ \frac{1}{x – 3}+ \frac{2x}{x^2 + 4}-\frac{6x + 4}{3x^2 + 4x + 5} \right]\\⇒\frac{dy}{dx}=\frac{1}{2}.\sqrt{\frac{(x – 3)(x^2 + 4)}{3x^2 + 4x + 5}}\left[ \frac{1}{x – 3}+ \frac{2x}{x^2 + 4}-\frac{6x + 4}{3x^2 + 4x + 5} \right]\\Ams:\ Ⓒ\ \frac{1}{2}\sqrt{\frac{(x – 3)(x^2 + 4)}{3x^2 + 4x + 5}}\left[ \frac{1}{x-3}+\frac{2x}{x^2+4}-\frac{6x+4}{3x^2 + 4x + 5} \right]\)
\(53.\ \frac{5x}{\sqrt[3]{1 – x^2}}+sin^2 (2x + 3)\) অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ – \(Ⓐ\ \frac{5(3 – x^2 )}{3(1 – x^2 )^{\frac{4}{3}}} + 2sin(4x + 6)\\Ⓑ\ \frac{(3 – x^2)}{(1 – x^2 )^4} + 2sin(4x + 6)\\Ⓒ\ \frac{x^2}{3x^{\frac{4}{3}}} + 2sin(4x + 6)\) Ⓓ এদের কোনোটিই নয়
\(Solution:\ \frac{d}{dx}\left[ \frac{5x}{\sqrt[3]{1 – x^2}}+sin^2 (2x + 3) \right]\\=\frac{d}{dx}\left[ \frac{5x}{(1-x^2 )^{\frac{1}{3}}}+sin^2 (2x + 3) \right]\\=5(1-x^2 )^{-\frac{1}{3}}+5x.\frac{-1}{3}.(1-x^2 )^{-\frac{4}{3}}.(-2x)+2sin(2x+3)cos(2x+3).2\\=\frac{5}{(1-x^2 )^{\frac{1}{3}}}+\frac{10x^2}{3(1-x^2 )^{\frac{4}{3}}}+2 sin⁡2(2x+3)\\=\frac{5.3(1 – x^2 )+ 10x^2}{3(1-x^2 )^{\frac{4}{3}}}+2 sin⁡(4x+6)\\= \frac{5(3 – x^2 )}{3(1 – x^2 )^{\frac{4}{3}}} + 2sin(4x + 6)\\Ans:\ Ⓐ\ \frac{5(3 – x^2 )}{3(1 – x^2 )^{\frac{4}{3}}} + 2sin(4x + 6)\) =
\(54.\ log\sqrt{\frac{1 +cos^2x}{1 – e^{2x}}}\) অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ -\(Ⓐ\ \frac{sin 2x}{2(1 + cos^2 x)} +\frac{e^{2x}}{1 – e^{2x}}\\Ⓑ\ -\frac{sin 2x}{2(1 + cos^2 x)} +\frac{e^{2x}}{1 – e^{2x}}\\Ⓒ\ \frac{sin 2x}{2(1 + cos^2 x)} +\frac{e^{2x}}{1 + e^{2x}}\\Ⓓ\ \frac{sin 2x}{2(1 + cos^2 x)} -\frac{e^{2x}}{1 – e^{2x}}\)
\(Solution:\ log\sqrt{\frac{1 +cos^2x}{1 – e^{2x}}}\\= \frac{1}{2}[log(1 + cos^2 x) – log(1 – e^{2x})]\\∴ \frac{dy}{dx}=\frac{1}{2}.\left[ \frac{1}{1 + cos^2 x}.2 cos⁡x.(-sinx)-\frac{1}{1 – e^{2x}}.(-2e^{2x}) \right]\\=\frac{1}{2}.\left[ \frac{-sin⁡2x}{1 + cos^2 x}+\frac{2e^{2x}}{1 – e^{2x}} \right]\\=-\frac{sin 2x}{2(1 + cos^2 x)} +\frac{e^{2x}}{1 – e^{2x}}\\Ans:\ Ⓑ\ -\frac{sin 2x}{2(1 + cos^2 x)} +\frac{e^{2x}}{1 – e^{2x}}\)
55. ^d/_dx [tan^-1 (^{cos x}/_{1 + sin x}) + sin logx] =
Ⓐ ¹/_x cos (logx)
Ⓑ ¹/₂ + ¹/_x cos (logx)
Ⓒ – ¹/₂ – ¹/_x cos (logx)
Ⓓ – ¹/₂ + ¹/_x cos (logx)
Solution: tan^-1 ^{cos x}/_{1 + sin x}
\(= tan^{-1} \frac{cos^2⁡\frac{x}{2}- sin^2\frac{x}{2}}{\left( cos\frac{x}{2}+sin\frac{x}{2} \right)^2}\\⁡= tan^{-1}\frac{(cos⁡⁡\frac{x}{2}+ sin⁡⁡\frac{x}{2})(cos⁡⁡\frac{x}{2}- sin⁡\frac{x}{2})}{(cos⁡\frac{x}{2}+sin⁡\frac{x}{2})^2}\\⁡= tan^{-1}\frac{cos⁡⁡\frac{x}{2}- sin⁡\frac{x}{2}}{cos⁡\frac{x}{2}+sin⁡\frac{x}{2}}\\=tan^{-1}\frac{1 – tan\frac{x}{2}}{1 – tan\frac{x}{2}}\\=tan^{-1}\frac{tan\frac{π}{4} – tan\frac{x}{2}}{1 – tan\frac{π}{4}.tan\frac{x}{2}}\)
= tan^-1 tan⁡(^π/₄ – ^x/₂) = ^π/₄ – ^x/₂
∴  ^d/_dx [tan^-1(^{cos x}/_{1 + sin x}) + sin logx]
= ^d/_dx [^π/₄ – ^x/₂ + sin logx]
= 0 – ¹/₂ + cos(log x).¹/_x = – ¹/₂ – ¹/_x cos (logx)
Ans: Ⓒ – ¹/₂ – ¹/_x cos (logx)
56. নীচের কোন সম্পর্কটির ক্ষেত্রে ^dy/_dx = ^{y(x – y)}/_x²
Ⓐ y = (1 + x)^x           Ⓑ y = x^{sin x}
Ⓒ x^cos²x                      Ⓓ x^y = e^x
Solution: Ⓓ x^y = e^x
∴ log x^y = log e^x
⇒ ylog x = x
∴ ^dy/_dx.log x + y.¹/_x = 1
⇒ ^dy/_dx . ^x/_y = 1 – ^y/_x = ^{x – y}/_x
⇒ ^dy/_dx = ^{y(x – y)}/_x²
Ans:Ⓓ x^y = e^x
57. নীচের কোন্ অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ sec x[1 + x + (1 + xtan x)(x + logx)] ?
Ⓐ x³logx                            Ⓑ √xlog √x
Ⓒ xsec xlog(xe^x)         Ⓓ ^tanx/_x log⁡(^{e^x}/_x^x)
Solution: y = xsec xlog(xe^x) = xsec x(log x + loge^x) = xsec x(log x + x)
∴ ^dy/_dx = 1.sec x(log x + x) + x.sec x.tan x(log x + x) + xsec x(¹/_x + 1)
          = sec x[(log x + x) + x.tan x(log x + x) + x(¹/_x + 1)]
= sec x[(log x + x)(1 + x.tan x) + 1 + x]
Ans: Ⓒ xsec xlog(xe^x)
58. যদি log(xy) = x² – y² হয়, তবে x = 1, y = 1 বিন্দুতে ^dy/_dx =
   Ⓐ 0 Ⓑ 1
Ⓒ ¹/₂ Ⓓ ¹/₃
Solution: log(xy) = x² – y²
∴ ¹/_xy (1.y + x.^dy/_dx) = 2x – 2y.^dy/_dx
x = 1, y = 1 বিন্দুতে,
¹/_1.1 (1.1 + 1.^dy/_dx) = 2.1 – 2.1.^dy/_dx
বা, ^dy/_dx + 2 ^dy/_dx = 2 – 1
বা, ^dy/_dx = ¹/₃Ans: Ⓓ ¹/₃
59. নীচের কোন সমীকরণটির ক্ষেত্রে ^dy/_dx = ^{y – 1 – x²y²}/_{1 – x + x²y²} হবে?
Ⓐ e^xy – 4xy = 4 Ⓑ xy = tan(x + y)
Ⓒ y^y = sin x Ⓓ log (xy) = e^{x + y} + 2
Solution: Ⓑ xy = tan(x + y)
⇒ tan^-1 xy = x + y
∴ ¹/_{1 + x²y²} .(1.y + x.^dy/_dx) = 1 + ^dy/_dx
⇒ (^x/_{1 + x²y²} – 1). ^dy/_dx = 1 – ^y/_{1 + x²y²}
⇒ ^{x – 1 – x²y²}/_{1 + x²y²} .^dy/_dx = ^{1 + x²y² – y}/_{1 + x²y²}
⇒ ^{x – 1 – x²y²}/_{1 + x²y²} .^dy/_dx = ^{y – 1 + x²y²}/_{1 – x + x²y²}
Ans: Ⓑ xy = tan(x + y)
60. যদি x = a(t – sin t) , y = a(1 – cos t) হয়, তবে t = ^π/₂ বিন্দুতে ^dy/_dx =
Ⓐ 0 Ⓑ 1
Ⓒ -1 Ⓓ π
Solution: x = a(t – sin t)
∴ ^dx/_dt = a(1 + cos t) = a.2sin² ^t/₂
     y = a(1 – cos t)
∴ ^dy/_dt = a(0 + sin t) = a.2sin ^t/₂ cos ^t/₂
∴ ^dy/_dx =(^dy/_dt)/(^dx/_dt)
      = ^{a.2sin ^t/₂ cos ^t/₂}/_{a.2sin ^t/₂ cos ^t/₂}
     = cot ^t/₂
       t = ^π/₂ বিন্দুতে,
^dy/_dx = cot ^π/₄ = 1
Ans:Ⓑ 1
61. যদি x = a(2t + sin 2t) , y = a(1 – cos 2t) হয়, তবে ^dy/_dx =
Ⓐ tant Ⓑ cosect
Ⓒ sect Ⓓ cot t
Solution: x = a(2t + sin 2t)
∴ ^dx/_dt = a(2 + 2cos 2t) = 2a(1 + cos 2t) = 2a.2cos² t
      y = a(1 – cos 2t)
∴ ^dy/_dt = a(0 + 2sin 2t) = 2a sin 2t) = 4a.sin t cos t
∴ ^dy/_dx =(^dy/_dt)/(^dx/_dt)
      = ^{4a.sin t cos t}/_{2a.2cos² t} = tan t
Ans:Ⓐ tant
62. যদি x = sec^-1 ^{1 + t²}/_{1 – t²} , y = sin^-1 ^{3t – t³}/_{1 – 3t²} হয়, তবে ^dy/_dx =
Ⓐ ¹/₂ Ⓑ 1
Ⓒ ³/₂ Ⓓ ¹/₃
Solution: x = sec^-1 ^{1 + t²}/_{1 – t²} = 2tan^-1 t
y = sin^-1 ^{3t – t³}/_{1 – 3t²} = 3tan^-1 t
∴ ^x/_y = ²/₃
বা, 2y = 3x
∴ 2.^dy/_dx = 3
বা, ^dy/_dx = ³/₂
Ans: Ⓒ ³/₂
63. যদি x = cos^-1(8t⁴ – 8t² + 1), y = sin^-1 (3t – 4t³) [0 < t < ¹/₂] হয়, তবে ^dy/_dx =
Ⓐ –¹/₂Ⓑ –²/₃
Ⓒ ³/₄ Ⓓ -1
Solution: ধরি, t = sin θ
x = cos^-1(8t⁴ – 8t² + 1)
= cos^-1[2(2t² – 1)² – 1]
= cos^-1[2(2Sin² θ – 1)² – 1]
= cos^-1[2(-cos 2θ)² – 1]
= cos^-1[2cos² 2θ – 1]
= cos^-1cos 4θ = 4θ
∴ ^dx/_dθ = 4
y = sin^-1 (3t – 4t³)
= sin^-1 (3 sin θ – 4sin³ θ)
= sin^-1sin 3θ = 3θ
∴ ^dy/_dθ = 3
∴ ^dy/_dx =(^dy/_dθ)/(^dx/_dθ)= ³/₄
Ans: Ⓒ ³/₄
64. যদি \(y = x^3\sqrt{\frac{x^2 + 4}{x^3+ 3}}\) হয়, তবে \(\frac{dy}{dx}=\ Ⓐ\ \frac{x^2 (3x^4+ 20x^2+ 36)}{(x + 3)\sqrt{(x^2+ 3)(x^2+ 4)}}\\Ⓑ\ \frac{x^2 (3x^4+20x^2+ 36)}{(x^2 + 3)\sqrt{(x^2+ 3)(x^2+ 4)}}\\Ⓒ\ \frac{3x^4+ 20x^2+ 36}{(x + 3)\sqrt{(x^2+3)(x^2+ 4)}}\) Ⓓ এদের কোনোটিই নয়
\(Solution:\ y = x^3\sqrt{\frac{x^2 + 4}{x^3+ 3}}\)
উভয় দিকে log নিয়ে পাই,
\(log y = log\left( x^3\sqrt{\frac{x^2 + 4}{x^3+3}} \right)\\⇒log y =log⁡x^3+\frac{1}{2}log⁡\frac{x^2 + 4}{x^3+ 3}\\⇒log y = 3log x + \frac{1}{2}[log (x^2 + 4) – log (x^2 + 3)\\∴ \frac{1}{y}.\frac{dy}{dx}=\frac{3}{x}+\frac{1}{2}.\left[ \frac{2x}{x^2 + 4}-\frac{2x}{x^2 + 3} \right]\\⇒ \frac{dy}{dx}=y.\left[\frac{3}{x}+ \frac{x}{x^2 + 4}-\frac{x}{x^2 + 3} \right]\\⇒ \frac{dy}{dx}= x^3\sqrt{\frac{x^2 +4}{x^3+ 3)}}.\frac{3(x^2 + 4)(x^2 + 3)+x^2 (x^2 + 3)-x^2 (x^2 + 4)}{x(x^2 + 4)(x^2 + 3)}\\⇒ \frac{dy}{dx} = x^3\sqrt{\frac{x^2 +4}{x^3+ 3)}}.\frac{3x^4 + 21x^2 +36 + x^4 + 3x^2 – x^(4 )- 4x^2)}{x(x^2 + 4)(x^2 + 3)}\\⇒ \frac{dy}{dx} = x^3\sqrt{\frac{x^2 +4}{x^3+ 3)}}.\frac{3x^4 + 20x^2 + 36}{x(x^2 + 4)(x^2 + 3}\\⇒ \frac{dy}{dx} =\frac{x^3\sqrt{x^2 + 4}(3x^4+ 20x^2+ 36)}{\sqrt{x^2 + 3}x(x^2+ 4)(x^2 + 3)} \\⇒ \frac{dy}{dx}= \frac{x^2 (3x^4+20x^2+ 36)}{(x^2 + 3)\sqrt{(x^2+ 3)(x^2+ 4)}}\\Ans: Ⓑ \frac{x^2 (3x^4+20x^2+ 36)}{(x^2 + 3)\sqrt{(x^2+ 3)(x^2+ 4)}}\)
65. যদি \(y=\sqrt{\frac{(x – a)(x – b)}{(x – c)(x – d)}}\) হয়, তবে \(\frac{dy}{dx}=\\Ⓐ\ \frac{1}{2}\sqrt{\frac{(x – a)(x – b)}{(x – c)(x – d)}}\left[ \frac{1}{x – a}+\frac{1}{x – b}-\frac{1}{x – c}-\frac{1}{x – d} \right]\\Ⓑ\ \frac{1}{2}\left[ \frac{1}{x – a}+\frac{1}{x – b}+\frac{1}{x – c}+\frac{1}{x – d} \right]\\Ⓒ\ \frac{1}{2}\sqrt{\frac{(x – a)(x – b)}{(x – c)(x – d)}}\left[ \frac{1}{x – a}+\frac{1}{x – b}-\frac{1}{x – c}-\frac{1}{x – d} \right]\)Ⓓএদের কোনোটিই নয়
\(Solution:\ y = \sqrt{\frac{(x – a)(x – b)}{(x – c)(x – d)}}\)
উভয় দিকে log নিয়ে পাই,
\(\quad logy = log\sqrt{\frac{(x – a)(x – b)}{(x – c)(x – d)}}\)
বা, log y = ¹/₂log⁡^{(x – a)(x – b)}/_{(x – c)(x – d)}
বা, log y = ¹/₂[log (x – a) + log (x – b) – log (x – c) – log (x – d)
∴ ¹/_y.^dy/_dx = ¹/₂.[¹/_{x – a} + ¹/_{x – b}– ¹/_{x – c} – ¹/_{x – d}]
⇒ ^dy/_dx = ^y/₂.[¹/_{x – a} + ¹/_{x – b}– ¹/_{x – c} – ¹/_{x – d}]
\(⇒\frac{dy}{dx}= \frac{1}{2}\sqrt{\frac{(x – a)(x – b)}{(x – c)(x – d)}}\left[ \frac{1}{x – a}+\frac{1}{x – b}-\frac{1}{x – c}-\frac{1}{x – d} \right]\\Ans:\ Ⓐ\ \frac{1}{2}\sqrt{\frac{(x – a)(x – b)}{(x – c)(x – d)}}\left[ \frac{1}{x – a}+\frac{1}{x – b}-\frac{1}{x – c}-\frac{1}{x – d} \right]\)
66. যদি \(y = log\sqrt{\frac{1 + sin x}{1 -sin x}}\)
হয়, তবে ^dy/_dx=
Ⓐ sinx      Ⓑ tanx
Ⓒ cotx      Ⓓ secx
\( Solution: y = log\sqrt{\frac{1 + sin x}{1 -sin x}}\\ = log\sqrt{\frac{(cos⁡\frac{x}{2}+sin⁡\frac{x}{2})^2}{(cos⁡\frac{x}{2}-sin⁡\frac{x}{2})^2}}\\= log\sqrt{\frac{(1 + tan⁡\frac{x}{2})^2}{(1 – tan⁡\frac{x}{2})^2}}\\=log⁡\frac{1 + tan⁡\frac{x}{2}}{1 – tan⁡\frac{x}{2}}=logtan(\frac{π}{4}+\frac{x}{2})\\∴ \frac{dy}{dx}=\frac{1}{tan(\frac{π}{4}+\frac{x}{2})}.sec^2(\frac{π}{4}+\frac{x}{2}).\frac{1}{2}\\=\frac{cos⁡(\frac{π}{4}+\frac{x}{2})}{2.sin(\frac{π}{4}+\frac{x}{2}).cos^2(\frac{π}{4}+\frac{x}{2})}\\=\frac{1}{2.sin(\frac{π}{4}+\frac{x}{2}).cos(\frac{π}{4}+\frac{x}{2})}\\=\frac{1}{sin2.(\frac{π}{4}+\frac{x}{2})}\\=\frac{1}{sin(\frac{π}{2}+x)}\\=\frac{1}{cosx}=sec⁡x\\Ans:\ Ⓓ secx\)
67. যদি y = ^{x + 2}/_{(x – 1)(x + 5)} হয়, তবে ^dy/_dx =
Ⓐ ^{x + 2}/_{(x – 1)(x + 5)}[¹/_{x + 2} – ¹/_{x – 1} + ¹/_{x + 5}]
Ⓑ ^{x + 2}/_{(x – 1)(x + 5)}[¹/_{x + 2} – ¹/_{x – 1} – ¹/_{x + 5})]
Ⓒ ^{(x+5)(x- 1)}/_{x + 2}[¹/_{x + 2} + ¹/_{x – 1} + ¹/_{x + 5}]
Ⓓ এদের কোনোটিই নয়
Solution: y = ^{x + 2}/_{(x – 1)(x + 5)}
উভয় দিকে log নিয়ে পাই,
      log y = log [^{x + 2}/_{(x – 1)(x + 5)}]
বা, log y = log (x + 2) – log (x – 1) – log (x + 5)
∴ ¹/_y.^dy/_dx = ¹/_{x + 2} – ¹/_{x – 1} – ¹/_{x + 5}
⇒ ^dy/_dx = y[¹/_{x + 2} – ¹/_{x – 1} – ¹/_{x + 5}]
⇒ ^dy/_dx = ^{x + 2}/_{(x – 1)(x + 5)}[¹/_{x + 2} – ¹/_{x – 1} – ¹/_{x + 5})]
Ans: Ⓑ ^{x + 2}/_{(x – 1)(x + 5)}[¹/_{x + 2} – ¹/_{x – 1} – ¹/_{x + 5})]
68. ^d/_dt (3^tcos t + sin t) =
Ⓐ 3^t(cos t log3 – sin t) + cos t
Ⓑ -3^t(cos t log3 – sin t) + cos t
Ⓒ 3^t(cos t log3 + sin t) + cos t
Ⓓ 3^t(cos t log3 – sin t) – cos t
Solution: ^d/_dt (3^tcos t + sin t)
= 3^t.log 3. cos t + 3^t.(-sin t) + cos t
= 3^t(3 cos t log 3 – sin t) + cos t
Ans: Ⓐ 3^t(cos t log3 – sin t) + cos t
69. ^d/_dt[(t ^{– 2} + t²)(2^t – 3^t)] =
Ⓐ (t^-2 + t² ) (2^tlog2 – 3^tlog3) + 2(2^t – 3^t)(t – t^{– 3})
Ⓑ (t² – t^{– 2})(2^tlog3 – 3^tlog2) + 2(3^t – 2^t)(t – t³)
Ⓒ(t² – t^{– 2})(2^tlog3 – 3^tlog2) + 2(3^t – 2^t)
Ⓓএদের কোনোটিই নয়
Solution: ^d/_dt[(t ^{– 2} + t²)(2^t – 3^t)]
= (-2t ^-3 + 2t)(2^t – 3^t) + (t ^{– 2} + t²)(2^t.log 2 – 3^t.log 3)
= 2(t – t^-3)(2^t – 3^t) + (t^-2 + t²)(2^t.log 2 – 3^t.log 3)
Ans: Ⓐ (t^-2 + t² ) (2^tlog2 – 3^tlog3) + 2(2^t – 3^t)(t – t^{– 3})
70. ^d/_dt (√t e^t sect) =
Ⓐ ^{e^t sect}/_2√t(1 + 2t – 2t tan t)
Ⓑ ^{e^t sect}/_√t(1 + 2t + 2t tan t)
Ⓒ ^{e^t sect}/_2√t(1 + 2t + 2t tan t)
Ⓓএদের কোনোটিই নয়
Solution: ^d/_dt (√t e^t sect)
= ¹/_2√t .e^t .sec t + √t.e^t .sec t + √t e^t sec t.tan t
= ^{e^t sect}/_2√t(1 + 2t + 2t tan t)
Ans: Ⓒ ^{e^t sect}/_2√t(1 + 2t + 2t tan t)
71. ^d/_du (^u/_{e^u – 1}) =
Ⓐ ^{e^u (1 – u) + 1}/_{(e^u – 1)²}     Ⓑ ^{e^u (1 – u) – 1}/_{(e^u – 1)²}
Ⓒ ^{e^u (1 – u) + 1}/_{(e^u + 1)²}     Ⓓ ^{e^u (1 – u) – 1}/_{e^u + 1}
Solution: ^d/_du (^u/_{e^u – 1})
= ^{1(e^u – 1) – u.e^u}/_{(e^u – 1)²}
= ^{e^u (1 – u) – 1}/_{(e^u – 1)²}
Ans: Ⓑ ^{e^u (1 – u) – 1}/_{(e^u – 1)²}
\(72.\ \frac{d}{du} \left( \frac{usin u + cos u}{u^2 logu} \right)=\\Ⓐ\ \frac{u^2 logucos u – (usin u + cos u)(1 + 2logu)}{u^3 (logu)^2}\\Ⓑ\ \frac{u^2 logucosu}{(logu)^2}-(1 + 2logu)\\Ⓒ\ \frac{(usin u + cos u)(1 + 2logu)}{u^3 (logu)^2}- u^2 logucos u\)Ⓓ এদের কোনোটিই নয়
\(Solution:\ \frac{d}{du} \left( \frac{usin u + cos u}{u^2 logu} \right)\\=\frac{(1.sin⁡u + u.cos⁡u – sin⁡u).u^2 logu – (usin u + cos u)(2u.log⁡u + u^2.\frac{1}{u})}{(u^2 log⁡u )^2}\\ = \frac{u^3 logu cos⁡u – (usin u + cos u).u.(2 log⁡u + 1)}{u^4 (log⁡u )^2}\\=\frac{u^2 logucos u – (usin u + cos u)(1 + 2logu)}{u^3 (logu)^2}\\Ans:\ Ⓐ\ \frac{u^2 logucos u – (usin u + cos u)(1 + 2logu)}{u^3 (logu)^2}\)
\(73.\ x = \frac{sin^3 t}{\sqrt{cos 2t}}\) এবং \(y = \frac{cos^3 t}{\sqrt{cos 2t}}\)
হলে t = ^π/₆ বিন্দুতে ^dy/_dx =
Ⓐ 0     Ⓑ 1     Ⓒ π
Ⓓএদের কোনোটিই নয়
\(Solution:\ x = \frac{sin^3 t}{\sqrt{cos 2t}}\\∴\frac{dx}{dt}=\frac{3sin^2⁡t.cos⁡t.\sqrt{cos 2t}-sin^3 t.\frac{1}{2\sqrt{cos 2t}}.(-2sin 2t)} {(\sqrt{cos 2t})^2}\\=\frac{6sin^2⁡t.cos⁡t.cos⁡2t+2sin^3t.sin 2t}{2cos 2t\sqrt{cos 2t}}\\\quad y = \frac{cos^3 t}{cos 2t}\\∴\frac{dy}{dt}=(\frac{3cos^2⁡t.(-sint)\sqrt{cos 2t}-cos^3t.\frac{1}{2\sqrt{cos 2t}}.(-2sin 2t)}{(\sqrt{cos 2t})^2}=\frac{-6cos^2⁡t.sin⁡t.cos⁡2t+2cos^3t.sin 2t}{2cos 2t\sqrt{cos 2t}}\\∴\frac{dy}{dx}=\frac{-6cos^2⁡t.sin⁡t.cos⁡2t+2cos^3t.sin 2t}{6sin^2⁡t.cos⁡t.cos⁡2t+2sin^3t.sin 2t}\)
t = ^π/₆ বিন্দুতে
\(∴\frac{dy}{dx}=\frac{-6cos^2⁡\frac{π}{6}.sin⁡\frac{π}{6}.cos⁡2.\frac{π}{6}+2cos^3\frac{π}{6}.sin 2.\frac{π}{6}}{6sin^2\frac{π}{6}t.cos⁡\frac{π}{6}.cos⁡2.\frac{π}{6}+2sin^3\frac{π}{6}.sin 2.\frac{π}{6}}\\=\frac{-6×\frac{3}{4}×\frac{1}{2}×\frac{1}{2}+ 2×\frac{3√3}{8}×\frac{√3}{2}}{6×\frac{1}{4}×\frac{√3}{2}×\frac{1}{2}+ 2×\frac{1}{8}×\frac{√3}{2}}\\=\frac{-\frac{18}{16}+\frac{18}{16}}{\frac{6√3}{16}+\frac{2√3}{16}}=0\\Ans:\ Ⓐ\ 0\)
\(74.\ \sqrt{1 – x^2}+\sqrt{1 – y^2}= a(x – y)\) হলে \(\frac{dy}{dx} =\\Ⓐ\ \frac{\sqrt{1 – x^2}}{\sqrt{1 – y^2}}\quad\quad Ⓑ\ \frac{\sqrt{1 + x^2}}{\sqrt{1 + y^2}}\\Ⓒ\ \frac{\sqrt{1 – y^2}}{\sqrt{1 – x^2}}\quad\quadⒹ\ \frac{\sqrt{1 + y^2}}{\sqrt{1 + x^2}}\)
Solution: ধরি, x = sin α,       y = sin β
\(\quad \sqrt{1 – x^2}+\sqrt{1 – y^2}= a(x – y)\\⇒\sqrt{1 – sin^2⁡α}+\sqrt{1 – sin^2⁡β}= a(sin⁡α – sin⁡β)\)
⇒ cos α + cos β = a(sin α – sin β)
⇒ 2.cos ^{α + β}/₂.cos^{α – β}/₂ = a.2.cos^{α + β}/₂.sin^{α – β}/₂
⇒cot ^{α – β}/₂ = a
⇒ ^{α – β}/₂ = cot^-1 a
⇒ α – β = 2cot^-1 a
⇒sin^-1 x – sin^-1 y = 2cot^-1 a
\(∴\frac{1}{\sqrt{1 – x^2}}- \frac{1} {\sqrt{1 – y^2}}.\frac{dy}{dx}= 0\\⇒\frac{1}{\sqrt{1 – y^2}}.\frac{dy}{dx}=\frac{1}{\sqrt{1 – x^2}}\\⇒ \frac{dy}{dx}=\frac{\sqrt{1 – y^2}}{\sqrt{1 – x^2}}\\Ans:\ Ⓒ \frac{\sqrt{1 – y^2}}{\sqrt{1 – x^2}}\)
75. y = e^{sin^-1x}⁡ এবং z = e^{-cos^-1x} হলে ^dy/_dz =
Ⓐ e^{tan^-1x}          Ⓑ tan^-1x
Ⓒ cot^-1x          Ⓓ e^π/2
\(Solution:\ y=e^{sin^{-1}x}\\∴\frac{dy}{dx}=e^{sin^{-1}x}.\frac{1}{\sqrt{1-x^2}}\\ z=e^{-cos^{-1}x}\\∴\frac{dz}{dx}=e^{-cos^{-1}x}.-\frac{1}{-\sqrt{1-x^2}}=e^{-cos^{-1}x}.\frac{1}{\sqrt{1-x^2}}\\∴\frac{dy}{dz}= \frac{\frac{dy}{dx}}{\frac{dz}{dx}}=\frac{e^{sin^{-1}x}.\frac{1}{\sqrt{1-x^2}}}{e^{-cos^{-1}x}.\frac{1}{\sqrt{1-x^2}}}\\=\frac{e^{sin^{-1}x}}{e^{-cos^{-1}x}}\\= e^{sin^{-1}x+cos^{-1}x}=e^{\frac{π}{2}}\\Ans:\ Ⓓ e^{\frac{π}{2}}\)
76. y = sin(^π/₆e^xy)হলে x = 0 -তে ^dy/_dx (y) =
Ⓐ ^√3π/₂₄       Ⓑ ^√3π/₁₂
Ⓒ ^π/₂₄            Ⓓ ^π/₆
Solution: y = sin(^π/₆e^xy)
∴ ^dy/_dx = cos(^π/₆.e^xy).^π/₆.e^xy.(1.y + x.^dy/_dx)
   x = 0 তে,
y = sin(^π/₆.e^0.y) = sin (^π/₆.e⁰) = sin ^π/₆ = ¹/₂
∴ x = 0 তে,
^dy/_dx = cos(^π/₆e^0.1/2).^π/₆e^0.1/2.(1. ¹/₂ + 0.^dy/_dx)
= cos(^π/₆).^π/₆.1.¹/₂ = ^√3/₂.^π/₁₂ = ^√3π/₂₄Ans: Ⓐ ^√3π/₂₄
77. নীচের কোন্ অপেক্ষকটির অন্তরকলজ (x-এর সাপেক্ষে) ^y/_{2y – x} ?
Ⓐ x^{x^{x^{. . . ∞}}}   Ⓑ y = √x^{√x^{√x^{. . . ∞}}}
Ⓒ y = x + ¹/_{x + ¹/_{x + . . . ∞}}
Ⓓ এদের কোনোটিই নয়
Solution: Ⓒ y = x + ¹/_{x + ¹/_{x + . . . ∞}}
∴ y = x + ¹/_y
⇒ y² = xy + 1
∴ 2y. ^dy/_dx = 1.y + x. ^dy/_dx + 0
⇒ (2y – x) ^dy/_dx = y
⇒ ^dy/_dx = ^y/_{2y – x}
Ans: Ⓒ y = x + ¹/_{x + ¹/_{x + . . . ∞}}
\(78.\ \sqrt{1 – x^4}\)-এর সাপেক্ষে \(\ \frac{\sqrt{1 + x^2}-\sqrt{1 – x^2}}{\sqrt{1 + x^2}+\sqrt{1 – x^2}}\) অন্তরকলজ \(\frac{\sqrt{(1- x^a )-1}}{x^b}\) হলে a + b=
Ⓐ 5         Ⓑ 7
Ⓒ 8         Ⓓ 10
Solution: ধরি, x² = cos 2θ
\(∴ y=\sqrt{1 – cos^2 2θ}=\sqrt{sin^2 2θ}=sin2θ\\∴ \frac{dy}{dθ}= cos 2θ.2 = 2x^2\\z=\frac{\sqrt{1 + x^2}-\sqrt{1 – x^2}}{\sqrt{1 + x^2}+\sqrt{1 – x^2}}\\=\frac{\sqrt{1 + cos 2θ}-\sqrt{1 – cos 2θ}}{\sqrt{1 + cos 2θ}+\sqrt{1 – cos 2θ}}\\=\frac{√2 cosθ-√2 sinθ}{√2 cosθ+√2 sinθ}\\=\frac{cosθ-sinθ}{cosθ+sinθ}\\=\frac{1-tanθ}{1+tanθ}=tan⁡(\frac{π}{4}-θ)\\∴ \frac{dz}{dθ}= sec^2(\frac{π}{4}- θ).(-1)\\=-\frac{2}{2cos^2⁡(\frac{π}{4} – θ)}\\= -\frac{2}{1+cos 2.⁡(\frac{π}{4} – θ)}\\= -\frac{2}{1+cos⁡(\frac{π}{2}- 2θ)}\\= \frac{-2}{1 + sin 2θ}=\frac{-2}{1 + \sqrt{1- x^4}}\\∴ \frac{dz}{dy}= \frac{\frac{dz}{dθ}}{\frac{dy}{dθ}}=\frac{\frac{-2}{1 + \sqrt{1- x^4}}}{2x^2}\\=\frac{-1}{x^2(1 + \sqrt{1- x^4})}\\=\frac{-(1 – \sqrt{1- x^4})}{x^2(1 + \sqrt{1- x^4})(1 – \sqrt{1- x^4})}\\=\frac{(\sqrt{1- x^4}-1)}{x^2(1-1+x^4)} \\=\frac{(\sqrt{1- x^4}-1)}{x^6}\)
∴ a = 4; b = 6
∴ a + b = 4 + 6 = 10
Ans: Ⓓ 10
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79. যদি y = log₂ (sin x³) হয়, তবে ^dy/_dx =
Ⓐ 3x²cot(x³)log₂ e
Ⓑ 3x²cot(x²log₂ e
Ⓒ x²cot(x³)log₂ e
Ⓓ 3x²cotx log_e 2
Solution: y = log₂ (sin x³) = log_e (sin x³).log₂ e
∴ ^dy/_dx = log₂ e.¹/_{sin x³}. ^d/_dx(sin x³)
= log₂ e .¹/_{sin x³}.cosx³.3x²
= 3x²cot(x³)log₂ e
Ans: Ⓐ 3x²cot(x³)log₂ e
৪0. যদি \(y = cos^{-1}\sqrt{2x-3}\) হয়, তবে \(\frac{dy}{dx}=\\Ⓐ\ -\frac{1}{2}.\frac{1}{\sqrt{(2x-3)(2-x)}}\quad Ⓑ\ -\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{(2x-3)(2-x)}}\\Ⓒ\ \frac{1}{\sqrt{2}}.\frac{1}{\sqrt{(2x-3)(2-x)}}\quad Ⓓ \frac{1}{\sqrt{2}}.\frac{1}{(2x-3)(2-x)}\)
\(Solution:\ y = cos^{-1}\sqrt{2x-3}\\∴\frac{dy}{dx}=-\frac{1}{\sqrt{1-(\sqrt{2x-3})^2}}×\frac{d}{dx}(\sqrt{2x-3})\\=-\frac{1}{1-2x+3}×\frac{2}{2\sqrt{2x-3}}=-\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{(2x-3)(2-x)}}\\Ans:\ Ⓑ\ -\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{(2x-3)(2-x)}}\)
81. বিকল্পগুলির মধ্যে কোন্ অপেক্ষকটির অন্তরকলজ
¹/_x cos(logx) + x^{cos x}(^{cos x}/_x – sin x.logx)
Ⓐ x^x² + a^x²
Ⓑ sin(logx) + x^{cos x}
Ⓒ (sin x)^cosx + e^3x
Ⓓ x^x + (sin x)^x
Solution: Ⓑ y = sin(logx) + x^{cos x} = sin(logx) + z
ধরি, z = x^{cos x}
∴ ^dy/_dx = coslogx.¹/_x + ^dz/_dx – – – (i)
আবার z = x^{cos x}
∴ log z = log xcos x = cosx logx
∴ ¹/_z.^dz/_dx = (-sinx).logx + cosx.¹/_x
⇒ ^dz/_dx = z(^cosx/_x – sinx.logx) = x^{cos x} (^cosx/_x – sinx.logx)
(i) থেকে পাই,
^dy/_dx = ^coslogx/_x + x^{cos x} (^cosx/_x – sinx.logx)
Ans: Ⓑ sin(logx) + x^{cos x}
82. ^d/_dx [(tan x)^{cot x} + (cot x)^{tan x}] =
    Ⓐ (tan x)^{cot x}[cosec² x(1 – logtan x)] + (cot x)^tanx[sec²x(logcot x – 1)]
    Ⓑ (tan x)^tan ^x[cosec² x(1 – logtan x)] + (cot x)^cotx[sec²x(logcot x – 1)]
    Ⓒ (tan x)^{cot x}[sec² x(1 – logcot x)] + (cot x)^{tan x}[cosec²x(logcot x – 1)]
    Ⓓ এদের কোনোটিই নয়
Solution: ধরি, (tan x)^{cot x} = u এবং (cot x)^{tan x} = v
      ^d/_dx [(tan x)^{cot x} + (cot x)^{tan x}]
= ^d/_dx (u + v) = ^du/_dx + ^dv/_dx – – – (i)
   u = (tan x)^{cot x}
∴ log u = log(tan x)^{cot x} = cotx log tanx
∴ ¹/_u.^du/_dx = (-cosec²x).logtanx + cotx.¹/_tanx.sec²x
⇒ ^du/_dx = u(cot²x.sec²x – cosec²x.logtanx) = cosec²x(tan x)^{cot x}(1 – logtanx)
    v = (cot x)^{tan x}
∴ log v = log (cot x)^{tan x} = tanx logcotx
∴ ¹/_v.^dv/_dx = sec²x.logcotx + tanx.¹/_cotx. (-cosec²x)
⇒ ^dv/_dx = v(sec²x.logcotx – tan²x.cosec²x) = sec²x(cot x)^{tan x}(logcotx – 1)
    (i) থেকে পাই,
^d/_dx [(tan x)^{cot x} + (cot x)^{tan x}]
= cosec²x(tan x)^{cot x}(1 – logtanx) + sec²x(cot x)^{tan x}(logcotx – 1)
Ans: Ⓐ (tan x)^{cot x}[cosec² x(1 – logtan x)] + (cot x)^tanx[sec²x(logcot x – 1)]
83. বিকল্পগুলির মধ্যে কোন্ অপেক্ষকটির অন্তরকলজ
     x^{sin x} [^{sin x}/_x + cos xlog x]+(sin x)^{cos x}[cos x cot x – sin xlog(sin x)]]
Ⓐ e^{cos^-1}⁡x + x^√x
Ⓑ (sin x)^{tan x} + (cos x)^secx
Ⓒ x^{sin x} + (sin x)^{cos x}
Ⓓ y^x + x^y + x^x = a^b
Solution: Ⓒ y = x^{sin x} + (sin x)^{cos x}
      y = u + v (ধরি)
∴ ^dy/_dx= ^du/_dx + ^dv/_dx – – – (i)
    u = x^{sin x}
∴ log u = logx^{sin x} = sin x log x
∴ ¹/_u.^du/_dx = cos x.logx + sin x.¹/_x
⇒ ^du/_dx = u(cos x.logx + ^{sin x}/_x) = x^{sin x}(cos x.logx + ^{sin x}/_x)
    v = (sin x)^{cos x}
∴ log v = log (sin x)^{cos x} = cosx log sin x
∴ ¹/_v.^dv/_dx = -sin x.logsinx + cos x.¹/_sinx.cos x
⇒ ^dv/_dx = v(cotx.cos x – sinx. logsinx) = (sin x)^{cos x}(cotx.cos x – sinx. logsinx)
(i) থেকে পাই,
^dy/_dx= x^{sin x}(cos x.logx + ^{sin x}/_x) + (sin x)^{cos x}(cotx.cos x – sinx. logsinx)
Ans: Ⓒ x^{sin x} + (sin x)^{cos x}
84. x^y + y^x = a হলে, ^dy/_dx =
\(Ⓐ\ -\frac{yx^{y-1}+y^x logy}{xy^{x-1}+x^y logx}\\Ⓑ\ \frac{yx^{y-1}+y^x logy}{xy^{x-1}+x^y logx}\\Ⓒ\ \frac{yx^{y-1}-y^x logy}{xy^{x-1}-x^y logx}\)Ⓓ এদের কোনোটিই নয়
Solution: x^y + y^x = a
      u + v = 1 (ধরি)
∴ ^du/_dx + ^dv/_dx = 0 – – – (i)
    u = x^y
∴ log u = logx^y = y log x
∴ ¹/_u.^du/_dx = ^dy/_dx logx + y.¹/_x
⇒ ^du/_dx = u(^dy/_dx logx + ^y/_x) = x^y(^dy/_dx logx + ^y/_x)
    v = y^x
∴ log v = log y^x = x log y
∴ ¹/_v.^dv/_dx = 1.logy + x.¹/_y.^dy/_dx
⇒ ^dv/_dx = v(logy + ^x/_y.^dy/_dx) = y^x(logy + ^x/_y.^dy/_dx)
(i) থেকে পাই,
   x^y(logx.^dy/_dx + ^y/_x) + y^x(logy + ^x/_y.^dy/_dx) = 0
⇒ x^y.logx.^dy/_dx + y^x. ^x/_y.^dy/_dx = -(x^y.^y/_x + y^x.logy)
⇒ ^dy/_dx(x^y.logx + y^x-1.x) = -(x^y-1.y + y^x.logy)
\(⇒\ \frac{dy}{dx}=-\frac{yx^{y-1}+y^x logy}{xy^{x-1}+x^y logx}\\Ans:\ Ⓐ\ -\frac{yx^{y-1}+y^x logy}{xy^{x-1}+x^y logx}\)
\(85.y= log\frac{1 – cos x}{1 + cos x} +a^x \) হলে,\(\frac{dy}{dx}=\)
Ⓐ cosec x + a^xlogx
Ⓑ sec x + a^xlog a
Ⓒ cosec x + x^xlog a
Ⓓ cosec x + a^xlog a
\(Solution:\ y= log\sqrt{\frac{1 – cos x}{1 + cos x}}+a^x\\= log\sqrt{tan^2\frac{x}{2}}+a^x= logtan\frac{x}{2}+a^x\\∴\frac{dy}{dx}=\frac{1}{tan\frac{x}{2}}×sec^2\frac{x}{2}×\frac{1}{2}+a^xlog_ea\\ =\frac{1}{2.sin\frac{x}{2}.cos⁡\frac{x}{2}}+a^xlog_ea\)
= ¹/_sin⁡x + a^xlog_ea
= cosecx + a^xlog_ea
Ans: Ⓓ cosec x + a^xlog a
\(86. e^y – \frac{a + b tan x}{a- b tan x} = 0\) হলে,\(\frac{dy}{dx}=\ Ⓐ\ \frac{2ab}{a^2cos^2x – b^2sin^2x}\quad Ⓑ\ \frac{-2ab}{a^2cos^2x – b^2sin^2x}\\Ⓒ\ \frac{2ab}{acosx – bsinx}\quad Ⓓ\ \frac{ab}{a^2cos^2x – b^2sin^2x}\)
Solution: e^y – ^{a + b tan x}/_{a- b tan x} = 0
⇒ e^y = ^{a + b tan x}/_{a- b tan x}
\(∴e^y.\frac{dy}{dx}=\frac{bsec^2⁡x(a- b tan x)-(a+ b tan x)(-bsec^2⁡x)}{(a- b tan x)^2}\\⇒ \frac{a + b tan x}{a- b tan x}.\frac{dy}{dx}=\frac{absec^2⁡x-b^2sec^2⁡xtan x+absec^2⁡x+b^2sec^2⁡xtan x}{(a- b tan x)^2}\\⇒(a + b tan x).\frac{dy}{dx}=\frac{2absec^2⁡x}{a- b tan⁡x}\\⇒\frac{dy}{dx}=\frac{2absec^2⁡x}{a^2- b^2tan^2⁡x}\\Ans:\ Ⓐ \frac{2ab}{a^2cos^2x – b^2sin^2x}\)
\(87.\ y =2tan^{-1}\frac{\sqrt{x-a}}{\sqrt{b-x}}\) হলে,\(\frac{dy}{dx}=\ Ⓐ\ \frac{1}{\sqrt{(x-a)(x-b)}}\quad Ⓑ\ \frac{1}{\sqrt{(x-a)(b-x}}\\Ⓒ\ \frac{1}{(x-a)(x-b)}\quad Ⓓ\ \frac{1}{(x-a)(b-x)}\)
\(Solution:\ y =2tan^{-1}\sqrt{\frac{x-a}{b-x}}\\∴\frac{dy}{dx}=2.\frac{1}{1+\sqrt{\left( \frac{x-a}{b-x} \right)^2}}.\frac{d}{dx}\sqrt{\left( \frac{x-a}{b-x} \right)}\\=\frac{2}{1+\frac{x-a}{b-x}}.\frac{1}{2\sqrt{\frac{x-a}{b-x}}}.\frac{1(b-x)-(x-a)(-1)}{(b-x)^2}\\=\frac{b-x}{b-x+x-a}.\frac{\sqrt{b-x}}{\sqrt{x-a}}.\frac{b-x+x-a}{(b-x)^2}\\=\frac{1}{b-a}.\frac{\sqrt{b-x}}{\sqrt{x-a}}.\frac{b-a}{b-x}\\=\frac{1}{\sqrt{x-a}}.\frac{1}{\sqrt{b-x}}\\=\frac{1}{\sqrt{(x-a)(b-x}}\\Ans:\ Ⓑ\ \frac{1}{\sqrt{(x-a)(b-x}}\)
\(88.\ y =cos^{-1}\frac{a+bcos x}{b+acosx}\ (b>a)\) হলে,\(\frac{dy}{dx}=\ Ⓐ\ \frac{\sqrt{a^2 – b^2}}{a + bcos x}\quad Ⓑ\ \frac{\sqrt{a^2 – b^2}}{b + acos x}\\Ⓒ\ \frac{\sqrt{b^2 – a^2}}{b + acos x}\quad Ⓓ\ \frac{\sqrt{b^2 – a^2}}{a + bcos x}\)
\(Solution:\ y =cos^{-1}\frac{a+bcos x}{b+acosx}\\∴\frac{dy}{dx}=-\frac{1}{\sqrt{1-(\frac{a+bcos x}{b+acosx})^2 }}.\frac{d}{dx}\left( \frac{a+bcos x}{b+acosx} \right)\\=-\frac{b+acosx}{\sqrt{(b+acosx)^2-(a+bcosx)^2}}.\frac{(b+acosx)(-bsinx)-(a+bcosx)(-asinx)}{(b+acosx)^2}\\=-\frac{1}{\sqrt{b^2+a^2cos^2⁡x-a^2-b^2cos^2⁡x}}.\frac{-b^2 sinx-absinxcosx+a^2 sinx+absinxcosx}{b+acosx}\\=-\frac{1}{\sqrt{(b^2-a^2 )(1-cos^2x)}}.\frac{-b^2 sinx+a^2 sinx}{b+acosx}\\= \frac{1}{\sqrt{b^2-a^2}sinx}.\frac{(b^2-a^2 )sinx}{(b+acosx)}\\=\frac{\sqrt{b^2-a^2}}{b+acosx}\\Ans:\ Ⓒ \ \frac{\sqrt{b^2 – a^2}}{b + acos x}\)
\(89.\ y =tan^{-1}\left[ \sqrt{\frac{a-b}{a+b}}tan\frac{x}{2} \right]\) হলে,\(\frac{dy}{dx}=\ Ⓐ\ \frac{\sqrt{a^2 – b^2}}{2(a + bcos x)}\\ Ⓑ\ \frac{\sqrt{a^2 – b^2}}{b + acos x}\\ Ⓒ\ \frac{\sqrt{b^2 – a^2}}{2(b + acos x)}\)Ⓓ এদের কোনোটিই নয়
\(Solution:\ y =tan^{-1}\left[ \sqrt{\frac{a-b}{a+b}}tan\frac{x}{2} \right]\\∴\frac{dy}{dx}=\frac{1}{1+\frac{a-b}{a+b}tan^2 \frac{x}{2}}.\sqrt{\frac{a-b}{a+b}}.sec^2\frac{⁡x}{2}.\frac{1}{2}\\= \frac{(a+b) cos^2\frac{⁡x}{2}}{(a+b) cos^2\frac{⁡x}{2}+(a-b) sin^2\frac{⁡x}{2}}.\sqrt{\frac{a-b}{a+b}}.\frac{1}{2cos^2\frac{⁡x}{2}}\\= \frac{a+b}{a(cos^2⁡\frac{⁡x}{2}+sin^2\frac{⁡x}{2})+b(cos^2⁡\frac{⁡x}{2}-sin^2\frac{⁡x}{2})}.\sqrt{\frac{a-b}{a+b}}.\frac{1}{2}\\=\frac{\sqrt{a+b}\sqrt{a-b}}{a+b(cos^2⁡\frac{⁡x}{2}-sin^2\frac{⁡x}{2})}.\frac{1}{2}\\=\frac{\sqrt{a^2-b^2}}{2(a+bcosx) }\\Ans:\ Ⓐ\ \frac{\sqrt{a^2-b^2}}{2(a+bcosx)}\)
\(90.\ y =sin^{-1}\frac{1}{\sqrt{1 + x^2}}+tan^{-1}\left(\frac{\sqrt{1 + x^2}-1}{x} \right)\)⁡হলে \( \frac{dy}{dx}= Ⓐ\ \frac{- 1}{1 + x^2}\\Ⓑ\ \frac{- 1}{2(1 + x^2 )}\\Ⓒ\ \frac{1}{2(1 + x^2 )}\\Ⓓ\ \frac{1}{1 + x^2}\)
Solution: ধরি, x = tan θ
\(∴ sin^{-1}\frac{1}{\sqrt{1 + x^2}}\\=sin^{-1}\frac{1}{\sqrt{1 + tan^2θ}}\)
=sin^-1 ¹/_sec⁡θ
= sin^-1 cosθ
=sin^-1 sin(^π/₂ – θ) = ^π/₂ – θ
\(tan^{-1}\left(\frac{\sqrt{1 + x^2}-1}{x} \right)\\=tan^{-1}\left(\frac{\sqrt{1 + tan^2θ}-1}{tan θ} \right)\\=tan^{-1}\left(\frac{sec⁡θ-1}{tan θ} \right)\\=tan^{-1}\left(\frac{1-cosθ}{sin θ} \right)\\=tan^{-1}\left(\frac{2sin^2⁡\frac{θ}{2}}{2sin⁡\frac{θ}{2}cos\frac{θ}{2}} \right)\\= tan^{-1} tan \frac{θ}{2}=\frac{θ}{2}\)
∴ y = ^π/₂ – θ + ^θ/₂ = ^π/₂ – ^θ/₂
         =^π/₂ – ¹/₂tan^-1x
∴ ^dy/_dx = 0 – ¹/₂.¹/_{1 + x²} = –¹/_{2(1 + x²)}
Ans: Ⓑ –¹/_{2(1 + x²)}
91. x^{sin y} + y^{sin x} = 1 হলে, ^dy/_dx =
\(Ⓐ -\frac{y}{x}.\frac{x^{siny} sin y + xy^{sinx} log y cos x}{x^{siny} y log x cos y + y^{sinx}.sin⁡x}\\Ⓑ -\frac{x}{y}.\frac{x^{siny}+xy^{sinx}log y cos x}{ylog x cos y}\\Ⓒ -\frac{x}{y}. \frac{sin⁡y+xy^{cos⁡x}}{x^{sin⁡x} + cos y}\)Ⓓ এদের কোনোটিই নয়
Solution: x^{sin y} + y^{sin x} = 1
    u + v = 1 (ধরি)
∴ ^du/_dx + ^dv/_dx = 0 – – – (i)
    u = x^{sin y}
∴ log u = logx^{sin y} = siny log x
∴ ¹/_u.^du/_dx = cosy. ^dy/_dx logx + siny.¹/_x
⇒ ^du/_dx = u(cosy. ^dy/_dx logx + siny.¹/_x) = x^{sin y}(cosy.logx.^dy/_dx + ^siny/_x)
    v = y^{sin x}
∴ log v = log y^{sin x} = sinx log y
∴ ¹/_v.^dv/_dx = cosx.logy + sinx.¹/_y.^dy/_dx
⇒ ^dv/_dx = v(cosx.logy + sinx.¹/_y.^dy/_dx) = y^{sin x}(cosx.logy + ^sinx/_y.^dy/_dx)
(i) থেকে পাই,
    x^{sin y}(cosy.logx.^dy/_dx + ^siny/_x) + y^{sin x}(cosx.logy + ^sinx/_y.^dy/_dx) = 0
⇒ x^{sin y}.cosy.logx.^dy/_dx + y^{sin x}. ^sinx/_y.^dy/_dx = -(x^{sin y}. ^siny/_x + y^{sin x}.cosx.logy)
⇒ ^dy/_dx(x^{sin y}.cosy.logx + y^{sin x}. ^sinx/_y) = -(x^{sin y}. ^siny/_x + y^{sin x}.cosx.logy)
⇒^dy/_dx. ¹/_y (yx^{sin y}.cosy.logx + y^{sin x}.sinx) = – ¹/_x.(x^{sin y}.siny + xy^{sin x}. cosx.logy)
\(⇒ \frac{dy}{dx}= -\frac{y}{x}.\frac{x^{siny} sin y + xy^{sinx} log y cos x}{x^{siny} y log x cos y + y^{sinx}.sin⁡x}\\Ans:\ Ⓐ\ -\frac{y}{x}.\frac{x^{siny} sin y + xy^{sinx} log y cos x}{x^{siny} y log x cos y + y^{sinx}.sin⁡x} \)
92. y = [(tanx)^tanx]^tanx হলে x = ^π/₄ -তে ^dy/_dx -এর মান =
Ⓐ 0           Ⓑ 2
Ⓒ 4           Ⓓ -3
Solution: y = [(tanx)^tanx]^tanx
উভয়দিকে log নিয়ে পাই,
log y = log[(tanx)^tanx]^tanx
         = tan x.log(tanx)^tanx
         = tan² x.log tanx
∴ ¹/_{y .}^dy/_dx = 2 tan x.sec² x.log tanx + tan² x. ¹/_{tan x}. sec² x
         = (2log tanx + 1)tan x.sec² x
⇒ ^dy/_dx = y(2log tanx + 1)tan x.sec² x
         = [(tanx)^tanx]^tanx (2log tanx + 1)tan x.sec² x
x = ^π/₄ -তে,
^dy/_dx =1.(2log 1 + 1).1.( √2)² = 2
Ans: Ⓑ 2
\(93.\ y =\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}log\left( x+\sqrt{x^2-a^2} \right)\) হলে \(\frac{dy}{dx}=Ⓐ\ \sqrt{x^2-2a}\quad Ⓑ\ \sqrt{x^2-2a}\\Ⓒ\ \sqrt{x^2-a^2}\quad\quad Ⓓ\ \sqrt{x^2-a^2}\)
\(Solution:\ y =\frac{x\sqrt{x^2-a^2}}{2}-\frac{a^2}{2}log\left( x+\sqrt{x^2-a^2} \right)\\∴ \frac{dy}{dx}=\frac{1}{2}\left(1.\sqrt{x^2-a^2}+x.\frac{2x}{2.\sqrt{x^2-a^2}}\right)-\frac{a^2}{2}.\frac{1}{x + \sqrt{x^2 – a^2}}.\left( 1+\frac{2x}{2\sqrt{x^2-a^2}}\right)\\=\frac{1}{2}\left( \sqrt{x^2-a^2}+\frac{x^2}{\sqrt{x^2-a^2}}\right)-\frac{a^2}{2}.\frac{1}{x+\sqrt{x^2-a^2}}.\frac{\sqrt{x^2-a^2}+x}{\sqrt{x^2-a^2}}\\=\frac{x^2 -a^2 + x^2}{2\sqrt{x^2-a^2}}-\frac{a^2}{2\sqrt{x^2-a^2}}\\=\frac{x^2-a^2+x^2-a^2}{2\sqrt{x^2-a^2}}\\=\frac{x^2-a^2}{\sqrt{x^2-a^2}}\\=\sqrt{x^2-a^2}\\Ans:\ Ⓒ\ \sqrt{x^2-a^2}\)
\(94.\ y=2sin^{-1}\frac{x – 2}{√6}-\sqrt{2 + 4x – x^2}\)
হলে x = 2 -এ ^dy/_dx =
Ⓐ ¹/_√6      Ⓑ ²/_√6
Ⓒ ²/_√3      Ⓓ ¹/_√3
\(Solution:\ y=2sin^{-1}\frac{x – 2}{√6}-\sqrt{2 + 4x – x^2}\\∴ \frac{dy}{dx}=2.\frac{1}{\sqrt{1-\left( \frac{x-2}{√6} \right)^2}} ).\frac{1}{√6}- \frac{1}{2\sqrt{2 + 4x – x^2}}.(4-2x)\\= 2.\frac{√6}{\sqrt{6 – (x – 2)^2}}.\frac{1}{√6}- \frac{2 – x}{\sqrt{2 + 4x – x^2}}\\= \frac{2}{\sqrt{6 – (x – 2)^2}}- \frac{2 – x}{\sqrt{2 + 4x – x^2}}\) x = 2-এ \(\frac{dy}{dx}=\frac{2}{\sqrt{6 – (2 – 2)^2}}- \frac{2 – 2}{\sqrt{2 + 4.2 – 2^2}}\\= \frac{2}{√6}\\Ans:\ Ⓑ\ \frac{2}{√6}\)
\(95.\ y=tan^{-1}⁡\frac{\sqrt{1 + t^2}+\sqrt{1 – t^2}}{\sqrt{1 + t^2}-\sqrt{1 – t^2}}\) হলে,\(\frac{dy}{dt}=\\Ⓐ\ \frac{t}{1 – t^4}\quad\quad Ⓑ\ -\frac{t}{1 – t^4}\\Ⓒ\ \frac{1}{1 – t^4}\quad\quad Ⓓ\ -\frac{1}{1 – t^4}\)
Solution: ধরি t² = cos 2θ
\(∴ y=tan^{-1}⁡\frac{\sqrt{1 + cos 2θ}+\sqrt{1 – cos 2θ}}{\sqrt{1 + cos 2θ}-\sqrt{1 – cos 2θ}}\\=tan^{-1}⁡\frac{√2 cosθ + √2 sinθ}{√2 cosθ – √2 sinθ}\\=tan^{-1}⁡\frac{cosθ + sinθ}{cosθ – sinθ}\\=tan^{-1}⁡\frac{1 + tanθ}{1 – tanθ}\\=tan^{-1}tan⁡(\frac{π}{4} + θ)\\=\frac{π}{4} + θ\\=\frac{π}{4} + \frac{1}{2}cos^{-1}⁡t^2\\∴ \frac{dy}{dx}= -\frac{1}{2}.\frac{1}{1- (t^2)^2}.2t= -\frac{t}{1 – t^4}\\Ans:\ Ⓑ\ -\frac{t}{1 – t^4}\)
\(96.\ y =cot^{-1}\frac{\sqrt{1 + x}-\sqrt{1 – x}}{\sqrt{1 + x}+\sqrt{1 – x}}\) হলে \(\left[ \frac{dy}{dx} \right]_{x=\frac{1}{2}}=\)
Ⓐ –¹/₂     Ⓑ –¹/_√2
Ⓒ – ¹/₃ Ⓓ – ¹/_√3
Solution: ধরি x = cos 2θ
∴ 1 + x = 1 + cos 2θ = 2cos² θ
এবং 1 – x = 1 – cos 2θ = 2sin² θ
\(y =cot^{-1}\frac{\sqrt{1 + x}-\sqrt{1 – x}}{\sqrt{1 + x}+\sqrt{1 – x}}\\=cot^{-1}\frac{√2 cosθ – √2 sinθ}{√2 cosθ – √2 sinθ}\\=cot^{-1}\frac{cosθ – sinθ}{cosθ – sinθ}\\=cot^{-1}\frac{1 – tanθ}{1 + tanθ}\\=cot^{-1}tan⁡\left( \frac{π}{4}- θ \right)\\=cot^{-1}tan⁡\left[\frac{π}{2}-\left(\frac{π}{4}+θ \right) \right]\\=\frac{π}{4}+θ\\=\frac{π}{4}+\frac{1}{2}cos^{-1}⁡x\)
\(∴ \frac{dy}{dx}=-\frac{1}{2}.\frac{1}{\sqrt{1- x^2}}\\∴ \left[ \frac{dy}{dx} \right]_{x=\frac{1}{2}}\\=-\frac{1}{2}.\frac{1}{\sqrt{1-\frac{1}{4}}}\\=-\frac{1}{2}.\frac{1}{\sqrt{\frac{3}{4}}}\\=-\frac{1}{2}.\frac{2}{√3}=-\frac{1}{√3}\\Ans:\ Ⓓ\ -\frac{1}{√3}\)
\(97.\ y\sqrt{x^2+ 1} = log(\sqrt{x^2+ 1}-x)\) হলে \((x^2+1)\frac{dy}{dx}+xy+1=\)
Ⓐ 0            Ⓑ 1
Ⓒ -1           Ⓓ এদের কোনোটিই নয়
\(Solution:\ y\sqrt{x^2+ 1}= log(\sqrt{x^2+ 1}-x)\\∴ y_1\sqrt{x^2+ 1}+y.\frac{2x}{2\sqrt{x^2+ 1}}= \frac{1}{\sqrt{x^2+ 1}-x}×\left( \frac{2x}{2\sqrt{x^2+ 1}}-1 \right)\\⇒ y_1\sqrt{x^2+ 1}+y.\frac{x}{\sqrt{x^2+ 1}}=\frac{1}{\sqrt{x^2+ 1}-x}×\frac{x-\sqrt{x^2+ 1}}{\sqrt{x^2+ 1}}\\⇒ y_1\sqrt{x^2+ 1} +\frac{xy}{\sqrt{x^2+ 1}}=\frac{-1}{\sqrt{x^2+ 1}}\\⇒ y_1 (x^2+ 1) + xy = -1\\⇒ (x^2+1) \frac{dy}{dx}+xy+1=0\\Ans:\ Ⓐ\ 0\)
98. y = 1 + ^a₁/_{x – a₁} + ^a₂x/_{(x – a₁)(x – a₂)} + ^a₃x²/_{(x – a₁)(x – a₂)(x – a₃)} হলে ^dy/_dx =
Ⓐ ^x/_y[^a₁/_{a₁ – x} + ^a₂/_{a₂ – x} + ^a₃/_{a₃ – x}]
Ⓑ ^y/_x[^a₁/_{x – a₁} + ^a₂/_{(x – a₂)} + ^a₃/_{(x – a₃)}
Ⓒ ^y/_x[^a₁/_{a₁ – x} + ^a₂/_{a₂ – x} + ^a₃/_{a₃ – x}]
Ⓓ ^x/_y[^a₁/_{x – a₁} + ^a₂/_{(x – a₂)} + ^a₃/_{(x – a₃)}]
Solution: y = 1 + ^a₁/_{x – a₁} + ^a₂x/_{(x – a₁)(x – a₂)} + ^a₃x²/_{(x – a₁)(x – a₂)(x – a₃)}
⇒ y = ^x/_{x – a₁} + ^a₂x/_{(x – a₁)(x – a₂)} + ^a₃x²/_{(x – a₁)(x – a₂)(x – a₃)}
⇒ y = ^x²/_{(x – a₁)(x – a₂)} + ^a₃x²/_{(x – a₁)(x – a₂)(x – a₃)}
⇒ y = ^x³/_{(x – a₁)(x – a₂)(x – a₃)}
উভয়দিকে log নিয়ে পাই,
     log y = 3log x – log(x – a₁) – log(x – a₂) – log(x – a₃)
∴ ¹/_y ^dy/_dx = 3.¹/_x – ¹/_{(x – a₁)} – ¹/_{(x – a₂)} – ¹/_{(x – a₃)}
⇒ ^dy/_dx = y[(¹/_x – ¹/_{(x – a₁)}) + (¹/_x – ¹/_{(x – a₂)}) + (¹/_x – ¹/_{(x – a₃)})]
⇒ ^dy/_dx = y[^-a_₁/_{x(x – a₁)} + ^-a_₂/_{x(x – a₂)} + ^-a₃/_{x(x – a₃)}]
⇒ ^dy/_dx = ^y/_x[^a₁/_{a₁ – x} + ^a₂/_{a₂ – x} + ^a₃/_{a₃ – x}]
Ans: Ⓒ ^y/_x[^a₁/_{a₁ – x} + ^a₂/_{a₂ – x} + ^a₃/_{a₃ – x}]
DERIVATIVE PART II →
← DIFFERENTIABILITY
October 4, 2026

Category: SEMESTER-3

CALCULAS 1ST ORDER DERIVATIVE Part II

CALCULAS 1ST ORDER DERIVATIVE Part II

1. d/dx cos(logx) = _______ Ⓐ – sin(logx) Ⓑ 1/x cos(logx)Ⓒ –1/x sin(logx) Ⓓ 1/x sin(logx)

6. y = 1/3x – 1 হলে, x = 0 বিন্দুতে dy/dx = _______ Ⓐ -3 Ⓑ 1/3Ⓒ –1/9 Ⓓ 3

9. যদি y = cosec³x হয়, তবে dy/dx = _______Ⓐ – 3cosecxcot xⒷ – 3cosec3xcot3 xⒸ – cosec3xcot xⒹ – 3cosec3xcot x

13. যদি y = f(x)n হয়, তবে dy/dx = ______Ⓐ n{f(x)}n – 1Ⓑ n{f(x)}n – 1.f'(x)Ⓒ n{f(x)}n .f'(x)Ⓓ nf'(x)

16. যদি y = 3u(x) হয়, তবে dy/dx = ______Ⓐ3u(x).log3Ⓑ 3u(x).u'(x)Ⓒ log3.u'(x)Ⓓ 3u(x).log3.u'(x)

20. d/dx(ex9) = ______ Ⓐ 9x8.ex9 Ⓑ x8.ex9Ⓒ 9ex8 Ⓓ 8ex8

21. d/dxlog(cot x) = ______Ⓐ secx cosecxⒷ – secx cosecxⒸ sinxcosxⒹ – sinxcosx

22. d/dxlog(logx) (x > 1) = ______ Ⓐ 1/logx Ⓑ 1/x2Ⓒ xlogx Ⓓ 1/xlogx

23. d/dx(eex) = ______ Ⓐ eex Ⓑ (ex)2 Ⓒ ex.eexⒹএদের কোনোটিই নয়

24. x-এর সাপেক্ষে log(cosecx + cot x) এর অন্তরকলজ kcosecx হলে k = ______Ⓐ 1 Ⓑ -1 Ⓒ 2 Ⓓ –1/2

28. tan-1(sin x/1 + cos x) অপেক্ষকটির x -এর সাপেক্ষে অন্তরকলজ ______ Ⓐ 1/2 Ⓑ 1 Ⓒ 0 Ⓓ 2

31. y = tanxsinx হলে, [dy/dx]x = π/4 = ______ Ⓐ 2 Ⓑ √2 Ⓒ √3 Ⓓ 1

32. যদি xy + 1 = cos(xy) হয়, তবে x = π/2, y = 0 বিন্দুতে dy/dx = ______ Ⓐ 0 Ⓑ 1 Ⓒ 2 Ⓓ 3

34. যদি tan y = 2t/1 – t2, sin x = 2t/1 + t2 হয়, তবে dy/dx = ______ Ⓐ 0 Ⓑ 1 Ⓒ -1 Ⓓ 2

35. যদি x = a(θ + sin θ) , y = a(1 + cos θ) হয়, তবে [dy/dx]θ = π/2 ______Ⓐ 0 Ⓑ -1 Ⓒ -2 Ⓓ 6

39. y = 1 + sin θ + cos θ/1 + sin θ – cos θ হলে dy/dθ + 1/1 – cos θ = ______Ⓐ 0 Ⓑ 1 Ⓒ -1 Ⓓ 2

42. নীচের ______ সম্পর্কটির ক্ষেত্রে dy/dx = y/x হবে।Ⓐ xy = yx Ⓑ xy.yx = 1Ⓒ (cos x)y = (sin y)xⒹ x3y4 = (x + y)7

45. x = t – e-t2/2t2 হলে t dx/dt + 2x – 1/2t = ______ Ⓐ e-t Ⓑ e-t2Ⓒ e-2t Ⓓ e-2t2

2. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

3. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

4. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।স্তম্ভ Aস্তম্ভ B

5. স্তম্ভ A-এর সাপেক্ষে স্তম্ভ B মেলাও।

Rearrangement of Sentences/Events

1. cos y = xcos(a + y) থেকে dy/dx নির্ণয়ের ধাপগুলি নীচে দেওয়া হল –

2. yz = ey – x থেকে dy/dx নির্ণয়ের ধাপগুলি নীচে দেওয়া হল –

Relationship between Statements ____

2. বিবৃতি-A: tan y = tan x + secx – 1/tan x – secx + 1 হলে dy/dx = 1/2বিবৃতি-B: y = tan-1(a cos x – b sin x/b cos x + a sin x) হলে dy/dx = – 1

True and False ___

2. বিবৃতি-I: d/dx(2x2 + 3)4 = 16x(2x2 + 3)3

4. বিবৃতি-I: d/dx (xx) = xx(1 + logx)বিবৃতি-II: d/dx(2ex – 3x + 10x ln(10) = (2ex – 3mxlogem + 10xloge10)বিবৃতি-III: d/dx (2x + 2 – ex + 1 + log10x = 2x + 2loge2 – ex + log10e/x

5. বিবৃতি-I: d/dx(10x .x10) = 10x.x9 (10 + xlog10)

10. বিবৃতি-I: tan-1x এর সাপেক্ষে tan-1x/1 + tan-1x এর অবকল সহগ 1/(1 – tan-1x)2

11. বিবৃতি-I: sin-1 2x/1 + x2 -এর সাপেক্ষে cos-1 1 – x2/1 + x2 এর অবকল সহগ 1/2

1. y = f(t) এবং x = Φ(t) হলে dy/dx = f’(t)/Φ’(t) যেখানে Φ’(t) ≠ 0

2. মনে করো, y = f(x) অপেক্ষকের বিপরীত অপেক্ষক x = g(y) এবং f'(x) = 1/1 + x2

4. মনে করো, (1 + x)n = C0 + C1x + C2x2 + . . . +Cnxn[i] C0 + 2C1 + 3C2 + . . . + (n +1)Cn =Ⓐ (n + 2)2n + 1Ⓑ (n + 2)2nⒸ (n + 1)2n + 1Ⓓ (n + 2)2n – 1

[ii] C1 – 2C2 + 3C3 – . . . + (- 1)n – 1nCn =Ⓐ 2 Ⓑ 0 Ⓒ 8 Ⓓ 4

DIFFERENTIATION SOLUTION OF SNDEY SEM-3

SOLUTION OF DIFFERENTIATION SNDEY SEMESTER-3

1. y = esin x হলে dy/dx =Ⓐ – esin x cos x Ⓑ esin x sin xⒸ esin x cos x Ⓓ ecos x

3. f(x) = 23x2 হলে f'(x) =Ⓐ 23x2.6x Ⓑ 23x2.log 2Ⓒ 23x2.3xlog 2 Ⓓ 23x2.6xlog 2

5. যদি dy/dx(tan-1 x)2 = k tan-1 x.1/1 + x2 হয় তবে k-এর মান হবে –Ⓐ 1 Ⓑ -2Ⓒ 2Ⓓ – 1

7. d/dx (5f(x))-এর মান হবে –Ⓐ 5f(x).f’(x)Ⓑ 5f(x).loge 5Ⓒ 5fx.f'(x)loge 5Ⓓ 5f(x).f’(x)loge 5

9. d/dx (xx)-এর মান হবে – Ⓐ x. xx – 1 Ⓑ xxlog x Ⓒ xx(1/x + log x) Ⓓ xx(1 + log x)

12. x = a cos θ, y = a sin θ হলে dy/dx এর মান হবে –Ⓐ tan θ Ⓑ – tan θⒸ -cot θ Ⓓ cot θ

13. x = 3 বিন্দুতে d/dx{|x – 1| + |x – 5|}-এর মান নিম্নের কোনটি হবে?Ⓐ -2 Ⓑ 0Ⓒ 2 Ⓓ 4

14. u ও v যদি x-এর অন্তরকলনযোগ্য অপেক্ষক হয়, তবে d/dx(tan-1 u/v) =

15. যদি d/dx(tan-1 x) = 1/1 + x2 হয়, তবে d/dx(cot-1 x) =Ⓐ -(1 + x2) Ⓑ – 1/1 + x2Ⓒ 2/1 + x2 Ⓓ 1/1 – x2

16. 2tan-1 x/a -এর x-এর সাপেক্ষে অন্তরকলজ করে পাই, k/(x2 + a2) হলে k =Ⓐ 1 Ⓑ 2Ⓒ 3 Ⓓ 4

17. log(cot-1 x) -এর x-এর সাপেক্ষে অন্তরকলজ করে পাই, 1/k(1 + x2) । এক্ষেত্রে k =Ⓐ x Ⓑ √xⒸ tan-1x Ⓓ – cot-1 x

19. d/dx (2sec-1 2x – 3sin-1 x + 3cos-1 x2 ) =

22. d/dx (1/px + q) =Ⓐ p/(px + q)2 Ⓑ q/(px + q)2Ⓒ – p/(px + q)2 Ⓓ – q/(px + q)2

24. d/dx(1010x)=Ⓐ 1010x + 1.log 10xⒷ 1010x + 1Ⓒ 1010x + 1.log 10Ⓓ 10x.log 10

25. d/dx(22x2 + 5x) =Ⓐ (4x + 5).22x2 + 5xⒷ 22x2 + 5x.log 2Ⓒ (4x+5) log⁡2Ⓓ (4x+5)22x2 + 5x.log 2

27. d/dxlog[log(logx)] =Ⓐ x/log(log x)Ⓑ x/log x.log(log x)Ⓒ x/xlog xⒹ এদের কোনোটিই নয়

28. d/dx(10log(cos x)) =Ⓐ tan x . 10log(cos x).log 10Ⓑ – tan x . 10log(cos x).log 10Ⓒ tan x . 10log(sin x).log 10Ⓓ – tan x . 10log(sin x).log 10

29. d/dx sin(cos x2) =Ⓐ 2x sin x2.cos(cos x2)Ⓑ -2x sin x2.cos(cos x2)Ⓒ x sin x2.cos(cos x)Ⓓ এদের কোনোটিই নয়

S N DEY SEMESTER-3 অবকলন বা অন্তরকলন (Differentiation)

34. d/dx(secn x) = nanb হলে a2 – b2 =Ⓐ 1 Ⓑ 0 Ⓒ -1Ⓓ এদের কোনোটিই নয়

36. d/dx(sin-1x)m =

41. x2 + xy + y2 = 100 হলে, dy/dx =Ⓐ – 2x + y/x + 2y Ⓑ – x + y/x + 2yⒸ 2x + y/x + 2y Ⓓ এদের কোনোটিই নয়

42. x3 + y3 = 3axy হলে, dy/dx =

45. x = ylog xy হলে, dy/dx =Ⓐ x – y/x + y Ⓑ y(x – y)/x + yⒸ x + y/x – y Ⓓ y(x – y)/x(x + y)

46. ax² + 2hxy + by² + 2gx + 2fy + c = 0 হলে, dy/dx =

47. (x2 + y2)2 = xy হলে dy/dx =

48. x = at2, y = 2at হলে dy/dx =Ⓐ t Ⓑ 1/tⒸ – 1/t Ⓓ1/t3

49. যদি x = t logt, y = logt/t হয়, তবে t = 1 বিন্দুতে dy/dx =Ⓐ 1 Ⓑ -1Ⓒ 5 Ⓓ -5

55. d/dx [tan-1 (cos x/1 + sin x) + sin logx] =Ⓐ 1/x cos (logx)Ⓑ 1/2 + 1/x cos (logx)Ⓒ – 1/2 – 1/x cos (logx)Ⓓ – 1/2 + 1/x cos (logx)

56. নীচের কোন সম্পর্কটির ক্ষেত্রে dy/dx = y(x – y)/x2Ⓐ y = (1 + x)x Ⓑ y = xsin xⒸ xcos2x Ⓓ xy = ex

57. নীচের কোন্ অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ sec x[1 + x + (1 + xtan x)(x + logx)] ?Ⓐ x3logx Ⓑ √xlog √xⒸ xsec xlog(xex) Ⓓ tanx/x log⁡(ex/xx)

58. যদি log(xy) = x2 – y2 হয়, তবে x = 1, y = 1 বিন্দুতে dy/dx = Ⓐ 0 Ⓑ 1Ⓒ 1/2 Ⓓ 1/3

59. নীচের কোন সমীকরণটির ক্ষেত্রে dy/dx = y – 1 – x2y2/1 – x + x2y2 হবে?Ⓐ exy – 4xy = 4 Ⓑ xy = tan(x + y)Ⓒ yy = sin x Ⓓ log (xy) = ex + y + 2

60. যদি x = a(t – sin t) , y = a(1 – cos t) হয়, তবে t = π/2 বিন্দুতে dy/dx =Ⓐ 0 Ⓑ 1Ⓒ -1 Ⓓ π

61. যদি x = a(2t + sin 2t) , y = a(1 – cos 2t) হয়, তবে dy/dx =Ⓐ tant Ⓑ cosectⒸ sect Ⓓ cot t

62. যদি x = sec-1 1 + t2/1 – t2 , y = sin-1 3t – t3/1 – 3t2 হয়, তবে dy/dx =Ⓐ 1/2 Ⓑ 1Ⓒ 3/2 Ⓓ 1/3

63. যদি x = cos-1(8t4 – 8t2 + 1), y = sin-1 (3t – 4t3) [0 < t < 1/2] হয়, তবে dy/dx =Ⓐ –1/2 Ⓑ –2/3 Ⓒ 3/4 Ⓓ -1

67. যদি y = x + 2/(x – 1)(x + 5) হয়, তবে dy/dx =Ⓐ x + 2/(x – 1)(x + 5)[1/x + 2 – 1/x – 1 + 1/x + 5]Ⓑ x + 2/(x – 1)(x + 5)[1/x + 2 – 1/x – 1 – 1/x + 5)]Ⓒ (x+5)(x- 1)/x + 2[1/x + 2 + 1/x – 1 + 1/x + 5]Ⓓ এদের কোনোটিই নয়

1. ^d/_dx cos(logx) = _______
Ⓐ – sin(logx) Ⓑ ¹/_x cos(logx)
Ⓒ –¹/_x sin(logx) Ⓓ ¹/_x sin(logx)

6. y = ¹/_{3x – 1} হলে, x = 0 বিন্দুতে ^dy/_dx = _______
Ⓐ -3 Ⓑ ¹/₃
Ⓒ –¹/₉Ⓓ 3

9. যদি y = cosec³x হয়, তবে ^dy/_dx = _______
Ⓐ – 3cosecxcot x
Ⓑ – 3cosec³xcot³ x
Ⓒ – cosec³xcot x
Ⓓ – 3cosec³xcot x

13. যদি y = f(x)ⁿ হয়, তবে ^dy/_dx = ______
Ⓐ n{f(x)}^{n – 1}
Ⓑ n{f(x)}^{n – 1}.f'(x)
Ⓒ n{f(x)}ⁿ .f'(x)
Ⓓ nf'(x)

16. যদি y = 3^u(x) হয়, তবে ^dy/_dx = ______
Ⓐ3^u(x).log3
Ⓑ 3^u(x).u'(x)
Ⓒ log3.u'(x)
Ⓓ 3^u(x).log3.u'(x)

20. ^d/_dx(e^x⁹) = ______
Ⓐ 9x⁸.e^x⁹ Ⓑ x⁸.e^x⁹
Ⓒ 9e^x⁸ Ⓓ 8e^x⁸

21. ^d/_dxlog(cot x) = ______
Ⓐ secx cosecx
Ⓑ – secx cosecx
Ⓒ sinxcosx
Ⓓ – sinxcosx

22. ^d/_dxlog(logx) (x > 1) = ______
Ⓐ ¹/_logx Ⓑ ¹/_x²
Ⓒ xlogx Ⓓ ¹/_xlogx

23. ^d/_dx(e^{e^x}) = ______
Ⓐ e^{e^x} Ⓑ (e^x)² Ⓒ e^x.e^{e^x}
Ⓓএদের কোনোটিই নয়

24. x-এর সাপেক্ষে log(cosecx + cot x) এর অন্তরকলজ kcosecx হলে k = ______
Ⓐ 1 Ⓑ -1 Ⓒ 2 Ⓓ –¹/₂

28. tan^-1(^{sin x}/_{1 + cos x}) অপেক্ষকটির x -এর সাপেক্ষে অন্তরকলজ ______
Ⓐ ¹/₂ Ⓑ 1 Ⓒ 0 Ⓓ 2

31. y = tanx^sinx হলে, [^dy/_dx]_x_{= ^π/₄} = ______
Ⓐ 2 Ⓑ √2 Ⓒ √3 Ⓓ 1

32. যদি xy + 1 = cos(xy) হয়, তবে x = ^π/₂, y = 0 বিন্দুতে ^dy/_dx = ______
Ⓐ 0 Ⓑ 1 Ⓒ 2 Ⓓ 3

34. যদি tan y = ^2t/_{1 – t²}, sin x = ^2t/_{1 + t²} হয়, তবে ^dy/_dx = ______
Ⓐ 0 Ⓑ 1 Ⓒ -1 Ⓓ 2

35. যদি x = a(θ + sin θ) , y = a(1 + cos θ) হয়, তবে [^dy/_dx]_θ_{= ^π/₂} ______
Ⓐ 0 Ⓑ -1 Ⓒ -2 Ⓓ 6

39. y = ^{1 + sin} ^θ ^{+ cos} ^θ/_{1 + sin} _θ _{– cos} _θ হলে ^dy/_dθ + ¹/_{1 – cos} _θ = ______
Ⓐ 0 Ⓑ 1 Ⓒ -1 Ⓓ 2

42. নীচের ______ সম্পর্কটির ক্ষেত্রে ^dy/_dx = ^y/_x হবে।
Ⓐ x^y = y^x Ⓑ x^y.y^x = 1
Ⓒ (cos x)^y = (sin y)^x
Ⓓ x³y⁴ = (x + y)⁷

45. x = ^{t – e^-t²}/_2t² হলে t ^dx/_dt + 2x – ¹/_2t = ______
Ⓐ e^-t Ⓑ e^{^-t²}
Ⓒ e^-2t Ⓓ e^-2t²

1. cos y = xcos(a + y) থেকে ^dy/_dx নির্ণয়ের ধাপগুলি নীচে দেওয়া হল –

2. y^z = e^{y – x} থেকে ^dy/_dx নির্ণয়ের ধাপগুলি নীচে দেওয়া হল –

2. বিবৃতি-A: tan y = ^{tan x + secx – 1}/_{tan x – secx + 1} হলে ^dy/_dx = ¹/₂
বিবৃতি-B: y = tan^-1(^{a cos x – b sin x}/_{b cos x + a sin x}) হলে ^dy/_dx = – 1

2. বিবৃতি-I: ^d/_dx(2x² + 3)⁴ = 16x(2x² + 3)³

4. বিবৃতি-I: ^d/_dx (x^x) = x^x(1 + logx)
বিবৃতি-II: ^d/_dx(2e^x – 3^x + 10^x ln(10) = (2e^x – 3m^xlog_e^m + 10^xlog_e¹⁰)
বিবৃতি-III: ^d/_dx (2^{x + 2} – e^{x + 1} + log₁₀x = 2^{x + 2}log_e² – e^x + ^log10e/_x

5. বিবৃতি-I: ^d/_dx(10^x .x¹⁰) = 10^x.x⁹ (10 + xlog10)

10. বিবৃতি-I: tan^-1x এর সাপেক্ষে ^{tan^-1x}/_{1 + tan^-1x} এর অবকল সহগ ¹/_{(1 – tan^-1x)²}

11. বিবৃতি-I: sin^-1 ^2x/_{1 + x²} -এর সাপেক্ষে cos^-1 ^{1 – x²}/_{1 + x²} এর অবকল সহগ ¹/₂

1. y = f(t) এবং x = Φ(t) হলে ^dy/_dx = ^f’(t)/_Φ’(t) যেখানে Φ’(t) ≠ 0

2. মনে করো, y = f(x) অপেক্ষকের বিপরীত অপেক্ষক x = g(y) এবং f'(x) = ¹/_{1 + x²}

4. মনে করো, (1 + x)ⁿ = C₀ + C₁x + C₂x² + . . . +C_nxⁿ
[i] C₀ + 2C₁ + 3C₂ + . . . + (n +1)C_n =
Ⓐ (n + 2)2^{n + 1}
Ⓑ (n + 2)2ⁿ
Ⓒ (n + 1)2^{n + 1}
Ⓓ (n + 2)2^{n – 1}

[ii] C₁ – 2C₂ + 3C₃ – . . . + (- 1)^{n – 1}nC_n =
Ⓐ 2 Ⓑ 0 Ⓒ 8 Ⓓ 4

1. y = e^{sin x} হলে ^dy/_dx =
Ⓐ – e^{sin x} cos x Ⓑ e^{sin x} sin x
Ⓒ e^{sin x} cos x Ⓓ e^{cos x}

3. f(x) = 2^3x² হলে f'(x) =
Ⓐ 2^3x².6x Ⓑ 2^3x².log 2
Ⓒ 2^3x².3xlog 2 Ⓓ 2^3x².6xlog 2

5. যদি ^dy/_dx(tan^-1 x)² = k tan^-1 x.¹/_{1 + x²} হয় তবে k-এর মান হবে –
Ⓐ 1 Ⓑ -2
Ⓒ 2Ⓓ – 1

7. ^d/_dx (5^f(x))-এর মান হবে –
Ⓐ 5^f(x).f’(x)
Ⓑ 5^f(x).log_e 5
Ⓒ 5fx.f'(x)loge 5
Ⓓ 5^f(x).f’(x)log_e 5

9. ^d/_dx (x^x)-এর মান হবে –
Ⓐ x. x^{x – 1} Ⓑ x^xlog x
Ⓒ x^x(¹/_x + log x) Ⓓ x^x(1 + log x)

12. x = a cos θ, y = a sin θ হলে ^dy/_dx এর মান হবে –
Ⓐ tan θ Ⓑ – tan θ
Ⓒ -cot θ Ⓓ cot θ

13. x = 3 বিন্দুতে ^d/_dx{|x – 1| + |x – 5|}-এর মান নিম্নের কোনটি হবে?
Ⓐ -2 Ⓑ 0
Ⓒ 2 Ⓓ 4

14. u ও v যদি x-এর অন্তরকলনযোগ্য অপেক্ষক হয়, তবে ^d/_dx(tan^-1 ^u/_v) =

15. যদি ^d/_dx(tan^-1 x) = ¹/_{1 + x²} হয়, তবে ^d/_dx(cot^-1 x) =
Ⓐ -(1 + x²) Ⓑ – ¹/_{1 + x²}
Ⓒ ²/_{1 + x²} Ⓓ ¹/_{1 – x²}

16. 2tan^-1 ^x/_a -এর x-এর সাপেক্ষে অন্তরকলজ করে পাই, ^k/_{(x² + a²)} হলে k =
Ⓐ 1 Ⓑ 2
Ⓒ 3 Ⓓ 4

17. log(cot^-1 x) -এর x-এর সাপেক্ষে অন্তরকলজ করে পাই, ¹/_{k(1 + x²)} । এক্ষেত্রে k =
Ⓐ x Ⓑ √x
Ⓒ tan^-1x Ⓓ – cot^-1 x

19. ^d/_dx (2sec^-1 2x – 3sin^-1 x + 3cos^-1 x² ) =

22. ^d/_dx (¹/_{px + q}) =
Ⓐ ^p/_{(px + q)²} Ⓑ ^q/_{(px + q)²}
Ⓒ – ^p/_{(px + q)²} Ⓓ – ^q/_{(px + q)²}

24. d/dx(10^10x)=
Ⓐ 10^{10x + 1}.log 10x
Ⓑ 10^{10x + 1}
Ⓒ 10^{10x + 1}.log 10
Ⓓ 10^x.log 10

25. ^d/_dx(2^2x² ^{+ 5x}) =
Ⓐ (4x + 5).2^2x² ^{+ 5x}
Ⓑ 2^2x² ^{+ 5x}.log 2
Ⓒ (4x+5) log⁡2
Ⓓ (4x+5)2^2x² ^{+ 5x}.log 2

27. ^d/_dxlog[log(logx)] =
Ⓐ ^x/_{log(log x)}
Ⓑ ^x/_{log x.log(log x)}
Ⓒ ^x/_{xlog x}
Ⓓ এদের কোনোটিই নয়

28. ^d/_dx(10^{log(cos x)}) =
Ⓐ tan x . 10^{log(cos x)}.log 10
Ⓑ – tan x . 10^{log(cos x)}.log 10
Ⓒ tan x . 10^{log(sin x)}.log 10
Ⓓ – tan x . 10^{log(sin x)}.log 10

29. ^d/_dx sin(cos x²) =
Ⓐ 2x sin x².cos(cos x²)
Ⓑ -2x sin x².cos(cos x²)
Ⓒ x sin x².cos(cos x)
Ⓓ এদের কোনোটিই নয়

34. ^d/_dx(secⁿ x) = naⁿb হলে a² – b² =
Ⓐ 1 Ⓑ 0 Ⓒ -1
Ⓓ এদের কোনোটিই নয়

36. ^d/_dx(sin^-1x)^m =

41. x² + xy + y² = 100 হলে, ^dy/_dx =
Ⓐ – ^{2x + y}/_{x + 2y} Ⓑ – ^{x + y}/_{x + 2y}
Ⓒ ^{2x + y}/_{x + 2y} Ⓓ এদের কোনোটিই নয়

42. x³ + y³ = 3axy হলে, ^dy/_dx =

45. x = ylog xy হলে, ^dy/_dx =
Ⓐ ^{x – y}/_{x + y} Ⓑ ^{y(x – y)}/_{x + y}
Ⓒ ^{x + y}/_{x – y} Ⓓ ^{y(x – y)}/_{x(x + y)}

46. ax² + 2hxy + by² + 2gx + 2fy + c = 0 হলে, ^dy/_dx =

47. (x² + y²)² = xy হলে ^dy/_dx =

48. x = at², y = 2at হলে ^dy/_dx =
Ⓐ t Ⓑ ¹/_t
Ⓒ – ¹/_t Ⓓ¹/_t3

49. যদি x = t logt, y = ^logt/_t হয়, তবে t = 1 বিন্দুতে ^dy/_dx =
Ⓐ 1 Ⓑ -1
Ⓒ 5 Ⓓ -5

55. ^d/_dx [tan^-1 (^{cos x}/_{1 + sin x}) + sin logx] =
Ⓐ ¹/_x cos (logx)
Ⓑ ¹/₂ + ¹/_x cos (logx)
Ⓒ – ¹/₂ – ¹/_x cos (logx)
Ⓓ – ¹/₂ + ¹/_x cos (logx)

56. নীচের কোন সম্পর্কটির ক্ষেত্রে ^dy/_dx = ^{y(x – y)}/_x²
Ⓐ y = (1 + x)^x Ⓑ y = x^{sin x}
Ⓒ x^cos²x Ⓓ x^y = e^x

57. নীচের কোন্ অপেক্ষকটির x-এর সাপেক্ষে অন্তরকলজ sec x[1 + x + (1 + xtan x)(x + logx)] ?
Ⓐ x³logx Ⓑ √xlog √x
Ⓒ xsec xlog(xe^x) Ⓓ ^tanx/_x log⁡(^{e^x}/_x^x)

58. যদি log(xy) = x² – y² হয়, তবে x = 1, y = 1 বিন্দুতে ^dy/_dx =
Ⓐ 0 Ⓑ 1
Ⓒ ¹/₂ Ⓓ ¹/₃

59. নীচের কোন সমীকরণটির ক্ষেত্রে ^dy/_dx = ^{y – 1 – x²y²}/_{1 – x + x²y²} হবে?
Ⓐ e^xy – 4xy = 4 Ⓑ xy = tan(x + y)
Ⓒ y^y = sin x Ⓓ log (xy) = e^{x + y} + 2

60. যদি x = a(t – sin t) , y = a(1 – cos t) হয়, তবে t = ^π/₂ বিন্দুতে ^dy/_dx =
Ⓐ 0 Ⓑ 1
Ⓒ -1 Ⓓ π

61. যদি x = a(2t + sin 2t) , y = a(1 – cos 2t) হয়, তবে ^dy/_dx =
Ⓐ tant Ⓑ cosect
Ⓒ sect Ⓓ cot t

62. যদি x = sec^-1 ^{1 + t²}/_{1 – t²} , y = sin^-1 ^{3t – t³}/_{1 – 3t²} হয়, তবে ^dy/_dx =
Ⓐ ¹/₂ Ⓑ 1
Ⓒ ³/₂ Ⓓ ¹/₃

63. যদি x = cos^-1(8t⁴ – 8t² + 1), y = sin^-1 (3t – 4t³) [0 < t < ¹/₂] হয়, তবে ^dy/_dx =
Ⓐ –¹/₂Ⓑ –²/₃
Ⓒ ³/₄ Ⓓ -1

68. ^d/_dt (3^tcos t + sin t) =
Ⓐ 3^t(cos t log3 – sin t) + cos t
Ⓑ -3^t(cos t log3 – sin t) + cos t
Ⓒ 3^t(cos t log3 + sin t) + cos t
Ⓓ 3^t(cos t log3 – sin t) – cos t

70. ^d/_dt (√t e^t sect) =
Ⓐ ^{e^t sect}/_2√t(1 + 2t – 2t tan t)
Ⓑ ^{e^t sect}/_√t(1 + 2t + 2t tan t)
Ⓒ ^{e^t sect}/_2√t(1 + 2t + 2t tan t)
Ⓓএদের কোনোটিই নয়

71. ^d/_du (^u/_{e^u – 1}) =
Ⓐ ^{e^u (1 – u) + 1}/_{(e^u – 1)²} Ⓑ ^{e^u (1 – u) – 1}/_{(e^u – 1)²}
Ⓒ ^{e^u (1 – u) + 1}/_{(e^u + 1)²} Ⓓ ^{e^u (1 – u) – 1}/_{e^u + 1}

75. y = e^{sin^-1x}⁡ এবং z = e^{-cos^-1x} হলে ^dy/_dz =
Ⓐ e^{tan^-1x} Ⓑ tan^-1x
Ⓒ cot^-1x Ⓓ e^π/2

76. y = sin(^π/₆e^xy)হলে x = 0 -তে ^dy/_dx (y) =
Ⓐ ^√3π/₂₄ Ⓑ ^√3π/₁₂
Ⓒ ^π/₂₄ Ⓓ ^π/₆

77. নীচের কোন্ অপেক্ষকটির অন্তরকলজ (x-এর সাপেক্ষে) ^y/_{2y – x} ?
Ⓐ x^{x^{x^{. . . ∞}}} Ⓑ y = √x^{√x^{√x^{. . . ∞}}}
Ⓒ y = x + ¹/_{x + ¹/_{x + . . . ∞}}
Ⓓ এদের কোনোটিই নয়

79. যদি y = log₂ (sin x³) হয়, তবে ^dy/_dx =
Ⓐ 3x²cot(x³)log₂ e
Ⓑ 3x²cot(x²log₂ e
Ⓒ x²cot(x³)log₂ e
Ⓓ 3x²cotx log_e 2

81. বিকল্পগুলির মধ্যে কোন্ অপেক্ষকটির অন্তরকলজ
¹/_x cos(logx) + x^{cos x}(^{cos x}/_x – sin x.logx)
Ⓐ x^x² + a^x²
Ⓑ sin(logx) + x^{cos x}
Ⓒ (sin x)^cosx + e^3x
Ⓓ x^x + (sin x)^x

84. x^y + y^x = a হলে, ^dy/_dx =

92. y = [(tanx)^tanx]^tanx হলে x = ^π/₄ -তে ^dy/_dx -এর মান =
Ⓐ 0 Ⓑ 2
Ⓒ 4 Ⓓ -3

হলে x = 2 -এ ^dy/_dx =
Ⓐ ¹/_√6 Ⓑ ²/_√6
Ⓒ ²/_√3 Ⓓ ¹/_√3