Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.3 | Set 3

Last Updated : 08 May, 2021

Question 33. Differentiate $y=tan^{-1}\left(\frac{x^{\frac{1}{3}}+a^\frac{1}{3}}{1-(ax)^{\frac{1}{3}}}\right)$ with respect to x.

Solution:

We have, $y=tan^{-1}\left(\frac{x^{\frac{1}{3}}+a^\frac{1}{3}}{1-(ax)^{\frac{1}{3}}}\right)$

= $tan^{-1}(x^\frac{1}{3})+tan^{-1}(a^\frac{1}{3})$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}\left(tan^{-1}(x^\frac{1}{3})+tan^{-1}(a^\frac{1}{3})\right)$

= $\frac{\frac{1}{3}x^{\frac{1}{3}-1}}{1+(x^{\frac{1}{3}})^2}$

= $\frac{\frac{1}{3}x^{\frac{-2}{3}}}{1+x^{\frac{2}{3}}}$

= $\frac{1}{3x^{\frac{2}{3}}(1+x^{\frac{2}{3}})}$

Question 34. Differentiate $y=sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$ with respect to x.

Solution:

We have, $y=sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$

On putting 2^x = tan θ, we get,

$y=sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$

= $sin^{-1}\left(\frac{2tanθ}{1+tan^2θ}\right)$

= $sin^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{1+\frac{sin^2θ}{cos^2θ}}\right)$

= $sin^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{sin^2θ+cos^2θ}{cos^2θ}}\right)$

= $sin^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{1}{cos^2θ}}\right)$

= $sin^{-1}\left(\frac{2sinθcos^2θ}{cosθ}\right)$

= $sin^{-1}\left(2sinθcosθ\right)$

= $sin^{-1}\left(sin2θ\right)$

= 2θ

= 2 tan⁻¹ (2^x)

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}\left(2 tan^{−1}(2x)\right)$

= $\frac{2×2^xlog2}{1+(2^x)^2}$

= $\frac{2^{x+1}log2}{1+4^x}$

Question 35. If $y=sin^{-1}\left(\frac{2x}{1+x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ , 0 < x < 1, prove that $\frac{dy}{dx}=\frac{4}{1+x^2}$ .

Solution:

We have, $y=sin^{-1}\left(\frac{2x}{1+x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$

= $sin^{-1}\left(\frac{2x}{1+x^2}\right)+cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

On putting x = tan θ, we get,

y = $sin^{-1}\left(\frac{2tanθ}{1+tan^2θ}\right)+cos^{-1}\left(\frac{1-tan^2θ}{1+tan^2θ}\right)$

= $sin^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{1+\frac{sin^2θ}{cos^2θ}}\right)+cos^{-1}\left(\frac{1-\frac{sin^2θ}{cos^2θ}}{1+\frac{sin^2θ}{cos^2θ}}\right)$

= $sin^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{sin^2θ+cos^2θ}{cos^2θ}}\right)+cos^{-1}\left(\frac{\frac{cos^2θ-sin^2θ}{cos^2θ}}{\frac{cos^2θ+sin^2θ}{cos^2θ}}\right)$

= $sin^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{1}{cos^2θ}}\right)+cos^{-1}\left(\frac{\frac{cos^2θ-(1-cos^2θ)}{cos^2θ}}{\frac{1}{cos^2θ}}\right)$

= $sin^{-1}\left(2sinθcosθ\right)+cos^{-1}\left(2cos^2θ-1\right)$

= $sin^{-1}\left(sin2θ\right)+cos^{-1}\left(cos2θ\right)$

Now, 0 < x < 1

=> 0 < tan θ < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ + 2θ

= 4θ

= 4 tan⁻¹ x

Now, L.H.S. = $\frac{dy}{dx} = \frac{d}{dx}\left(4tan^{−1}x\right)$

= $\frac{4}{1+x^2}$

= R.H.S.

Hence proved.

Question 36. If $y=sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)+cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$ , 0 < x < ∞, prove that $\frac{dy}{dx}=\frac{2}{1+x^2}$ .

Solution:

We have, $y=sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)+cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$

On putting x = tan θ, we get,

$y=sin^{-1}\left(\frac{tanθ}{\sqrt{1+tan^2θ}}\right)+cos^{-1}\left(\frac{1}{\sqrt{1+tan^2θ}}\right)$

= $sin^{-1}\left(\frac{tanθ}{\sqrt{sec^2θ}}\right)+cos^{-1}\left(\frac{1}{\sqrt{sec^2θ}}\right)$

= $sin^{-1}\left(\frac{tanθ}{secθ}\right)+cos^{-1}\left(\frac{1}{secθ}\right)$

= $sin^{-1}\left(\frac{\frac{sinθ}{cosθ}}{\frac{1}{cosθ}}\right)+cos^{-1}\left(cosθ\right)$

= $sin^{-1}\left(sinθ\right)+cos^{-1}\left(cosθ\right)$

Now, 0 < x < ∞

=> 0 < tan θ < ∞

=> 0 < θ < π/2

So, y = θ + θ

= 2θ

= 2 tan⁻¹ x

Now, L.H.S. = $\frac{dy}{dx} = \frac{d}{dx}\left(2tan^{−1}x\right)$

= $\frac{2}{1+x^2}$

= R.H.S.

Hence proved.

Question 37 Differentiate the following with respect to x :

(i) cos⁻¹ (sin x)

Solution:

We have, y = cos⁻¹ (sin x)

= $cos^{−1}(cos(\frac{π}{2}-x))$

= $\frac{π}{2}-x$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(\frac{π}{2}-x)$

= 0 − 1

= −1

(ii) $cot^{−1}\left(\frac{1-x}{1+x}\right)$

Solution:

We have, y = $cot^{−1}\left(\frac{1-x}{1+x}\right)$

On putting x = tan θ, we get,

$y=cot^{−1}\left(\frac{1-tanθ}{1+tanθ}\right)$

= $cot^{−1}\left(\frac{tan\frac{π}{4}-tanθ}{1+tan\frac{π}{4}tanθ}\right)$

= $cot^{−1}\left(tan(\frac{π}{4}-θ)\right)$

= $cot^{−1}\left(cot(\frac{π}{2}-\frac{π}{4}+θ)\right)$

= $cot^{−1}\left(cot(\frac{π}{4}+θ)\right)$

= $\frac{π}{4}+θ$

= $\frac{π}{4}+tan^{-1}x$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(\frac{π}{4}+tan^{-1}x)$

= 0 + $\frac{1}{1+x^2}$

= $\frac{1}{1+x^2}$

Question 38. Differentiate $y=cot^{-1}\left[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$ , 0 < x < π/2 with respect to x.

Solution:

We have, $y=cot^{-1}\left[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$

= $cot^{-1}\left[\frac{(\sqrt{1+sinx}+\sqrt{1-sinx})^2}{(\sqrt{1+sinx}-\sqrt{1-sinx})(\sqrt{1+sinx}+\sqrt{1-sinx})}\right]$

= $cot^{-1}\left[\frac{1+sinx+1-sinx+2(\sqrt{1+sinx})(\sqrt{1-sinx})}{1+sinx-1+sinx}\right]$

= $cot^{-1}\left[\frac{2+2(\sqrt{1-sin^2x})}{2sinx}\right]$

= $cot^{-1}\left[\frac{2+2cosx}{2sinx}\right]$

= $cot^{-1}\left[\frac{1+cosx}{sinx}\right]$

= $cot^{-1}\left[\frac{2cos^2\frac{x}{2}}{2sin\frac{x}{2}cos\frac{x}{2}}\right]$

= $cot^{-1}\left[cot\frac{x}{2}\right]$

= $\frac{x}{2}$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2})$

= $\frac{1}{2}$

Question 39. If $y=tan^{-1}\left(\frac{2x}{1-x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ , x > 0, prove that $\frac{dy}{dx}=\frac{4}{1+x^2}$ .

Solution:

We have, $y=tan^{-1}\left(\frac{2x}{1-x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$

= $tan^{-1}\left(\frac{2x}{1-x^2}\right)+cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

On putting x = tan θ, we get,

y = $tan^{-1}\left(\frac{2tanθ}{1-tan^2θ}\right)+cos^{-1}\left(\frac{1-tan^2θ}{1+tan^2θ}\right)$

= $tan^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{1-\frac{sin^2θ}{cos^2θ}}\right)+cos^{-1}\left(\frac{1-\frac{sin^2θ}{cos^2θ}}{1+\frac{sin^2θ}{cos^2θ}}\right)$

= $tan^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{cos^2θ-sin^2θ}{cos^2θ}}\right)+cos^{-1}\left(\frac{\frac{cos^2θ-sin^2θ}{cos^2θ}}{\frac{cos^2θ+sin^2θ}{cos^2θ}}\right)$

= $tan^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{cos^2θ-(1-cos^2θ)}{cos^2θ}}\right)+cos^{-1}\left(\frac{\frac{cos^2θ-(1-cos^2θ)}{cos^2θ}}{\frac{1}{cos^2θ}}\right)$

= $tan^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{2cos^2θ-1}{cos^2θ}}\right)+cos^{-1}\left(\frac{\frac{2cos^2θ-1}{cos^2θ}}{\frac{1}{cos^2θ}}\right)$

= $tan^{-1}\left(\frac{\frac{2sinθ}{cosθ}}{\frac{cos2θ}{cos^2θ}}\right)+cos^{-1}\left(\frac{\frac{cos2θ}{cos^2θ}}{\frac{1}{cos^2θ}}\right)$

= $tan^{-1}\left(\frac{2sinθcosθ}{cos2θ}\right)+cos^{-1}\left(cos2θ\right)$

= $tan^{-1}\left(\frac{sin2θ}{cos2θ}\right)+cos^{-1}\left(cos2θ\right)$

= $tan^{-1}\left(tan2θ\right)+cos^{-1}\left(cos2θ\right)$

Here, 0 < x < ∞

=> 0 < tan θ < ∞

=> 0 < θ < π/2

=> 0 < 2θ < π

So, y = 2θ + 2θ

= 4θ

= 4 tan⁻¹ x

Now, L.H.S. = $\frac{dy}{dx} = \frac{d}{dx}\left(4tan^{−1}x\right)$

= $\frac{4}{1+x^2}$

= R.H.S.

Hence proved.

Question 40. If $y=sec^{-1}\left(\frac{x+1}{x-1}\right)+sin^{-1}\left(\frac{x-1}{x+1}\right)$ , x > 0, find $\frac{dy}{dx}$ .

Solution:

We have, $y=sec^{-1}\left(\frac{x+1}{x-1}\right)+sin^{-1}\left(\frac{x-1}{x+1}\right)$

= $cos^{-1}\left(\frac{x-1}{x+1}\right)+sin^{-1}\left(\frac{x-1}{x+1}\right)$

= $\frac{π}{2}$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(\frac{π}{2})$

= 0

Question 41. If $y=sin\left[2tan^{-1}\sqrt{(\frac{1-x}{1+x})}\right]$ , find $\frac{dy}{dx}$ .

Solution:

We have, $y=sin\left[2tan^{-1}\sqrt{(\frac{1-x}{1+x})}\right]$

On putting x = cos 2θ, we get,

$y=sin\left[2tan^{-1}\sqrt{(\frac{1-cos2θ}{1+cos2θ})}\right]$

= $sin\left[2tan^{-1}\sqrt{(\frac{2sin^2θ}{2cos^2θ})}\right]$

= $sin\left[2tan^{-1}(\sqrt{tan^2θ})\right]$

= $sin\left[2tan^{-1}(tanθ)\right]$

= $sin\left(2θ\right)$

= $sin\left(2×\frac{1}{2}cos^{-1}x\right)$

= $sin\left(cos^{-1}x\right)$

= $sin\left(sin^{-1}(\sqrt{1-x^2})\right)$

= $\sqrt{1-x^2}$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1-x^2})$

= $\frac{-2x}{2\sqrt{1-x^2}}$

= $\frac{-x}{\sqrt{1-x^2}}$

Question 42. If $y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$ , 0 < x < 1/2, find $\frac{dy}{dx}$ .

Solution:

We have, $y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$

On putting 2x = cos θ, we get,

$y=cos^{-1}(cosθ)+2cos^{-1}(\sqrt{1-cos^2θ})$

= $cos^{-1}(cosθ)+2cos^{-1}(sinθ)$

= $cos^{-1}(cosθ)+2cos^{-1}(cos(\frac{π}{2}-θ))$

Now, 0 < x < 1/2

=> 0 < 2x < 1

=> 0 < cos θ < 1

=> 0 < θ < π/2

and 0 > −θ > −π/2

=> π/2 > (π/2 −θ) > 0

So, y = $θ+2(\frac{π}{2}-θ)$

= π − θ

= π − cos⁻¹ (2x)

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(π−cos^{−1}(2x))$

= $0-\left(\frac{-2}{\sqrt{1-(2x)^2}}\right)$

= $\frac{2}{\sqrt{1-4x^2}}$

Question 43. If the derivative of tan⁻¹ (a + bx) takes the value of 1 at x = 0, prove that 1 + a² = b.

Solution:

We have, y = tan⁻¹ (a + bx)

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(tan^{−1}(a + bx))$

= $\frac{b}{1+(a+bx)^2}$

At x = 0, we have,

=> $\frac{b}{1+(a+b(0))^2}$ = 1

=> $\frac{b}{1+a^2}$ = 1

=> 1 + a² = b

Hence proved.

Question 44. If $y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$ , −1/2 < x < 0, find $\frac{dy}{dx}$ .

Solution:

We have, $y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$

On putting 2x = cos θ, we get,

$y=cos^{-1}(cosθ)+2cos^{-1}(\sqrt{1-cos^2θ})$

= $cos^{-1}(cosθ)+2cos^{-1}(sinθ)$

= $cos^{-1}(cosθ)+2cos^{-1}(cos(\frac{π}{2}-θ))$

Now, −1/2 < x < 0

=> −1 < 2x < 0

=> −1 < cos θ < 0

=> π/2 < θ < π

and −π/2 > −θ > −π

=> 0 > (π/2 −θ) > −π/2

So, y = $θ+2(-\frac{π}{2}+θ)$

= −π + 3θ

= −π + 3 cos⁻¹ (2x)

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(−π+3cos^{−1}(2x))$

= 0 + $\frac{-6}{\sqrt{1-(2x)^2}}$

= $\frac{-6}{\sqrt{1-4x^2}}$

Question 45. If $y=tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ , find $\frac{dy}{dx}$ .

Solution:

We have, $y=tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$

On putting x = cos 2θ, we get,

$y=tan^{-1}\left(\frac{\sqrt{1+cos2θ}-\sqrt{1-cos2θ}}{\sqrt{1+cos2θ}+\sqrt{1-cos2θ}}\right)$

= $tan^{-1}\left(\frac{\sqrt{2cos^2θ}-\sqrt{2sin^2θ}}{\sqrt{2cos^2θ}+\sqrt{2sin^2θ}}\right)$

= $tan^{-1}\left(\frac{\sqrt{2}(cosθ-sinθ)}{\sqrt{2}(cosθ+sinθ)}\right)$

= $tan^{-1}\left(\frac{\frac{cosθ-sinθ}{cosθ}}{\frac{cosθ+sinθ}{cosθ}}\right)$

= $tan^{-1}\left(\frac{1-tanθ}{1+tanθ}\right)$

= $tan^{-1}\left(\frac{tan\frac{π}{4}-tanθ}{1+tan\frac{π}{4}tanθ}\right)$

= $tan^{-1}\left(tan(\frac{π}{4}-θ)\right)$

= $\frac{π}{4}-θ$

= $\frac{π}{4}-\frac{1}{2}cos^{-1}x$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(\frac{π}{4}-\frac{1}{2}cos^{-1}x)$

= 0 − $\left(\frac{-1}{2\sqrt{1-x^2}}\right)$

= $\frac{1}{2\sqrt{1-x^2}}$

Question 46. If $y=cos^{-1}\left(\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right)$ , find $\frac{dy}{dx}$ .

Solution:

We have, $y=cos^{-1}\left(\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right)$

On putting x = cos θ, we get,

$y=cos^{-1}\left(\frac{2cosθ-3\sqrt{1-cos^2θ}}{\sqrt{13}}\right)$

= $cos^{-1}\left(\frac{2cosθ-3sinθ}{\sqrt{13}}\right)$

= $cos^{-1}\left(\frac{2}{\sqrt{13}}cosθ-\frac{3}{\sqrt{13}}sinθ\right)$

Let $cosØ=\frac{2}{\sqrt{13}}$

=> sin Ø = $\sqrt{1-cos^2Ø}$

=> sin Ø = $\sqrt{1-\left(\frac{2}{\sqrt{13}}\right)^2}$

=> sin Ø = $\sqrt{1-\frac{4}{13}}$

=> sin Ø = $\sqrt{\frac{9}{13}}$

=> sin Ø = $\frac{3}{\sqrt{13}}$

So, y = $cos^{-1}\left(cosØcosθ-sinØsinθ\right)$

= $cos^{-1}\left(cos(Ø+θ)\right)$

= Ø + θ

= $cos^{-1}\left(\frac{2}{\sqrt{13}}\right)+cos^{-1}x$

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(cos^{-1}\left(\frac{2}{\sqrt{13}}\right)+cos^{-1}x)$

= 0 + $\left(\frac{-1}{\sqrt{1-x^2}}\right)$

= $\frac{-1}{\sqrt{1-x^2}}$

Question 47. Differentiate $y=sin^{-1}\left[\frac{2^{x+1}×3^x}{1+(36)^x}\right]$ with respect to x.

Solution:

We have, $y=sin^{-1}\left[\frac{2^{x+1}×3^x}{1+(36)^x}\right]$

= $sin^{-1}\left[\frac{2×2^x×3^x}{1+(36)^x}\right]$

= $sin^{-1}\left[\frac{2×6^x}{1+(6)^{2x}}\right]$

On putting 6^x = tan θ, we get,

= $sin^{-1}\left[\frac{2tanθ}{1+tan^2θ}\right]$

= $sin^{-1}\left[\frac{\frac{2sinθ}{cosθ}}{1+\frac{sin^2θ}{cos^2θ}}\right]$

= $sin^{-1}\left[\frac{\frac{2sinθ}{cosθ}}{\frac{cos^2θ+sin^2θ}{cos^2θ}}\right]$

= $sin^{-1}\left[\frac{\frac{2sinθ}{cosθ}}{\frac{1}{cos^2θ}}\right]$

= $sin^{-1}\left[{\frac{2sinθcos^2θ}{cosθ}}\right]$

= $sin^{-1}\left(2sinθcosθ\right)$

= $sin^{-1}\left(sin2θ\right)$

= 2θ

= 2 tan⁻¹ (6^x)

Differentiating with respect to x, we get,

$\frac{dy}{dx} = \frac{d}{dx}(2tan^{−1}(6x))$

= $\frac{2×6^xlog6}{1+(6^x)^2}$

= $\frac{2×6^xlog6}{1+36^x}$

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Class 12 RD Sharma Solutions - Chapter 11 Differentiation - Exercise 11.3 | Set 2

Class 12 RD Sharma Solutions- Chapter 11 Differentiation - Exercise 11.4 | Set 1

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Chapter 1: Relations

Chapter 2: Functions

Chapter 3: Binary Operations

Chapter 4: Inverse Trigonometric Functions

Chapter 5: Algebra of Matrices

Chapter 6: Determinants

Chapter 7: Adjoint and Inverse of a Matrix

Chapter 8: Solutions of Simultaneous Linear Equations

Chapter 9: Continuity

Chapter 10: Differentiability

Chapter 11: Differentiation

Chapter 12: Higher Order Derivatives

Chapter 13: Derivative as a Rate Measurer

Chapter 14: Differentials, Errors, and Approximations

Chapter 15: Mean Value Theorems

Chapter 16: Tangents and Normals

Chapter 17: Increasing and Decreasing Functions

Chapter 18: Maxima and Minima

Chapter 19: Indefinite Integrals

Chapter 20: Definite Integrals

Chapter 21: Areas of Bounded Regions

Chapter 22: Differential Equations

Chapter 23: Algebra of Vectors

Chapter 24: Scalar Or Dot Product

Chapter 25: Vector or Cross Product

Chapter 26: Scalar Triple Product

Chapter 27: Direction Cosines and Direction Ratios

Chapter 28: Straight Line in Space

Chapter 29: The Plane

Chapter 30: Linear programming

Chapter 31: Probability

Chapter 32: Mean and Variance of a Random Variable

Chapter 33: Binomial Distribution

Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.3 | Set 3

Question 33. Differentiatewith respect to x.

Question 34. Differentiatewith respect to x.

Question 35. If, 0 < x < 1, prove that.

Question 36. If, 0 < x < ∞, prove that.

Question 37 Differentiate the following with respect to x :

(i) cos−1 (sin x)

(ii)

Question 38. Differentiate, 0 < x < π/2 with respect to x.

Question 39. If, x > 0, prove that.

Question 40. If, x > 0, find.

Question 41. If, find.

Question 42. If , 0 < x < 1/2, find.

Question 43. If the derivative of tan−1 (a + bx) takes the value of 1 at x = 0, prove that 1 + a2 = b.

Question 44. If, −1/2 < x < 0, find.

Question 45. If, find.

Question 46. If , find.

Question 47. Differentiatewith respect to x.

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Question 33. Differentiate $y=tan^{-1}\left(\frac{x^{\frac{1}{3}}+a^\frac{1}{3}}{1-(ax)^{\frac{1}{3}}}\right)$ with respect to x.

Question 34. Differentiate $y=sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$ with respect to x.

Question 35. If $y=sin^{-1}\left(\frac{2x}{1+x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ , 0 < x < 1, prove that $\frac{dy}{dx}=\frac{4}{1+x^2}$ .

Question 36. If $y=sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)+cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)$ , 0 < x < ∞, prove that $\frac{dy}{dx}=\frac{2}{1+x^2}$ .

(i) cos⁻¹ (sin x)

(ii) $cot^{−1}\left(\frac{1-x}{1+x}\right)$

Question 38. Differentiate $y=cot^{-1}\left[\frac{\sqrt{1+sinx}+\sqrt{1-sinx}}{\sqrt{1+sinx}-\sqrt{1-sinx}}\right]$ , 0 < x < π/2 with respect to x.

Question 39. If $y=tan^{-1}\left(\frac{2x}{1-x^2}\right)+sec^{-1}\left(\frac{1+x^2}{1-x^2}\right)$ , x > 0, prove that $\frac{dy}{dx}=\frac{4}{1+x^2}$ .

Question 40. If $y=sec^{-1}\left(\frac{x+1}{x-1}\right)+sin^{-1}\left(\frac{x-1}{x+1}\right)$ , x > 0, find $\frac{dy}{dx}$ .

Question 41. If $y=sin\left[2tan^{-1}\sqrt{(\frac{1-x}{1+x})}\right]$ , find $\frac{dy}{dx}$ .

Question 42. If $y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$ , 0 < x < 1/2, find $\frac{dy}{dx}$ .

Question 43. If the derivative of tan⁻¹ (a + bx) takes the value of 1 at x = 0, prove that 1 + a² = b.

Question 44. If $y=cos^{-1}(2x)+2cos^{-1}(\sqrt{1-4x^2})$ , −1/2 < x < 0, find $\frac{dy}{dx}$ .

Question 45. If $y=tan^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)$ , find $\frac{dy}{dx}$ .

Question 46. If $y=cos^{-1}\left(\frac{2x-3\sqrt{1-x^2}}{\sqrt{13}}\right)$ , find $\frac{dy}{dx}$ .

Question 47. Differentiate $y=sin^{-1}\left[\frac{2^{x+1}×3^x}{1+(36)^x}\right]$ with respect to x.