Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.3 | Set 2 (original) (raw)

Last Updated : 23 Aug, 2024

Question 17. Differentiatey=tan^{-1}\left(\frac{2^{x+1}}{1-4^x}\right) , −∞ < x < 0 with respect to x.

**Solution:

We have,y=tan^{-1}\left(\frac{2^{x+1}}{1-4^x}\right) , −∞ < x < 0

On putting 2x = tan θ, we get,

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

=tan^{-1}\left(tan2θ\right)

Now, −∞ < x < 0

=> 0 < 2x < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan−1 (2x)

Differentiating with respect to x, we get,

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

=\frac{2.2^xlog2}{1+(2^x)^2}

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

Question 18. Differentiatey=tan^{-1}\left(\frac{2a^{x}}{1-a^{2x}}\right) , a > 1, −∞ < x < 0 with respect to x.

**Solution:

We have,y=tan^{-1}\left(\frac{2a^{x}}{1-a^{2x}}\right) , −∞ < x < 0

On putting ax = tan θ, we get,

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

=tan^{-1}\left(tan2θ\right)

Now, −∞ < x < 0

=> 0 < ax < 1

=> 0 < θ < π/4

=> 0 < 2θ < π/2

So, y = 2θ

= 2 tan−1 (ax)

Differentiating with respect to x, we get,

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

=\frac{2a^xloga}{1+(a^x)^2}

=\frac{2a^{x}loga}{1+a^{2x}}

Question 19. Differentiatey=sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) , 0 < x < 1 with respect to x.

**Solution:

We have,y=sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) , 0 < x < 1

On putting x = cos 2θ, we get,

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

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

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

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

=sin^{-1}\left(sin(θ+\frac{π}{4})\right)

Now, 0 < x < 1

=> 0 < cos 2θ < 1

=> 0 < 2θ < π/2

=> 0 < θ < π/4

=> π/4 < (θ+π/4) < π/2

So, y =θ+\frac{π}{4}

=\frac{1}{2}cos^{-1}x+\frac{π}{4}

Differentiating with respect to x, we get,

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

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

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

Question 20. Differentiatey=tan^{-1}\left(\frac{\sqrt{1+a^2x^2}-1}{ax}\right) , x ≠ **0 with respect to x.

**Solution:

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

On putting ax = tan θ, we get,

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

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

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

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

=tan^{-1}\left(tan\frac{θ}{2}\right)

=\frac{θ}{2}

=\frac{1}{2}tan^{-1}\left(ax\right)

Differentiating with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}\left(\frac{1}{2}tan^{-1}\left(ax\right)\right)

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

Question 21. Differentiatey=tan^{-1}\left(\frac{sinx}{1+cosx}\right) , −π < x < π with respect to x.

**Solution:

We have,y=tan^{-1}\left(\frac{sinx}{1+cosx}\right) , −π < x < π

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

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

=\frac{x}{2}

Differentiating with respect to x, we get,

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

=\frac{1}{2}

Question 22. Differentiatey=sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) with respect to x.

**Solution:

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

On putting x = cot θ, we get,

y=sin^{-1}\left(\frac{1}{\sqrt{1+cot^2θ}}\right)

=sin^{-1}\left(\frac{1}{cosecθ}\right)

=sin^{-1}\left(sinθ\right)

= θ

= cot−1 x

Differentiating with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}\left(cot^{-1}x\right)

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

Question 23. Differentiatey=cos^{-1}\left(\frac{1-x^{2n}}{1+x^{2n}}\right) , 0 < x < ∞ with respect to x.

**Solution:

We have,y=cos^{-1}\left(\frac{1-x^{2n}}{1+x^{2n}}\right) ,0 < x < ∞

On putting xn = tan θ, we get,

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

=cos^{-1}\left(cos2θ\right)

Now, 0 < x < ∞

=> 0 < xn < ∞

=> 0 < θ < π/2

=> 0 < 2θ < π

So, y = 2θ

= 2 tan–1 (xn)

Differentiating with respect to x, we get,

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

=\frac{2}{1+x^{2n}}×(nx^{n-1})

=\frac{2nx^{n-1}}{1+x^{2n}}

Question 24. Differentiatey=sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+sec^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) , x ∈ R with respect to x.

**Solution:

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

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

=\frac{π}{2}

Differentiating with respect to x, we get,

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

= 0

Question 25. Differentiatey=tan^{-1}\left(\frac{a+x}{1-ax}\right) with respect to x.

**Solution:

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

=tan^{-1}a+tan^{-1}x

Differentiating with respect to x, we get,

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

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

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

Question 26. Differentiatey=tan^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right) with respect to x.

**Solution:

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

=tan^{-1}\sqrt{x}+tan^{-1}\sqrt{a}

Differentiating with respect to x, we get,

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

=\frac{1}{1+(\sqrt{x})^2}×\frac{1}{2\sqrt{x}} + 0

=\frac{1}{2\sqrt{x(}1+x)}

Question 27. Differentiatey=tan^{-1}\left(\frac{a+btanx}{b-atanx}\right) with respect to x.

**Solution:

We have,y=tan^{-1}\left(\frac{a+btanx}{b-atanx}\right)

=tan^{-1}\left(\frac{\frac{a+btanx}{b}}{\frac{b-atanx}{b}}\right)

=tan^{-1}\left(\frac{\frac{a}{b}+tanx}{1-\frac{atanx}{b}}\right)

=tan^{-1}(\frac{a}{b})+tan^{-1}(tanx)

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

Differentiating with respect to x, we get,

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

= 0 + 1

= 1

**Question 28. Differentiatey=tan^{-1}\left(\frac{a+bx}{b-ax}\right) **with respect to x.

**Solution:

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

=tan^{-1}\left(\frac{\frac{a+bx}{b}}{\frac{b-ax}{b}}\right)

=tan^{-1}\left(\frac{\frac{a}{b}+x}{1-\frac{ax}{b}}\right)

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

Differentiating with respect to x, we get,

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

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

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

Question 29. Differentiatey=tan^{-1}\left(\frac{x-a}{x+a}\right) with respect to x.

**Solution:

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

=tan^{-1}\left(\frac{\frac{x-a}{a}}{\frac{x+a}{a}}\right)

=tan^{-1}\left(\frac{\frac{x}{a}-1}{1+\frac{x}{a}}\right)

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

Differentiating with respect to x, we get,

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

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

=\frac{a^2}{a(a^2+x^2)}

=\frac{a}{a^2+x^2}

Question 30. Differentiatey=tan^{-1}\left(\frac{x}{1+6x^2}\right) with respect to x.

**Solution:

We have,y=tan^{-1}\left(\frac{x}{1+6x^2}\right)

=tan^{-1}\left(\frac{3x-2x}{1+(3x)(2x)}\right)

=tan^{-1}(3x)-tan^{-1}(2x)

Differentiating with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}\left((tan^{-1}(3x)-tan^{-1}(2x)\right)

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

=\frac{3}{1+9x^2}-\frac{2}{1+4x^2}

Question 31. Differentiatey=tan^{-1}\left(\frac{5x}{1-6x^2}\right) with respect to x.

**Solution:

We have,y=tan^{-1}\left(\frac{5x}{1-6x^2}\right)

=tan^{-1}\left(\frac{3x+2x}{1-(3x)(2x)}\right)

=tan^{-1}(3x)+tan^{-1}(2x)

Differentiating with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}\left((tan^{-1}(3x)+tan^{-1}(2x)\right)

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

=\frac{3}{1+9x^2}+\frac{2}{1+4x^2}

Question 32. Differentiatey=tan^{-1}\left(\frac{cosx+sinx}{cosx-sinx}\right) , −π/4 < x < π/4 with respect to x.

**Solution:

We have,y=tan^{-1}\left(\frac{cosx+sinx}{cosx-sinx}\right) , −π/4 < x < π/4

=tan^{-1}\left(\frac{\frac{cosx+sinx}{cosx}}{\frac{cosx-sinx}{cosx}}\right)

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

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

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

=\frac{π}{4}+x

Differentiating with respect to x, we get,

\frac{dy}{dx}=\frac{d}{dx}\left(\frac{π}{4}+x\right)

= 0 + 1

= 1

Summary

This section likely covers more advanced topics such as:

Derivatives of inverse trigonometric functions
Derivatives involving exponential and logarithmic functions
Implicit differentiation of complex equations
Application of differentiation in related rates problems
Differentiating functions defined in terms of other functions

Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.3 | Set 2 (original) (raw)

**Question 17. Differentiatey=tan^{-1}\left(\frac{2^{x+1}}{1-4^x}\right) ****, −∞ < x < 0 with respect to x.**

**Question 18. Differentiatey=tan^{-1}\left(\frac{2a^{x}}{1-a^{2x}}\right) ****, a > 1, −∞ < x < 0 with respect to x.**

**Question 19. Differentiatey=sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) ****, 0 < x < 1 with respect to x.**

**Question 20. Differentiatey=tan^{-1}\left(\frac{\sqrt{1+a^2x^2}-1}{ax}\right) ****, x** ≠ **0 with respect to x.

**Question 21. Differentiatey=tan^{-1}\left(\frac{sinx}{1+cosx}\right) ****, −π < x < π with respect to x.**

**Question 22. Differentiatey=sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) **with respect to x.

**Question 23. Differentiatey=cos^{-1}\left(\frac{1-x^{2n}}{1+x^{2n}}\right) ****, 0 < x < ∞ with respect to x.**

**Question 24. Differentiatey=sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+sec^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) ****, x ∈ R with respect to x.**

**Question 25. Differentiatey=tan^{-1}\left(\frac{a+x}{1-ax}\right) **with respect to x.

**Question 26. Differentiatey=tan^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right) **with respect to x.

**Question 27. Differentiatey=tan^{-1}\left(\frac{a+btanx}{b-atanx}\right) **with respect to x.

**Question 29. Differentiatey=tan^{-1}\left(\frac{x-a}{x+a}\right) **with respect to x.

**Question 30. Differentiatey=tan^{-1}\left(\frac{x}{1+6x^2}\right) **with respect to x.

**Question 31. Differentiatey=tan^{-1}\left(\frac{5x}{1-6x^2}\right) **with respect to x.

**Question 32. Differentiatey=tan^{-1}\left(\frac{cosx+sinx}{cosx-sinx}\right) ****, −π/4 < x < π/4 with respect to x.**

Summary

Question 17. Differentiatey=tan^{-1}\left(\frac{2^{x+1}}{1-4^x}\right) , −∞ < x < 0 with respect to x.

Question 18. Differentiatey=tan^{-1}\left(\frac{2a^{x}}{1-a^{2x}}\right) , a > 1, −∞ < x < 0 with respect to x.

Question 19. Differentiatey=sin^{-1}\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{2}\right) , 0 < x < 1 with respect to x.

Question 20. Differentiatey=tan^{-1}\left(\frac{\sqrt{1+a^2x^2}-1}{ax}\right) , x ≠ **0 with respect to x.

Question 21. Differentiatey=tan^{-1}\left(\frac{sinx}{1+cosx}\right) , −π < x < π with respect to x.

Question 22. Differentiatey=sin^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) with respect to x.

Question 23. Differentiatey=cos^{-1}\left(\frac{1-x^{2n}}{1+x^{2n}}\right) , 0 < x < ∞ with respect to x.

Question 24. Differentiatey=sin^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+sec^{-1}\left(\frac{1+x^{2}}{1-x^{2}}\right) , x ∈ R with respect to x.

Question 25. Differentiatey=tan^{-1}\left(\frac{a+x}{1-ax}\right) with respect to x.

Question 26. Differentiatey=tan^{-1}\left(\frac{\sqrt{x}+\sqrt{a}}{1-\sqrt{xa}}\right) with respect to x.

Question 27. Differentiatey=tan^{-1}\left(\frac{a+btanx}{b-atanx}\right) with respect to x.

Question 29. Differentiatey=tan^{-1}\left(\frac{x-a}{x+a}\right) with respect to x.

Question 30. Differentiatey=tan^{-1}\left(\frac{x}{1+6x^2}\right) with respect to x.

Question 31. Differentiatey=tan^{-1}\left(\frac{5x}{1-6x^2}\right) with respect to x.

Question 32. Differentiatey=tan^{-1}\left(\frac{cosx+sinx}{cosx-sinx}\right) , −π/4 < x < π/4 with respect to x.