Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 08 May, 2021

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

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :