#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 23 Maths Textbook Solution.

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

Hint:

\begin{aligned} &\frac{\mathrm{d}}{\mathrm{dx}}(\text { constsant })=0 \\ &\frac{d}{d \mathrm{x}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1} \end{aligned}

Given:

\begin{aligned} &\cos ^{-1}\left\{\frac{1-x^{2 n}}{1+x^{2 n}}\right\} \\ &0

Solution:

$y=cos ^{-1}\left\{\frac{1-x^{2 n}}{1+x^{2 n}}\right\}$

Let,

$x^{n}=\tan \theta$

$\theta =\tan ^{-1}\left ( x^{n} \right )$

$y=\cos ^{-1}\left \{ \frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta } \right \}$

$y=\cos ^{-1}\left \{ \cos 2\theta \right \}$

Using,

$\frac{1-\tan ^{2}\theta }{1+\tan ^{2}\theta }=\cos 2\theta$

Considering the limits,

\begin{aligned} &0

Now,                                                                                                $\left \{ \cos ^{-1}\left ( \cos \theta \right )=\theta ,if\: \theta \varepsilon \left [ 0,\pi \right ] \right \}$

$y=\cos ^{-1}\left ( \cos 2\theta \right )$

$y=2\theta$                                                                                            $Since\hspace{0.2cm}x^{n}=\tan \theta$

Differentiating with respect to x we get

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

Using

\begin{aligned} &\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}} \\ &\frac{d y}{d x}=\frac{2 \times 1}{1+\left(x^{n}\right)^{2}} \times n x^{n-1} \\ &\frac{d y}{d x}=\frac{2 n x^{n-1}}{1+x^{2 n}} \end{aligned}