#### Need Solution for R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 24 Maths Textbook Solution.

Answer:$\frac{dy}{dx}=0$

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:

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

Solution:

Let,

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

Using

\begin{aligned} &\sec ^{-1} x=\cos ^{-1}\left(\frac{1}{x}\right) \\ &y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \end{aligned}

Using

\begin{aligned} &\sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2} \\ &\mathrm{y}=\frac{\pi}{2} \end{aligned}

Differentiating with respect to x we get

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

$\frac{dy}{dx}=0$                                                                        $\left \{ \frac{d}{dx}\left ( constant \right )=0 \right \}$