#### Provide solution for RD Sharma maths class 12 chapter 10 Differentiation exercise Multiple choice question 10

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

Hint:

Differentiate y function w.r.t x

Given:

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

Solution:

Let

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

Differentiate y w.r.t x then

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

$=\frac{1}{\sqrt{1-\left(\frac{1-x^{2}}{1+x^{2}}\right)^{2}}}\left\{\frac{\left(1+x^{2}\right) \frac{d}{d x}\left(1-x^{2}\right)-\left(1-x^{2}\right) \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\right\}$        $\left[\because \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}, \frac{d}{d x}\left(\frac{u}{v}\right)=v \frac{d u}{d x}-u \frac{d v}{d x}\right]$

$=\frac{1}{\sqrt{\frac{\left(1+x^{2}\right)^{2}-\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}}}\left[\frac{\left(1+x^{2}\right)\left\{\frac{d}{d x}(1)-\frac{d}{d x}\left(x^{2}\right)\right\}-\left(1-x^{2}\right)\left\{\frac{d}{d x}(1)+\frac{d}{d x}\left(x^{2}\right)\right\}}{\left(1+x^{2}\right)^{2}}\right]$                $\left[\because \frac{d}{d x}\{f(x)+g(x)\}=\frac{d}{d x} f(x)+\frac{d}{d x} g(x)\right]$

$=\frac{\sqrt{\left(1+x^{2}\right)^{2}}}{\sqrt{\left(1+x^{4}+2 x^{2}\right)-\left(1+x^{4}-2 x^{2}\right)}}\left[\frac{\left(1+x^{2}\right)(0-2 x)-\left(1-x^{2}\right)(0+2 x)}{\left(1+x^{2}\right)^{2}}\right]$

$\left[\frac{d}{d x}\left(x^{n}\right)=n x^{n-1},\left(a^{2}+b^{2}+2 a b\right)=(a+b)^{2} \&(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$

$=\frac{\left(1+x^{2}\right)}{\sqrt{1+x^{4}+2 x^{2}-1-x^{4}+2 x^{2}}}\left[\frac{-2 x-2 x^{3}-\left(2 x-2 x^{3}\right)}{\left(1+x^{2}\right)^{2}}\right]$

$=\frac{1+x^{2}}{\sqrt{4 x^{2}}}\left[\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{\left(1+x^{2}\right)^{2}}\right]$

\begin{aligned} &\left.=\frac{1+x^{2}}{\sqrt{(2 x)^{2}}\left[\frac{-4 x}{\left(1+x^{2}\right)^{2}}\right.}\right]=\frac{\left(1+x^{2}\right)}{2 x} \times \frac{-4 x}{\left(1+x^{2}\right)^{2}}=\frac{-2}{\left(1+x^{2}\right)} \\\\ &\therefore \frac{d y}{d x}=\frac{-2}{1+x^{2}} \end{aligned}