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Solution:

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

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

$\Rightarrow \sin y=\frac{1-x^{2}}{1+x^{2}}$

Differentiating this relationship with respect to x, we obtain

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

Using chain rule, we obtain

$\frac{d}{d x}(\sin y)=\cos y \cdot \frac{d y}{d x}$

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

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

$\therefore \frac{d}{d x}(\sin y)=\frac{2 x}{1+x^{2}} \frac{d y}{d x}$   ...(2)

$\frac{d}{d x}\left(\frac{1-x^{2}}{1+x^{2}}\right)=\frac{\left(1+x^{2}\right) \cdot\left(1-x^{2}\right)^{\prime}-\left(1-x^{2}\right) \cdot\left(1+x^{2}\right)^{\prime}}{\left(1+x^{2}\right)^{2}}$   [Using quotient rule]

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

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

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

From (1), (2), and (3), we obtain

$\frac{2 x}{1+x^{2}} \frac{d y}{d x}=\frac{-4 x}{\left(1+x^{2}\right)^{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}}$

Alternate method

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

$\Rightarrow \sin y=\frac{1-x^{2}}{1+x^{2}}$

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

$\Rightarrow(1+\sin y) x^{2}=1-\sin y$

$\Rightarrow x^{2}=\frac{1-\sin y}{1+\sin y}$

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

$\Rightarrow x=\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}$

$\Rightarrow x=\frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}}$

$\Rightarrow x=\tan \left(\frac{\pi}{4}-\frac{y}{2}\right)$'

Differentiating this relationship with respect to x, we obtain

$\frac{d}{d x}(x)=\frac{d}{d x} \cdot\left[\tan \left(\frac{\pi}{4}-\frac{y}{2}\right)\right]$

$\Rightarrow 1=\sec ^{2}\left(\frac{\pi}{4}-\frac{y}{2}\right) \cdot \frac{d}{d x}\left(\frac{\pi}{4}-\frac{y}{2}\right)$

$\Rightarrow 1=\left[1+\tan ^{2}\left(\frac{\pi}{4}-\frac{y}{2}\right)\right] \cdot\left(-\frac{1}{2} \frac{d y}{d x}\right)$

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

$\Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}}$