Question

Differentiate: $y={\sin }^{-1}\left(\frac{{1-x}^2}{{1+x}^2}\right),0<x<1$

Answer 1

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

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

$\Rightarrow \cos y\frac{dy}{dx}=\frac{(1+x^2)\frac{d}{dx}(1-x^2)-(1-x^2)\frac{d}{dx}(1+x^2)}{{(1+x^2)}^2}$

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

$\Rightarrow \cos y\frac{dy}{dx}=\frac{2x(-1-x^2-1+x^2)}{{(1+x^2)}^2}$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{2x(-2)}{{(1+x^2)}^2}$

$\Rightarrow \cos y\frac{dy}{dx}=\frac{-4x}{{(1+x^2)}^2}$

$\Rightarrow \sqrt{1-{\sin }^{2}y}\frac{dy}{dx}=\frac{-4x}{{(1+x^2)}^2}$

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

$\Rightarrow \sqrt{\frac{{(1+x^2)}^2-{(1-x^2)}^2}{{(1+x^2)}^2}}\frac{dy}{dx}=\frac{-4x}{{(1+x^2)}^2}$

$\Rightarrow \frac{\sqrt{{(1+x^2)}^2-{(1-x^2)}^2}}{1+x^2}\frac{dy}{dx}=\frac{-4x}{{(1+x^2)}^2}$

$\Rightarrow \sqrt{{(1+x^2)}^2-{(1-x^2)}^2}\frac{dy}{dx}=\frac{-4x}{1+x^2}$

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

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

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

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