Question

$y=si{n}^1\left[\frac{2x}{1+x^2}\right],then\frac{dy}{dx}$ is

Answer 1

$lety=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$
$lett=\frac{2x}{1+x^2}$

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

$\frac{dt}{dx}=\frac{2(1-x^2)}{(1+x^2{)}^2}$

$y=si{n}^{-1}t$

Differentiating on both sides w.r.t x

$\frac{dy}{dx}=\frac{1}{\sqrt{1-t^2}}.\frac{dt}{dx}$

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

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