Question

$y=\frac{1}{2}{\sin }^{-1}\left(\frac{2x}{x^2+1}\right)$ then find $\frac{dy}{dx}$.

Answer 1

Given,

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

supposing $x=\tan \theta$ we get

or, $y=\frac{1}{2}\times 2{\tan }^{-1}x$

or, $y={\tan }^{-1}x$.

Now differentiating both sides with respect to $x$ we get,

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