Question

Diff: ${\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$ w.r.t.$x$.

Answer 1

We have

${\cos }^{-1}\left(\frac{2x}{1+x^2}\right)=2{\tan }^{-1}x$ [ Direct formula]

Then the derivative of ${\cos }^{-1}\left(\frac{2x}{1+x^2}\right)$ w.r. to $x$ is $\frac{d}{dx}\left(2{\tan }^{-1}x\right)=\frac{2}{1+x^2}$.