Question

Differentiate ${\tan }^{-1}\frac{x}{\sqrt{1-x^2}}$ with respect to ${\sin }^{-1}(2x\sqrt{1-x^2}$).

Answer 1

Let $P={\tan }^{-1}\frac{x}{\sqrt{1-x^2}}$ and $Q={\tan }^{-1}\left(2x\sqrt{1-x^2}\right)$

Put $x=\sin Q$, we get

$P={\tan }^{-1}\left(\frac{\sin Q}{\cos Q}\right)=Q={\sin }^{-1}x$

and $Q={\sin }^{-1}\left(2\sin Q\cos Q\right)=2Q=2{\sin }^{-1}x$

so $\frac{dP}{dQ}=\frac{\frac{dP}{dx}}{\frac{dQ}{dx}}=\frac{\frac{d\left({\sin }^{-1}x\right)}{dx}}{\frac{2d\left({\sin }^{-1}x\right)}{dx}}=\frac{1}{2}$