Question

Differentiate ${\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ w.r.t ${\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$.

Answer 1

Let $f\left(x\right)={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ and $g\left(x\right)={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

Let $x=\sin \theta$

$f\left(\theta \right)={\tan }^{-1}\left(\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}\right)={\tan }^{-1}\left(\frac{\sin \theta }{\cos \theta }\right)={\tan }^{-1}\left(\tan \theta \right)\Rightarrow f\left(\theta \right)=\theta$

$g\left(\theta \right)={\sin }^{-1}\left(2\sin \theta \sqrt{1-{\sin }^2\theta }\right)={\sin }^{-1}\left(2\sin \theta \cos \theta \right)={\sin }^{-1}\left(\sin 2\theta \right)\Rightarrow g\left(\theta \right)=2\theta$

$\therefore$ we have to differentiate $f\left(\theta \right)$ w.r.t. $g\left(\theta \right)$

$\Rightarrow \frac{d\left(f\right(\theta \left)\right)}{d\left(g\right(\theta \left)\right)}=\frac{\frac{d\left(f\right(\theta \left)\right)}{d\theta }}{\frac{d\left(g\right(\theta \left)\right)}{d\theta }}=\frac{\frac{d}{d\theta }\left(\theta \right)}{\frac{d}{d\theta }\left(2\theta \right)}=\frac{1}{2}$