Question

Find the derivative of $f\left(x\right)={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ w.r.t. $g\left(x\right)={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$.

Answer 1

$f\left(x\right)=ta{n}^{-1}\left(\frac{2x}{1-x^2}\right).g\left(x\right)=si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$

Let, $x=tan\theta ,\theta =ta{n}^{-1}x$

$f\left(tan\theta \right)-ta{n}^{-1}\left(\frac{2tan\theta }{1-ta{n}^2\theta }\right)=ta{n}^{-1}\left(tan2\theta \right)$

$=2\theta$

$\therefore f\left(x\right)=2ta{n}^{-1}x$

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

Now, $g\left(tan\theta \right)=si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$

$=2\theta$

$g\left(x\right)=2ta{n}^{-1}x$

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

$\therefore \frac{df\left(x\right)}{dg\left(x\right)}=\frac{df\left(x\right)}{dx}\times \frac{dx}{dg\left(x\right)}$

$=\frac{2}{1+x^2}\times \frac{1+x^2}{2}$

$=1$