Question

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

Answer 1

We have,

$\begin{array}{c}y={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\\ putx=\tan \theta \\ y={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)\\ y={\cos }^{-1}\left(\cos 2\theta \right)\\ y=2\theta \to \left(i\right)\end{array}$

Put the value of $\theta$ in equation $\left(i\right)$

$\begin{array}{c}y=2{\tan }^{-1}x\\ \frac{dy}{dx}=\frac{2}{1+x^2}\end{array}$

Hence, this is the answer.