Find :
Question:

Find $\frac{d y}{d x}$ :

$y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right), 0<x<1$

Solution:

The given relationship is,

$y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)$

$\Rightarrow \cos y=\frac{1-x^{2}}{1+x^{2}}$

$\Rightarrow \frac{1-\tan ^{2} \frac{y}{2}}{1+\tan ^{2} \frac{y}{2}}=\frac{1-x^{2}}{1+x^{2}}$

On comparing L.H.S. and R.H.S. of the above relationship, we obtain

$\tan \frac{y}{2}=x$

Differentiating this relationship with respect to x, we obtain

$\sec ^{2}\left(\frac{\mathrm{y}}{2}\right) \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}}{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})$

$\Rightarrow \sec ^{2} \frac{y}{2} \times \frac{1}{2} \frac{\mathrm{dy}}{\mathrm{dx}}=1$

$\Rightarrow \frac{d y}{d x}=\frac{2}{\sec ^{2} \frac{y}{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{2}{1+\tan ^{2} \frac{y}{2}}$

$\therefore \frac{d y}{d x}=\frac{2}{1+x^{2}}$