#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 20 maths

Answer: $\frac{4 x}{\left(1-x^{2}\right)^{2}} \cos \left(\frac{1+x^{2}}{1-x^{2}}\right)$

Hint: You must know the rules of solving derivative of trigonometric functions.

Given: $\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)$

Solution:

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

Differentiating with respect to x

$\frac{d y}{d x}=\frac{d}{d x}\left[\sin \left(\frac{1+x^{2}}{1-x^{2}}\right)\right]$

$\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right) \frac{d}{d x}\left(\frac{1+x^{2}}{1-x^{2}}\right)$                  [  using chain rule]

$\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right)\left[\frac{\left(1-x^{2}\right) \frac{d}{d x}\left(1+x^{2}\right)-\left(1+x^{2}\right) \frac{d}{d x}\left(1-x^{2}\right)}{\left(1-x^{2}\right)^{2}}\right]$$...\frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

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

$\frac{d y}{d x}=\cos \left(\frac{1+x^{2}}{1-x^{2}}\right)\left[\frac{2 x-2 x^{3}+2 x+2 x^{3}}{\left(1-x^{2}\right)^{2}}\right]$

$\frac{d y}{d x}=\frac{4 x}{\left(1-x^{2}\right)^{2}} \cos \left(\frac{1+x^{2}}{1-x^{2}}\right)$