#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 19

Answer: $\frac{1}{\sqrt{1+x}(1-x)^{\frac{3}{2}}}$

Hint: You must know the rule of solving derivative of polynomial function.

Given: $\sqrt{\frac{1+x}{1-x}}$

Solution:

Let  $y=\sqrt{\frac{1+x}{1-x}}$

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

Differentiating with respect to x

\begin{aligned} &\frac{d y}{d x}=\frac{d}{d x}\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}} \\ &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1+x}{1-x}\right)^{\frac{1}{2}-1} \times \frac{d}{d x}\left(\frac{1+x}{1-x}\right) \end{aligned}              [  using chain rule]

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

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

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

\begin{aligned} &\frac{d y}{d x}=\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}} \times\left\{\frac{2}{(1-x)^{2}}\right\} \\ &\frac{d y}{d x}=\frac{1}{\sqrt{1+x}(1-x)^{\frac{3}{2}}} \end{aligned}