#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 17 maths textbook solution

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

Hint: You must know the rules of solving derivative of polynomial function

Given: $\sqrt{\frac{1-x^{2}}{1+x^{2}}}$

Solution:

Let  $y=\sqrt{\frac{1-x^{2}}{1+x^{2}}}$

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

Differentiating with respect to x

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

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

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

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

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

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

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