#### Please solve RD Sharma class 12 chapter Differentiation exercise 10.2 question 52 maths textbook solution

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

Hint: you must know the rule of solving derivative of trigonometric functions

Given: $\frac{x^{2}\left(1-x^{2}\right)^{3}}{\cos 2 x}$

Solution:

Let  $y=\frac{x^{2}\left(1-x^{2}\right)^{3}}{\cos 2 x}$

Differentiate with respect to x

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\cos 2 x \frac{\mathrm{d}}{\mathrm{dx}}\left\{x^{2}\left(1-x^{2}\right)^{3}-x^{2}\left(1-x^{2}\right)^{3} \frac{\mathrm{d}}{\mathrm{dx}} \cos 2 x\right\}}{\cos ^{2} 2 x} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

$\Rightarrow \frac{\cos 2 x\left\{x^{2} \frac{\mathrm{d}}{\mathrm{dx}}\left(1-x^{2}\right)^{3}+\left(1-x^{2}\right)^{3} \frac{\mathrm{d}}{\mathrm{dx}} x^{2}\right\}-x^{2}\left(1-x^{2}\right)^{3}(-2 \sin 2 x)}{\cos ^{2} 2 x}$

$\Rightarrow \frac{\cos 2 x\left\{-6 x^{3}\left(1-x^{2}\right)^{2}+\left(1-x^{2}\right)^{3} 2 x\right\}+2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos ^{2} 2 x}$

$\Rightarrow-\frac{6 x^{3}\left(1-x^{2}\right)^{2}}{\cos 2 x}+\frac{2 x\left(1-x^{2}\right)^{3}}{\cos 2 x}+\frac{2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos ^{2} 2 x}$

$\Rightarrow 2 x\left(1-x^{2}\right)^{2} \sec 2 x\left\{\left(1-x^{2}\right)-3 x^{2}+x\left(1-x^{2}\right) \tan 2 x\right\}$