#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.2 question 42

Answer: $\left[\frac{6 x \sin x+3 x^{2} \cos x}{\sqrt{\left(7-x^{2}\right)}}+\frac{3 x^{3} \sin x}{\left(7-x^{2}\right)^{\frac{3}{2}}}\right]$

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

Given: $y=\frac{3 x^{2} \sin x}{\sqrt{\left(7-x^{2}\right)}}$

Solution:

Let  $y=\frac{3 x^{2} \sin x}{\sqrt{\left(7-x^{2}\right)}}$

Differentiate with respect to x,

$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{3 x^{2} \sin x}{\left(7-x^{2}\right)^{\frac{1}{2}}}\right]$

$\frac{dy}{dx}=\frac{\left(7-x^{2}\right)^{\frac{1}{2}} \times \frac{\mathrm{d}}{\mathrm{dx}}\left(3 x^{2} \sin x\right)-\left(3 x^{2} \sin x\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(7-x^{2}\right)^{\frac{1}{2}}}{\left[\left(7-x^{2}\right)^{\frac{1}{2}}\right]^{2}} \ldots \frac{d}{d x}\left(\frac{u}{v}\right)=$   $\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{\left(7-x^{2}\right)^{\frac{1}{2}}\left(3 x^{2} \cos x+6 x \sin x\right)-3 x^{2} \sin x \times \frac{1}{2}\left(7-x^{2}\right)^{\frac{1}{2}-1}(-2 x)}{7-x^{2}}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{\left(7-x^{2}\right)^{\frac{1}{2}}\left(3 x^{2} \cos x+6 x \sin x\right)-3 x^{2} \sin x \times \frac{1}{2}\left(7-x^{2}\right)^{-\frac{1}{2}}(-2 x)}{7-x^{2}}\right]$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{6 x \sin x+3 x^{2} \cos x}{\sqrt{\left(7-x^{2}\right)}}+\frac{3 x^{3} \sin x}{\left(7-x^{2}\right)^{\frac{3}{2}}}\right]$