# Get Answers to all your Questions

#### explain solution RD Sharma class 12 chapter Differentiation exercise 10.2 question 39 maths

Answer:  $\frac{2^{x}}{\left(x^{2}+3\right)^{2}}\left[\cos x \log _{e} 2-\sin x-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right]$

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

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

Solution:

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

Differentiate with respect to x,

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

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

$\frac{d y}{d x}=\left[\frac{\left(x^{2}+3\right)^{2}\left\{2^{x} \frac{d}{d x} \cos x+\cos x \frac{d}{d x} 2^{x}\right\}-\left(2^{x} \cos x\right) 2\left(x^{2}+3\right) \frac{d}{d x}\left(x^{2}+3\right)}{\left[x^{2}+3\right]^{4}}\right]$

$\frac{d y}{d x}=\left[\frac{\left(x^{2}+3\right)^{2}\left\{-2^{x} \sin x+\cos x 2^{x} \log _{e} 2\right\}-2\left(2^{x} \cos x\right)\left(x^{2}+3\right)(2 x)}{\left[x^{2}+3\right]^{4}}\right]$

$\frac{d y}{d x}=\left[\frac{2^{x}\left(x^{2}+3\right)^{2}\left\{\left(\cos x \log _{e} 2-\sin x\right\}-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right.}{\left[x^{2}+3\right]^{4}}\right]$

$\frac{d y}{d x}=\frac{2^{x}}{\left(x^{2}+3\right)^{2}}\left[\cos x \log _{e} 2-\sin x-\frac{4 x \cos x}{\left(x^{2}+3\right)}\right]$