#### provide solution for RD Sharma maths class 12 chapter Differentiation exercise  10.2 question 49

Answer: $\frac{e^{x} \sin x+e^{x} \cos x}{\left(x^{2}+2\right)^{3}}-\frac{6 x e^{x} \sin x}{\left(x^{2}+2\right)^{4}}$

Hint: you must know the rules of solving exponential derivative

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

Solution:

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

Differentiate with respect to x

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(x^{2}+2\right)^{3} \frac{\mathrm{d}}{\mathrm{dx}}\left(e^{x} \sin x\right)-e^{x} \sin x \frac{\mathrm{d}}{\mathrm{dx}}\left(x^{2}+2\right)^{3}}{\left[\left(x^{2}+2\right)^{3}\right]^{2}} \cdot \cdot \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x}-u \frac{d v}{d x}}{v^{2}}$

$=\frac{\left(x^{2}+2\right)^{3}\left[e^{x} \cos x+\sin x e^{x}\right]-e^{x} \sin x 3\left(x^{2}+2\right)^{2}(2 x)}{\left(x^{2}+2\right)^{6}}$

$=\frac{\left(x^{2}+2\right)^{2}\left[\left(x^{2}+2\right)\left[e^{x} \cos x+e^{x} \sin x\right]-6 x e^{x} \sin x\right]}{\left(x^{2}+2\right)^{6}}$

$\Rightarrow \frac{\left(x^{2}+2\right)\left(e^{x} \cos x+e^{x} \sin x\right)-6 x e^{x} \sin x}{\left(x^{2}+2\right)^{4}}$

$\Rightarrow \frac{e^{x} \sin x+e^{x} \cos x}{\left(x^{2}+2\right)^{3}}-\frac{6 x e^{x} \sin x}{\left(x^{2}+2\right)^{4}}$

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