Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

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

Solution:

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

On differentiating y with respect to $x$, we get

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

Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule)

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

We have $(\text { uv })^{\prime}=$ vu' $+$ uv' (product rule)

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

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\mathrm{x}}\right)=\mathrm{e}^{\mathrm{x}}, \frac{\mathrm{d}}{\mathrm{dx}}(\sin \mathrm{x})=\cos \mathrm{x}$ and $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1}$

$\Rightarrow \frac{d y}{d x}=\frac{\left(x^{2}+2\right)^{3}\left[\sin x\left(e^{x}\right)+e^{x}(\cos x)\right]-\left(e^{x} \sin x\right)\left[3\left(x^{2}+2\right)^{3-1} \frac{d}{d x}\left(x^{2}+2\right)\right]}{\left(x^{2}+2\right)^{6}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

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

However, $\frac{d}{d x}\left(x^{2}\right)=2 x$ and derivative of a constant is 0 .

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\mathrm{x}^{2}+2\right)^{3} \mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})}{\left(\mathrm{x}^{2}+2\right)^{6}}-\frac{6 \mathrm{xe}^{\mathrm{x}} \sin \mathrm{x}\left(\mathrm{x}^{2}+2\right)^{2}}{\left(\mathrm{x}^{2}+2\right)^{6}}$

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

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

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