Differentiate the following functions with respect to x :

Solution:

Let $y=e^{3 x} \cos (2 x)$

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

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

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(e^{3 x} \times \cos 2 x\right)$

Recall that (uv) $^{\prime}=v u^{\prime}+u v^{\prime}$ (product rule)

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

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

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

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

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

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

We have $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})=1$

$\Rightarrow \frac{d y}{d x}=3 e^{3 x} \cos 2 x \times 1-2 e^{3 x} \sin 2 x \times 1$

$\Rightarrow \frac{d y}{d x}=3 e^{3 x} \cos 2 x-2 e^{3 x} \sin 2 x$

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

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