Differentiate the following functions with respect to $x$ :

Let $y=\sin (3 x+5)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}[\sin (3 \mathrm{x}+5)]$

We know $\frac{d}{d x}(\sin x)=\cos x$

$\Rightarrow \frac{d y}{d x}=\cos (3 x+5) \frac{d}{d x}(3 x+5)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\cos (3 x+5)\left[\frac{d}{d x}(3 x)+\frac{d}{d x}(5)\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos (3 \mathrm{x}+5)\left[3 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(5)\right]$

However, $\frac{d}{d x}(x)=1$ and derivative of a constant is 0 .

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos (3 \mathrm{x}+5)[3 \times 1+0]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=3 \cos (3 \mathrm{x}+5)$

Thus, $\frac{d}{d x}[\sin (3 x+5)]=3 \cos (3 x+5)$