Differentiate the following functions with respect to x :

Let $y=\sin ^{2}(2 x+1)$

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

$\frac{d y}{d x}=\frac{d}{d x}\left[\sin ^{2}(2 x+1)\right]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \frac{d y}{d x}=2 \sin ^{2-1}(2 x+1) \frac{d}{d x}[\sin (2 x+1)]$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=2 \sin (2 x+1) \frac{d}{d x}[\sin (2 x+1)]$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \sin (2 \mathrm{x}+1) \cos (2 \mathrm{x}+1) \frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x}+1)$ [using chain rule]

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin [2(2 \mathrm{x}+1)] \frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x}+1)[\because \sin (2 \theta)=2 \sin \theta \cos \theta]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (4 \mathrm{x}+2)\left[\frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(1)\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (4 \mathrm{x}+2)\left[2 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}}(1)\right]$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin (4 \mathrm{x}+2)[2 \times 1+0]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=2 \sin (4 \mathrm{x}+2)$

Thus, $\frac{d}{d x}\left[\sin ^{2}(2 x+1)\right]=2 \sin (4 x+2)$