Differentiate the following functions with respect to x :

Solution:

Let $y=\sin ^{2}\{\log (2 x+3)\}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\sin ^{2}\{\log (2 \mathrm{x}+3)\}\right]$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \sin ^{2-1}\{\log (2 \mathrm{x}+3)\} \frac{\mathrm{d}}{\mathrm{dx}}[\sin \{\log (2 \mathrm{x}+3)\}]$ [chain rule]

$\Rightarrow \frac{d y}{d x}=2 \sin \{\log (2 x+3)\} \frac{d}{d x}[\sin \{\log (2 x+3)\}]$

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

$\Rightarrow \frac{d y}{d x}=2 \sin \{\log (2 x+3)\} \cos \{\log (2 x+3)\} \frac{d}{d x}[\log (2 x+3)]$

As $\sin (2 \theta)=2 \sin \theta \cos \theta$, we have

$\frac{d y}{d x}=\sin \{2 \times \log (2 x+3)\} \frac{d}{d x}[\log (2 x+3)]$

$\frac{d y}{d x}=\sin \{2 \log (2 x+3)\} \frac{d}{d x}[\log (2 x+3)]$

We know $\frac{d}{d x}(\log x)=\frac{1}{x}$

$\Rightarrow \frac{d y}{d x}=\sin \{2 \log (2 x+3)\}\left[\frac{1}{(2 x+3)} \frac{d}{d x}(2 x+3)\right]$

$\Rightarrow \frac{d y}{d x}=\frac{\sin \{2 \log (2 x+3)\}}{2 x+3} \frac{d}{d x}(2 x+3)$

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin \{2 \log (2 \mathrm{x}+3)\}}{2 \mathrm{x}+3}[2 \times 1+0]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\sin (2 \log (2 \mathrm{x}+3)\}}{2 \mathrm{x}+3} \times 2$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \sin \{2 \log (2 \mathrm{x}+3)\}}{2 \mathrm{x}+3}$

Thus, $\frac{\mathrm{d}}{\mathrm{dx}}\left[\sin ^{2}\{\log (2 \mathrm{x}+3)\}\right]=\frac{2 \sin (2 \log (2 \mathrm{x}+3))}{2 \mathrm{x}+3}$