Differentiate the following functions with respect to x :

Let $y=\sin (\log x)$

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

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

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

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

However, $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})=\frac{1}{\mathrm{x}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\cos (\log \mathrm{x}) \times \frac{1}{\mathrm{x}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}} \cos (\log \mathrm{x})$

Thus, $\frac{d}{d x}[\sin (\log x)]=\frac{1}{x} \cos (\log x)$