Differentiate the following functions with respect to x :

Let $y=\log _{7}(2 x-3)$

Recall that $\log _{a} b=\frac{\log b}{\log a}$

$\Rightarrow \log _{7}(2 x-3)=\frac{\log (2 x-3)}{\log 7}$

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

$\Rightarrow \frac{d y}{d x}=\left(\frac{1}{\log 7}\right)\left(\frac{1}{2 x-3}\right) \frac{d}{d x}(2 x-3)$ [using chain rule]

$\Rightarrow \frac{d y}{d x}=\frac{1}{(2 x-3) \log 7}\left[\frac{d}{d x}(2 x)-\frac{d}{d x}(3)\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{(2 \mathrm{x}-3) \log 7}\left[2 \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\frac{\mathrm{d}}{\mathrm{dx}}(3)\right]$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{(2 \mathrm{x}-3) \log 7}[2 \times 1-0]$

$\therefore \frac{d y}{d x}=\frac{2}{(2 x-3) \log 7}$

Thus, $\frac{\mathrm{d}}{\mathrm{dx}}\left[\log _{7}(2 \mathrm{x}-3)\right]=\frac{2}{(2 \mathrm{x}-3) \log 7}$