# Differentiate the following functions with respect to x :

Question:

Differentiate the following functions with respect to $x$ :

$\log \sqrt{\frac{x-1}{x+1}}$

Solution:

Let $y=\log \sqrt{\frac{x-1}{x+1}}$

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

$\frac{d y}{d x}=\frac{d}{d x}\left(\log \sqrt{\frac{x-1}{x+1}}\right)$

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)^{\frac{1}{2}}} \frac{\mathrm{d}}{\mathrm{dx}}\left[\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)^{\frac{1}{2}}\right]$ [using chain rule]

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)^{-\frac{1}{2}} \frac{1}{2}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)^{\frac{1}{2}-1} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$ [using chain rule]

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)^{-\frac{1}{2}}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)^{-\frac{1}{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)^{-1} \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right) \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$

Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule)

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

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

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{2}\left(\frac{x+1}{x-1}\right)\left[\frac{(x+1)(1-0)-(x-1)(1+0)}{(x+1)^{2}}\right]$

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{(\mathrm{x}-1)(\mathrm{x}+1)}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\mathrm{x}^{2}-1}$

Thus, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\log \sqrt{\frac{\mathrm{x}-1}{\mathrm{x}+1}}\right)=\frac{1}{\mathrm{x}^{2}-1}$