Solve this

Given $y=\log (\sqrt{x-1}-\sqrt{x+1})$

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

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

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

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

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

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

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

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

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

$\left.-(x+1)^{-\frac{1}{2}}\left\{\frac{d}{d x}(x)+\frac{d}{d x}(1)\right\}\right]$

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

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

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

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

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

$\Rightarrow \frac{d y}{d x}=-\frac{1}{2 \sqrt{x+1} \sqrt{x-1}}$

$\therefore \frac{d y}{d x}=-\frac{1}{2 \sqrt{x^{2}-1}}$

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