Solve this

Given $\mathrm{y}=\sqrt{\mathrm{x}+1}+\sqrt{\mathrm{x}-1}$

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

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

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

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

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

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}(\mathrm{x}+1)^{-\frac{1}{2}}[1+0]+\frac{1}{2}(\mathrm{x}-1)^{-\frac{1}{2}}[1-0]$

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

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

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

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

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

But, $y=\sqrt{x+1}+\sqrt{x-1}$

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

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

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