# Solve this

Question:

If $\mathrm{y}=\sqrt{\mathrm{x}+1}+\sqrt{\mathrm{x}-1}$, prove that $\sqrt{\mathrm{x}^{2}-1} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2} \mathrm{y}$.

Solution:

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

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

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