Differentiate the following functions with respect to x :

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

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

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

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

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

$\Rightarrow y=\frac{2 x^{2}+2 \sqrt{x^{4}-1}}{2}$

$\Rightarrow y=x^{2}+\sqrt{x^{4}-1}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+\sqrt{\mathrm{x}^{4}-1}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{\mathrm{x}^{4}-1}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left[\left(\mathrm{x}^{4}-1\right)^{\frac{1}{2}}\right]$

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+\frac{1}{2}\left(\mathrm{x}^{4}-1\right)^{\frac{1}{2}-1} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{4}-1\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+\frac{1}{2}\left(\mathrm{x}^{4}-1\right)^{-\frac{1}{2}} \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{4}-1\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+\frac{1}{2 \sqrt{\mathrm{x}^{4}-1}}\left[\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{4}\right)-\frac{\mathrm{d}}{\mathrm{dx}}(1)\right]$

We have $\frac{d}{d x}\left(x^{4}\right)=4 x^{3}$ and derivative of a constant is 0 .

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+\frac{1}{2 \sqrt{\mathrm{x}^{4}-1}}\left[4 \mathrm{x}^{3}-0\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}+\frac{1}{2 \sqrt{\mathrm{x}^{4}-1}} \times 4 \mathrm{x}^{3}$

$\therefore \frac{d y}{d x}=2 x+\frac{2 x^{3}}{\sqrt{x^{4}-1}}$

Thus, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\sqrt{\mathrm{x}^{2}+1}+\sqrt{\mathrm{x}^{2}-1}}{\sqrt{\mathrm{x}^{2}+1}-\sqrt{\mathrm{x}^{2}-1}}\right)=2 \mathrm{x}+\frac{2 \mathrm{x}^{3}}{\sqrt{\mathrm{x}^{4}-1}}$