Differentiate the following functions with respect to x :

Solution:

Let $y=\log \left(x+\sqrt{x^{2}+1}\right)$

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

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

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

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

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

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

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

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

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

However, $\frac{d}{d x}\left(x^{2}\right)=2 x$ and derivative of a constant is 0 .

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

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

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

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

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

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

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