Differentiate the following functions with respect to x :

Solution:

Let $y=\log \left(x+2+\sqrt{x^{2}+4 x+1}\right)$

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

$\frac{d y}{d x}=\frac{d}{d x}\left[\log \left(x+2+\sqrt{x^{2}+4 x+1}\right)\right]$

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

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

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

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

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

Also the derivative of a constant is 0 .

$\Rightarrow \frac{d y}{d x}=\frac{1}{x+2+\sqrt{x^{2}+4 x+1}}\left[1+0+\frac{1}{2}\left(x^{2}+4 x+1\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(x^{2}+4 x+1\right)\right]$

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

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

$\left.+\frac{1}{2 \sqrt{x^{2}+4 x+1}}\left(\frac{d}{d x}\left(x^{2}\right)+4 \frac{d}{d x}(x)+\frac{d}{d x}(1)\right)\right]$

However, $\frac{d}{d x}\left(x^{2}\right)=2 x$ and $\frac{d}{d x}(x)=1$

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

$\Rightarrow \frac{d y}{d x}=\frac{1}{x+2+\sqrt{x^{2}+4 x+1}}\left[1+\frac{2 x+4}{2 \sqrt{x^{2}+4 x+1}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{x+2+\sqrt{x^{2}+4 x+1}}\left[1+\frac{x+2}{\sqrt{x^{2}+4 x+1}}\right]$

$\Rightarrow \frac{d y}{d x}=\frac{1}{x+2+\sqrt{x^{2}+4 x+1}}\left[\frac{\sqrt{x^{2}+4 x+1}+x+2}{\sqrt{x^{2}+4 x+1}}\right]$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{\mathrm{x}^{2}+4 \mathrm{x}+1}}$

Thus, $\frac{d}{d x}\left[\log \left(x+2+\sqrt{x^{2}+4 x+1}\right)\right]=\frac{1}{\sqrt{x^{2}+4 x+1}}$