Question

Differentiate

$\log \left\{x+2+\sqrt{x^2+4x+1}\right\}$

Answer 1

$\mathrm{Let}y=\log \left[x+2+\sqrt{x^2+4x+1}\right]$
Differentiate it with respect to x we get,
$$\begin{gathered} \frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}=\frac{d}{dx}\log \left[x+2+\sqrt{x^2+4x+1}\right] \\ =\frac{1}{\left[x+2+\sqrt{x^4+4x+1}\right]}\frac{d}{dx}\left[x+2+{\left(x^2+4x+1\right)}^{\frac{1}{2}}\right]\left[\mathrm{Using}\mathrm{chain}\mathrm{rule}\right] \\ =\frac{1}{\left[x+2+\sqrt{x^4+4x+1}\right]}\times \left[1+0+\frac{1}{2}{\left(x^2+4x+1\right)}^{\frac{-1}{2}}\frac{d}{dx}\left(x^2+4x+1\right)\right] \\ =\frac{1+{\displaystyle \frac{\left(2x+4\right)}{2\left(\sqrt{x^2+4x+1}\right)}}}{\left[x+2+\sqrt{x^4+4x+1}\right]} \\ =\frac{\sqrt{x^2+4x+1}+x+2}{\left[x+2+\sqrt{x^2+4x+1}\right]\times \sqrt{x^2+4x+1}} \\ =\frac{1}{\sqrt{x^2+4x+1}} \\ \mathrm{So},\frac{d}{dx}\left[\log \left\{x+2+\sqrt{x^2+4x+1}\right\}\right]=\frac{1}{\sqrt{x^2+4x+1}} \end{gathered}$$