Question

Differentiate

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

Answer 1

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