Question

Differentiate $\log (x+\sqrt{a^2+x^2})$

Answer 1

Let $y=log(x+\sqrt{a^2+x^2})$

On diffrently wrt 'x' we get

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

$\frac{d}{dx}=\frac{1}{x+\sqrt{a^2+x^2}}[1+\frac{2x}{2\sqrt{a^2+x^2}}]$

$\frac{d}{dx}=\frac{1}{\sqrt{a^2+x^2}}\times \frac{x+\sqrt{a^2+x^2}}{\sqrt{a^2+x^2}}=\frac{1}{\sqrt{a^2+x^2}}$

$\frac{d}{dx}=\frac{1}{\sqrt{a^2+x^2}}$