Question

Differentiate w.r.t $x$

$y=x+\sqrt{x^2+a^2}$

Answer 1

$y=x+\sqrt{x^2+a^2}$

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

$=1+\frac{1}{2\sqrt{x^2+a^2}}\frac{d}{dx}\left(x^2\right)$

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

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