Question

Differentiate with respect to $x$

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

Answer 1

given,

$y=\ln (x+\sqrt{1+x^2})$

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

$[\because \frac{\ln x}{dx}=\frac{1}{x}]$

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

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