Question

If $y=\sqrt{x^2+1}$ find $\frac{dy}{dx}$.

Answer 1

$y=\sqrt{x^2+1}\frac{dy}{dx}=\frac{d}{dx}\sqrt{x^2+1}\frac{dy}{dx}=\frac{1}{2\sqrt{x^2+1}}\frac{d}{dx}(x^2+1)\frac{dy}{dx}=\frac{1}{2\sqrt{x^2+1}}2x\frac{dy}{dx}=\frac{x}{\sqrt{x^2+1}}$