Question

Find $\frac{dy}{dx}$ where
$y=\sqrt{\frac{x^2+1}{x^2+2}}$

Answer 1

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

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

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

$=\sqrt{x^2+1}[-x({x^2+2)}^{\frac{-3}{2}}]+\frac{x}{\left(\sqrt{x^2+2}\right)\left(\sqrt{x^2+1}\right)}$