Question

If $y=\frac{1}{\sqrt{a^2-x^2}}$, find $\frac{dy}{dx}$

Answer 1

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

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

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

$=\frac{0-\frac{-2x}{2\sqrt{a^2-x^2}}}{(a^2-x^2)}$

$=\frac{x}{(a^2-x^2{)}^{3/2}}$.