Question

If $y=\frac{1}{\sqrt{x}}=(x{)}^{-1/2}$, then find $\frac{dy}{dx}$.

Answer 1

Differentiating both sides w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{d\left[\right(x{)}^{-1/2}]}{dx}=-\frac{1}{2}{x}^{-\left(1/2\right)-1}=-\frac{1}{2}{x}^{-3/2}=-\frac{1}{2x^{3/2}}$.