Question

How do you find the derivative of $y=\frac{1}{\sqrt{x}}$ ?

Answer 1

Step 1: Convert radical to exponent form

Given: $y=\frac{1}{\sqrt{x}}$

$$\begin{gathered} y=\frac{1}{\sqrt{x}} \\ y=x^{-\frac{1}{2}} \end{gathered}$$

Step 2: Solve for differentiation

We know that $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}{x}^{-\frac{1}{2}}\\ & =& -\frac{1}{2}\left(x^{-\frac{1}{2}-1}\right)\\ & =& -\frac{1}{2}{x}^{-\frac{3}{2}}\\ & =& -\frac{1}{2\sqrt{x^3}}\end{array}$

Hence, the derivative of $y=\frac{1}{\sqrt{x}}$is $\frac{dy}{dx}=-\frac{1}{2\sqrt{x^3}}$.