# How do you differentiate g(y) =xsqrtx  using the product rule?

Jan 25, 2016

#### Answer:

For $g \left(x\right) = x \sqrt{x}$, we get $g ' \left(x\right) = \frac{3 \sqrt{x}}{2}$

#### Explanation:

For $g \left(x\right) = x \sqrt{x} = {x}^{\frac{3}{2}}$, but the question specifies the product rule, so we think of it as:

$g \left(x\right) = F S$.

The product rule gives $g ' \left(x\right) = F ' G + F G '$.

Here, we have $F = x$, so F'=1#

and $G = \sqrt{x} = {x}^{\frac{1}{2}}$, so $G ' = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \sqrt{x}}$.

Applying the product rule we get

$g ' \left(x\right) = \left(1\right) \left(\sqrt{x}\right) + \left(x\right) \left(\frac{1}{2 \sqrt{x}}\right)$

$= \sqrt{x} + \frac{x}{2} \sqrt{x} = \sqrt{x} + \frac{\sqrt{x}}{2} = \frac{3}{2} \sqrt{x}$