# How do you differentiate f(x) =x^2sqrtx  using the product rule?

Dec 17, 2016

The answer is $= \frac{5}{2} x \sqrt{x}$

#### Explanation:

The product rule is

$\left(u v\right) ' = u ' v + u v '$

Here, we use

$\left({x}^{n}\right) ' = n {x}^{n - 1}$

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

So,

$u = {x}^{2}$, $\implies$, $u ' = 2 x$

$v = \sqrt{x}$, $\implies$, $v ' = \frac{1}{2 \sqrt{x}}$

Therefore,

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

$= \frac{4 {x}^{2} + {x}^{2}}{2 \sqrt{x}}$

$= \frac{5 {x}^{2}}{2 \sqrt{x}} = \frac{5}{2} x \sqrt{x}$