# How do you find the derivative of x*x^(1/2)?

Aug 8, 2015

$\frac{d}{\mathrm{dx}} \left(x \cdot {x}^{\frac{1}{2}}\right) = \frac{d}{\mathrm{dx}} {x}^{\frac{3}{2}} = \frac{3}{2} {x}^{\frac{1}{2}}$

or

$\frac{d}{\mathrm{dx}} \left(x \cdot {x}^{\frac{1}{2}}\right) = \left(\frac{d}{\mathrm{dx}} x\right) \cdot {x}^{\frac{1}{2}} + x \cdot \left(\frac{d}{\mathrm{dx}} {x}^{\frac{1}{2}}\right)$

$= 1 \cdot {x}^{\frac{1}{2}} + x \cdot \left(\frac{1}{2}\right) {x}^{- \frac{1}{2}} = {x}^{\frac{1}{2}} + \frac{1}{2} {x}^{\frac{1}{2}} = \frac{3}{2} {x}^{\frac{1}{2}}$

#### Explanation:

We can either multiply $x \cdot {x}^{\frac{1}{2}} = {x}^{\frac{3}{2}}$ first then use the power rule, or we can use the product rule, using the power rule on each part.

Power Rule
$\frac{d}{\mathrm{dx}} {x}^{k} = k \cdot {x}^{k - 1}$

Product Rule
$\frac{d}{\mathrm{dx}} \left(f \left(x\right) \cdot g \left(x\right)\right) = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$