# How do you differentiate f(x)=xsqrt(x-1) using the product rule?

Dec 23, 2015

The point here is seeing that a root can be converted into a power, as stated by the following power rule: ${a}^{\frac{n}{m}} = \sqrt[m]{{a}^{n}}$

#### Explanation:

Using such concept, we can rewrite the expression: $f \left(x\right) = x {\left(x - 1\right)}^{\frac{1}{2}}$

The product rule states that for a function $f \left(x\right) = g \left(x\right) h \left(x\right)$, then $f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

Before, let's also remember that the chain rule states $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$ and rename $u = x - 1$ so we can derivate the term ${\left(x - 1\right)}^{\frac{1}{2}}$

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

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

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