# How do you use the Product Rule to find the derivative of  y = (x^2 - 1) sqrtx?

Aug 16, 2015

$\textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{2} - 1}{2 \sqrt{x}}}$

#### Explanation:

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

The Product Rule, states that if $y = u v$, then

$\frac{\mathrm{dy}}{\mathrm{dx}} = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

Let $u = {x}^{2} - 1$ and $v = {x}^{\frac{1}{2}}$

Then

$\frac{\mathrm{du}}{\mathrm{dx}} = 2 x$

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$

dy/dx=(x^2-1)× 1/2x^(-1/2)+ x^(1/2)×2x

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} - 1}{2 {x}^{\frac{1}{2}}} + 2 {x}^{\frac{3}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} - 1}{2 {x}^{\frac{1}{2}}} + \frac{4 {x}^{2}}{2 {x}^{\frac{1}{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{5 {x}^{2} - 1}{2 \sqrt{x}}$