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

Aug 16, 2015

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

#### Explanation:

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

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 $u = x$ and $v = {\left(1 - {x}^{2}\right)}^{\frac{1}{2}}$

Then

$\frac{\mathrm{du}}{\mathrm{dx}} = 1$

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{2} {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right) = - x {\left(1 - {x}^{2}\right)}^{- \frac{1}{2}}$

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

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

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