# What is the derivative of f(x) = x(sqrt( 1 - x^2))?

Jan 19, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = \sqrt{1 - {x}^{2}} - {x}^{2} / \left(\sqrt{1 - {x}^{2}}\right)$.

#### Explanation:

We will require the use of two rules: the product rule and the chain rule. The product rule states that:

$\frac{d \left(f g\right)}{\mathrm{dx}}$ = $\frac{\mathrm{df}}{\mathrm{dx}} \cdot g \left(x\right) + f \left(x\right) \cdot \frac{\mathrm{dg}}{\mathrm{dx}}$.

The chain rule states that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$, where $u$ is a function of $x$ and $y$ is a function of $u$.

Therefore,

$\frac{\mathrm{df}}{\mathrm{dx}} = \left(x\right) ' \cdot \left(\sqrt{1 - {x}^{2}}\right) + x \cdot \left(\sqrt{1 - {x}^{2}}\right) '$

To find the derivative of $\sqrt{1 - {x}^{2}}$, use the chain rule, with

$u = 1 - {x}^{2} : \left(\sqrt{u}\right) ' = \frac{1}{2 \sqrt{u}} \cdot u '$

= -(2x)/(2(sqrt(1-x^2)) $= - \frac{x}{\sqrt{1 - {x}^{2}}}$.

Substituting this result into the original equation:

$\frac{\mathrm{df}}{\mathrm{dx}} = \sqrt{1 - {x}^{2}} - {x}^{2} / \left(\sqrt{1 - {x}^{2}}\right)$.