# How do you differentiate f(x)= x * (4-x^2)^(1/2) using the product rule?

Mar 5, 2016

$f ' \left(x\right) = \frac{4 - 2 {x}^{2}}{\sqrt{4 - {x}^{2}}}$

#### Explanation:

The Product Rule:

frac{"d"}{"d"x}(uv) = v frac{"d"u}{"d"x} + u frac{"d"v}{"d"x}

In this question,

• $u = x$

• $v = {\left(4 - {x}^{2}\right)}^{\frac{1}{2}}$

$f ' \left(x\right) = \frac{\text{d"}{"d} x}{x {\left(4 - {x}^{2}\right)}^{\frac{1}{2}}}$

$= {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} \frac{\text{d"}{"d"x}(x) + x frac{"d"}{"d} x}{{\left(4 - {x}^{2}\right)}^{\frac{1}{2}}}$

$= {\left(4 - {x}^{2}\right)}^{\frac{1}{2}} \left(1\right) + x \left(\frac{1}{2}\right) {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \frac{\text{d"}{"d} x}{4 - {x}^{2}}$

$= \sqrt{4 - {x}^{2}} + x \left(\frac{1}{2}\right) {\left(4 - {x}^{2}\right)}^{- \frac{1}{2}} \left(- 2 x\right)$

$= \sqrt{4 - {x}^{2}} - {x}^{2} / \sqrt{4 - {x}^{2}}$

$= \frac{{\left(\sqrt{4 - {x}^{2}}\right)}^{2} - {x}^{2}}{\sqrt{4 - {x}^{2}}}$

$= \frac{4 - 2 {x}^{2}}{\sqrt{4 - {x}^{2}}}$