# How do you find the intervals of increasing and decreasing using the first derivative given y=xsqrt(16-x^2)?

Jul 30, 2017

The interval of increasing is $\left(- \sqrt{8} , \sqrt{8}\right)$
The intervals of decreasing are $\left(- 4 , - \sqrt{8}\right) \cup \left(\sqrt{8} , 4\right)$

#### Explanation:

The function is $y = x \sqrt{16 - {x}^{2}}$

We need the domain of $y$

$16 - {x}^{2} \ge 0$, $\implies$, ${x}^{2} \ge 16$

Therefore the domain is $x \in \left(- 4 , 4\right)$

We need the derivative is

$\left(u v\right) ' = u ' v + u v '$

$u \left(x\right) = x$, $\implies$, $u ' \left(x\right) = 1$

$v \left(x\right) = \sqrt{16 - {x}^{2}}$, $\implies$, $v ' \left(x\right) = \frac{- 2 x}{2 \sqrt{16 - {x}^{2}}} = - \frac{x}{\sqrt{16 - {x}^{2}}}$

So,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \sqrt{16 - {x}^{2}} - {x}^{2} / \left(\sqrt{16 - {x}^{2}}\right) = \frac{16 - {x}^{2} - {x}^{2}}{\sqrt{16 - {x}^{2}}}$

$= \frac{16 - 2 {x}^{2}}{\sqrt{16 - {x}^{2}}}$

The critical points are when $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\frac{16 - 2 {x}^{2}}{\sqrt{16 - {x}^{2}}} = 0$

That is

$16 - {x}^{2} = 0$, $\implies$, ${x}^{2} = 8$, $\implies$, $x = \pm \sqrt{8}$

We can build the variation chart

$\textcolor{w h i t e}{a a a a}$$I n t e r v a l$$\textcolor{w h i t e}{a a a a}$$\left(- 4 , - \sqrt{8}\right)$$\textcolor{w h i t e}{a a a a}$$\left(- \sqrt{8} , \sqrt{8}\right)$$\textcolor{w h i t e}{a a a a}$$\left(\sqrt{8} , 4\right)$

$\textcolor{w h i t e}{a a a a}$$\frac{\mathrm{dy}}{\mathrm{dx}}$$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a a a}$$-$

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a a a a a a a a a a a a}$↘$\textcolor{w h i t e}{a a a a a a a a a a a a}$↗$\textcolor{w h i t e}{a a a a a a a a a}$↘

graph{xsqrt(16-x^2) [-16.02, 16.01, -8.01, 8.01]}