# What are the critical points of  f(x) = e^(x^(1/3))*sqrt(x+8)?

Apr 8, 2018

$f {\left(x\right)}^{/} =$$\left(3\right)$$\left(1\right)$$+$$\left(2\right)$$\left(4\right)$

#### Explanation:

$f \left(x\right) = {e}^{{x}^{\frac{1}{3}}} \cdot \sqrt{x + 8}$

$y = \sqrt{x + 8}$ $\left(1\right)$

$z = {e}^{{x}^{\frac{1}{3}}}$ $\left(2\right)$

$f \left(x\right) = z \cdot y$

$f {\left(x\right)}^{/} = {z}^{/} \cdot y + z \cdot {y}^{/}$

$z = {e}^{{x}^{\frac{1}{3}}}$

${x}^{\frac{1}{3}} = n$

$z = {e}^{n}$

${x}^{\frac{1}{3}} = n$

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

$z = {e}^{n}$

$\frac{\mathrm{dz}}{\mathrm{dn}} = {e}^{n}$

$\frac{\mathrm{dz}}{\mathrm{dn}} \cdot \frac{\mathrm{dn}}{\mathrm{dx}}$ = $\frac{\mathrm{dz}}{\mathrm{dx}}$

$\frac{\mathrm{dz}}{\mathrm{dx}} =$ $\frac{1}{3} {x}^{- \frac{2}{3}} \cdot {e}^{{x}^{\frac{1}{3}}}$ $\left(3\right)$

$y = \sqrt{x + 8}$

$u = x + 8$

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

$y = \sqrt{u}$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2} {u}^{- \frac{1}{2}}$

$\frac{\mathrm{dy}}{\mathrm{du}} \cdot$$\frac{\mathrm{du}}{\mathrm{dx}}$$= \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {\left(x + 8\right)}^{- \frac{1}{2}}$ $\left(4\right)$

$f {\left(x\right)}^{/} = {z}^{/} \cdot y + z \cdot {y}^{/}$

$f {\left(x\right)}^{/} =$$\left(3\right)$$\left(1\right)$$+$$\left(2\right)$$\left(4\right)$