# What is the first derivative and second derivative of 4x^(1/3)+2x^(4/3)?

May 6, 2016

$\frac{d y}{d x} = \frac{4}{3} \cdot {x}^{- \frac{2}{3}} + \frac{8}{3} \cdot {x}^{\frac{1}{3}} \text{ (the first derivative)}$
$\frac{{d}^{2} y}{d {t}^{2}} = \frac{8}{9} \cdot {x}^{- \frac{2}{3}} \left(- {x}^{-} 1 + 1\right) \text{ (the second derivative)}$

#### Explanation:

$y = 4 {x}^{\frac{1}{3}} + 2 {x}^{\frac{4}{3}}$

$\frac{d y}{d x} = \frac{1}{3} \cdot 4 \cdot {x}^{\left(\frac{1}{3} - 1\right)} + \frac{4}{3} \cdot 2 {x}^{\left(\frac{4}{3} - 1\right)}$

$\frac{d y}{d x} = \frac{4}{3} \cdot {x}^{- \frac{2}{3}} + \frac{8}{3} \cdot {x}^{\frac{1}{3}} \text{ (the first derivative)}$

$\frac{{d}^{2} y}{d {t}^{2}} = - \frac{2}{3} \cdot \frac{4}{3} \cdot {x}^{\left(- \frac{2}{3} - 1\right)} + \frac{8}{3} \cdot \frac{1}{3} \cdot {x}^{\left(\frac{1}{3} - 1\right)}$

(d^2 y)/(d t^2)=-8/9*x^((-5/3))+8/9*x^((-2/3)

$\frac{{d}^{2} y}{d {t}^{2}} = \frac{8}{9} \cdot {x}^{- \frac{2}{3}} \left(- {x}^{-} 1 + 1\right) \text{ (the second derivative)}$