# What is the second derivative of f(x)=(x^2-x^3)^(1/3)?

Jan 30, 2016

$\frac{1}{3} \left(2 - 6 x\right) {\left({x}^{2} - {x}^{3}\right)}^{- \frac{2}{3}} \left\{1 - \frac{\left(4 - 6 x\right)}{\left(3 x - 3 {x}^{2}\right)}\right\}$

#### Explanation:

here,
$f \left(x\right) = {\left({x}^{2} - {x}^{3}\right)}^{\frac{1}{3}}$

so, the first derivative is,

$f p r i m e \left(x\right) = \frac{d}{\mathrm{dx}} \left(f \left(x\right)\right)$

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

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

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

so the second derivative is,

$f p r i m e p r i m e \left(x\right) = \frac{d}{\mathrm{dx}} \left(f p r i m e \left(x\right)\right)$

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

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

$= \frac{1}{3} \left\{\left(2 x - 3 {x}^{2}\right) \frac{d}{\mathrm{dx}} {\left({x}^{2} - {x}^{3}\right)}^{- \frac{2}{3}} + {\left({x}^{2} - {x}^{3}\right)}^{- \frac{2}{3}} \frac{d}{\mathrm{dx}} \left(2 x - 3 {x}^{2}\right)\right\}$

$= \frac{1}{3} \left\{- \frac{2}{3} \left(2 x - 3 {x}^{2}\right) {\left({x}^{2} - {x}^{3}\right)}^{- \frac{5}{3}} \left(2 - 6 x\right) + {\left({x}^{2} - {x}^{3}\right)}^{- \frac{2}{3}} \left(2 - 6 x\right)\right\}$

$= \frac{1}{3} \left(2 - 6 x\right) {\left({x}^{2} - {x}^{3}\right)}^{- \frac{2}{3}} \left\{- \frac{2}{3} \left(\frac{2 x - 3 {x}^{2}}{{x}^{2} - {x}^{3}}\right) + 1\right\}$

=1/3(2-6x)(x^2-x^3)^(-2/3){-2/3((cancel(x)(2-3x))/(cancel(x)(x-x^2))+1}

$= \frac{1}{3} \left(2 - 6 x\right) {\left({x}^{2} - {x}^{3}\right)}^{- \frac{2}{3}} \left\{1 - \frac{\left(4 - 6 x\right)}{\left(3 x - 3 {x}^{2}\right)}\right\}$