# How do you find the derivative for f(x) = 1 / [root(3)(3 - x^3)]?

May 27, 2015

We will need to use the chain rule plus some basic properties of exponents and simple derivatives for this.

$f \left(x\right) = \frac{1}{\sqrt[3]{3 - {x}^{3}}} = {\left(3 - {x}^{3}\right)}^{- \frac{1}{3}}$

Let $h \left(x\right) = \left(3 - {x}^{3}\right)$
and $g \left(x\right) = {x}^{- \frac{1}{3}}$

so $f \left(x\right) = g \left(h \left(x\right)\right)$

Note that $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{d g \left(h \left(x\right)\right)}{d h \left(x\right)} \cdot \frac{d h \left(x\right)}{\mathrm{dx}}$

$\frac{d g \left(h \left(x\right)\right)}{d h \left(x\right)} = - \frac{1}{3} {\left(h \left(x\right)\right)}^{- \frac{4}{3}}$

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

$\frac{d f \left(x\right)}{\mathrm{dx}} = \left(- \frac{1}{3} {\left(h \left(x\right)\right)}^{- \frac{4}{3}}\right) \cdot \left(- 3 {x}^{2}\right)$

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

$= \frac{{x}^{2}}{\left(3 - {x}^{3}\right) \sqrt[3]{3 - {x}^{3}}}$