# Question #3b457

Jan 16, 2018

$\frac{9 \left(1 - {x}^{2}\right)}{{\left(7 - 3 x + {x}^{3}\right)}^{4}}$

#### Explanation:

First let's rewrite the given function:

$f \left(x\right) = \frac{1}{7 - 3 x + {x}^{3}} ^ 3 = {\left(7 - 3 x + {x}^{3}\right)}^{- 3}$.

By the chain rule $\frac{d}{\mathrm{dx}} \left(g \left(h \left(x\right)\right)\right) = g ' \left(h \left(x\right)\right) \cdot h ' \left(x\right)$. For this function we have $g \left(x\right) = {x}^{-} 3$ and $h \left(x\right) = 7 - 3 x + {x}^{3}$.

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = - 3 {\left(7 - 3 x + {x}^{3}\right)}^{-} 4 \cdot \left(- 3 + 3 {x}^{2}\right)$

$= \frac{9 \left(1 - {x}^{2}\right)}{{\left(7 - 3 x + {x}^{3}\right)}^{4}}$