# How do you find the derivative of f(x) = (4x^3 + 2x) ^ - 4?

Apr 17, 2016

$\frac{d f \left(x\right)}{d x} = - 4 {\left(4 {x}^{3} + 2 x\right)}^{- 5} \cdot \left(12 {x}^{2} + 2\right)$

#### Explanation:

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

$u = 4 {x}^{3} + 2 x$

$f \left(x\right) = {u}^{-} 4$

$\frac{d f \left(x\right)}{d u} = - 4 {u}^{-} 5$

$\frac{d u}{d x} = 12 {x}^{2} + 2$

$\frac{d f \left(x\right)}{d x} = \frac{d f \left(x\right)}{d u} \cdot \frac{d u}{d x}$

$\frac{d f \left(x\right)}{d x} = - 4 {u}^{-} 5 \cdot \left(12 {x}^{2} + 2\right)$

$\frac{d f \left(x\right)}{d x} = - 4 {\left(4 {x}^{3} + 2 x\right)}^{- 5} \cdot \left(12 {x}^{2} + 2\right)$