# How do you differentiate f(x) = (4x^3 + 2x) ^ - 4?

Jun 9, 2016

$\frac{- 8 \left(6 {x}^{2} + 1\right)}{4 {x}^{3} + 2 x} ^ 5$

#### Explanation:

differentiate f(x) using the $\textcolor{b l u e}{\text{chain rule}}$

f(x) = g(h(x)) then f'(x) = g'(h(x)).h'(x)
$\text{-------------------------------------------}$

$g \left(h \left(x\right)\right) = {\left(4 {x}^{3} + 2 x\right)}^{-} 4 \Rightarrow g ' \left(h \left(x\right)\right) = - 4 {\left(4 {x}^{3} + 2 x\right)}^{-} 5$

$h \left(x\right) = 4 {x}^{3} + 2 x \Rightarrow h ' \left(x\right) = 12 {x}^{2} + 2 = 2 \left(6 {x}^{2} + 1\right)$
$\text{-------------------------------------------------------------------------}$
Substitute these values into f'(x)

$f ' \left(x\right) = - 4 {\left(4 {x}^{3} + 2 x\right)}^{-} 5 .2 \left(6 {x}^{2} + 1\right)$

$= \frac{- 8 \left(6 {x}^{2} + 1\right)}{4 {x}^{3} + 2 x} ^ 5$