# How do you use the chain rule to differentiate f(x)=(3x-9)^2(4x^3+2x^-9)^-9?

Oct 21, 2017

see below

#### Explanation:

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

$f ' \left(x\right) = {\left(3 x - 9\right)}^{2} \frac{d}{\mathrm{dy}} {\left(4 {x}^{3} + 2 {x}^{-} 9\right)}^{-} 9 + {\left(4 {x}^{3} + 2 {x}^{-} 9\right)}^{-} 9 \frac{d}{\mathrm{dx}} {\left(3 x - 9\right)}^{2}$

$= {\left(3 x - 9\right)}^{2} \left(- 9\right) {\left(4 {x}^{3} + 2 {x}^{-} 9\right)}^{-} 10 \frac{d}{\mathrm{dy}} \left(4 {x}^{3} + 2 {x}^{-} 9\right) + {\left(4 {x}^{3} + 2 {x}^{-} 9\right)}^{-} 9 \left(2\right) \left(3 x - 9\right) \frac{d}{\mathrm{dx}} \left(3 x - 9\right)$

$= {\left(3 x - 9\right)}^{2} \left(- 9\right) {\left(4 {x}^{3} + 2 {x}^{-} 9\right)}^{-} 10 \left(12 {x}^{2} - 18 {x}^{-} 10\right) + {\left(4 {x}^{3} + 2 {x}^{-} 9\right)}^{-} 9 \left(2\right) \left(3 x - 9\right) \left(3\right)$

$= \left(- 9\right) {\left(3 x - 9\right)}^{2} {\left(4 {x}^{3} + 2 {x}^{-} 9\right)}^{-} 10 \left(12 {x}^{2} - 18 {x}^{-} 10\right) + \left(6\right) {\left(4 {x}^{3} + 2 {x}^{-} 9\right)}^{-} 9 \left(3 x - 9\right)$