# How do you differentiate  f(x)=(3x^5 - 4x^3 + 2)^23  using the chain rule.?

Dec 9, 2015

$f ' \left(x\right) = 69 {x}^{2} {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{22} \left(5 {x}^{2} - 4\right)$

#### Explanation:

Remember: Chain rule:

$\text{Derivative of } f \left(g \left(x\right)\right) = f ' \left(x\right) g \left(x\right) \cdot g ' \left(x\right)$

Derivative of Power and chain rule: $f \left(x\right) = {\left(g \left(x\right)\right)}^{n} = f ' \left(x\right) = n \left(g {\left(x\right)}^{n - 1}\right) \cdot g ' \left(x\right)$

Given $f \left(x\right) {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{23}$
f'(x) = 23(3x^5-4x^3+2)^(23-1) * color(red)(d/(dx)(3x^5 -4x^3+2)

=23(3x^5 -4x^3+2)^22 color(red)((15x^4 -12x^2+0)
$= 23 {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{22} \textcolor{red}{15 {x}^{4} - 12 {x}^{2}}$ or

by factor out the greatest common factor $\textcolor{b l u e}{3 {x}^{2}}$from $15 {x}^{4} - 12 {x}^{2}$
$f ' \left(x\right) = 23 \cdot \textcolor{b l u e}{3 {x}^{2}} {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{22} \left(5 {x}^{2} - 4\right)$

Simplify:
$f ' \left(x\right) = 69 {x}^{2} {\left(3 {x}^{5} - 4 {x}^{3} + 2\right)}^{22} \left(5 {x}^{2} - 4\right)$