# Thread: derivative of the power of a function

1. ## derivative of the power of a function

Find the derivative of s= 2x^2 (3-x^3)^6

using product rule
ds/dx = (3-x^3)^6(6x^2) + 2x^3(6)(-3x^2)(3-x^3)^5
ds/dx= (3-x^3)^5(6x^2(3-x^3) -36x^5)
ds/dx= (3-x^3)^5(18x^2-6x^5-36x^5)
(3-x^3)^5(18x^2-42x^5)

I can't get the answer (3-x^3)^5 (12x-40x^4)

2. Hello, shinn90!

You seem to be making some silly errors . . .

$\text{Differentiate: }\;s\:=\:\underbrace{2x^2}_{f(x)} \underbrace{(3-x^3)^6}_{g(x)}$
Product Rule:

. . $\frac{ds}{dx} \;=\;\overbrace{4x}^{f'(x)}\cdot\overbrace{(3-x^3)^6}^{g(x)} + \overbrace{2x^2}^{f(x)}\cdot\overbrace{6(3-x^3)^5\cdot(\text{-}3x^2)}^{g'(x)}$

. . . . . $= \;4x(3-x^3)^6 - 36x^4(3-x^3)^5$

Factor: . $4x(3-x^3)^5\,\bigg[(3-x^3) - 9x^3\bigg] \;=\;4x(3-x^3)^5(3-10x^3)$

This is a factored (better) form of their answer.