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Math Help - derivative of the power of a function

  1. #1
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    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)
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  2. #2
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  3. #3
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    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.

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