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Math Help - Derivative of Sin

  1. #1
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    Derivative of Sin

    I have a problem and to check that I have it correct as I do not have answers.

    y=8x sin(3x^4-2)^3 - the product rule kicks in here
    y'=8 sin(3x^4-2)^3 + 8x cos(3x^4-2)^3 (3)(3x^4-2)^2 (12x^3)

    Can someone check to see if I understand this please?

    Thanks
    Joanne
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  2. #2
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    sin(3x^4-2)^3 can mean

    \sin^3(3x^4 - 2)\ \rm{i.e.}\m\ [\sin(3x^4 - 2)]^3

    and if that's what you mean then your working is good except for a missing 'sin' between the (3) and the subsequent (3x^4-2). Edit: sorry, no you need to remove the power 3 from the cos, as well. In that case I'll do a pic, as well.

    If sin(3x^4-2)^3 means

    \sin[(3x^4-2)^3] then that's another matter. [Oh... you did mean that, and it's all good.]
    Last edited by tom@ballooncalculus; February 19th 2011 at 02:31 PM. Reason: Oh
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  3. #3
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    u = 8x \rightarrow u' = 8

    v = \sin(3x^4-2)^3 \rightarrow v' = \cos(3x^4-2)^3 \cdot 36x^3(3x^4-2)^2


    y' = vu' + uv' = 8 \cdot (3x^4-2)^3 + 8x \cdot \cos(3x^4-2)^3 \cdot 36x^3(3x^4-2)^2

    y' = 8(3x^4-2)^2 [(3x^4-2) + 36x^4(3x^4-2) \cos(3x^4-2)^3

    I do get the same answer as you although I imagine you'd be expected to simplify your equation
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  4. #4
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    Thank you to you both for the reply. It helped.
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