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Math Help - Derivative of 3cos^5x.

  1. #1
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    Derivative of 3cos^5x.

    hi, my first post so pls be gentle!
    HNC maths question driving me mad as must be doing something stupidly wrong.
    3cos^5x is the question and result should be -15sinx cos^4x
    i must be doing something fundamentally wrong as i cant match the answer given. any help greatly appreciated.
    Last edited by mr fantastic; May 8th 2010 at 03:08 AM. Reason: Re-titled.
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    Quote Originally Posted by neilj View Post
    hi, my first post so pls be gentle!
    HNC maths question driving me mad as must be doing something stupidly wrong.
    3cos^5x is the question and result should be -15sinx cos^4x
    i must be doing something fundamentally wrong as i cant match the answer given. any help greatly appreciated.
    derivative of 3\cos^5{x} = -15\sin{x}cos^4{x}
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  3. #3
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    Just in case a picture helps...



    ... where



    ... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
    _________________________________________
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  4. #4
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    Quote Originally Posted by neilj View Post
    hi, my first post so pls be gentle!
    HNC maths question driving me mad as must be doing something stupidly wrong.
    3cos^5x is the question and result should be -15sinx cos^4x
    i must be doing something fundamentally wrong as i cant match the answer given. any help greatly appreciated.
    Put simpler still...

    \frac{dy}{dx} = \frac{du}{dx}\cdot\frac{dy}{du}.


    If y = 3\cos^5{x}

    Let u = \cos{x} so that y = 3u^5.


    \frac{du}{dx} = -\sin{x}


    \frac{dy}{du} = 15u^4 = 15\cos^4{x}.


    Therefore \frac{dy}{dx} = -15\sin{x}\cos^4{x}.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Put simpler still...

    \frac{dy}{dx} = \frac{du}{dx}\cdot\frac{dy}{du}.


    If y = 3\cos^5{x}

    Let u = \cos{x} so that y = 3u^5.


    \frac{du}{dx} = -\sin{x}


    \frac{dy}{du} = 15u^4 = 15\cos^4{x}.


    Therefore \frac{dy}{dx} = -15\sin{x}\cos^4{x}.
    thanks for that. must have tried every variation bar that one!! splitting out into u and y parts was my downfall.
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