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Math Help - Basic Chain Rule Problem with Trig Functions

  1. #1
    Newbie Tatsuya's Avatar
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    [SOLVED] Basic Chain Rule Problem with Trig Functions

    Hi, everyone. First post but I have a feeling I'll be coming back often.

    Problem: For what values of x in the interval [0,2\Pi] does the graph of f(x)=\cos^2(x)+\sin(x) have a horizontal tangent?

    Answer: f'(x)=-2\cos(x)\sin(x)+\cos(x)

    f(x) has horizontal tangents at x=\frac{\Pi}{2}, \frac{3\Pi}{2}


    I don't have a problem getting the x-values for the horizontal tangents but I can't get the right derivative. Here's my work:


    f(x)=\cos^2(x)+\sin(x)

    f(x)=u^2+\sin(x), u=\cos(x)

    \frac{dy}{du}=2u+\cos(x), \frac{du}{dx}=-\sin(x)

    \frac{dy}{dx}=(2u+\cos(x))(-\sin(x))

    \frac{dy}{dx}=-2u\sin(x)-\sin(x)\cos(x)

    Substitute \frac{du}{dx} back into the equation.

    \frac{dy}{dx}=-2\sin(x)\cos(x)-\sin(x)\cos(x)

    \frac{dy}{dx}=-3\sin(x)\cos(x)

    Where is my error?

    Thank you for your help,
    Tim
    Last edited by Tatsuya; October 23rd 2010 at 09:31 PM.
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  2. #2
    Senior Member Educated's Avatar
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    \frac{dy}{dx}=(2u+\cos(x))(-\sin(x))

    This is incorrect. You are only using the chain rule for the \cos^2 x part, so leave the sin(x) part out of it.

    It should be: \frac{dy}{dx}=(2u)(-\sin(x))+\cos(x)
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  3. #3
    Newbie Tatsuya's Avatar
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    Thank you SO much! That was too simple.


    0=-2\cos(x)\sin(x)+\cos(x)

    0=\cos(x) at \frac{\pi}{2} and \frac{3\pi}{2}

    Don't ask me why I put \frac{5\pi}{6} up there, it's wrong and I edited that out. Must have copied that down in error.

    Thanks again, Educated!
    Tim
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