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Math Help - derivative of trig function and horizontal tangent lines

  1. #1
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    derivative of trig function and horizontal tangent lines

    I think I am going to start helping people in algebra or something to earn my keep here. Anyways, here is the one I am workin on.

    Find where graph of y= cosx/ (1+sinx) has horizontal tangent lines on the interval (0, pi/2)

    I have simplified the derivative down to:

    (-sinx-cos^2x)/ (1+ sinx) and I know that i need to find where x = 0 i am just at a loss as to how to find it. Any help is greatly appreciated, as always.

    Thanks

    Jordan
    Last edited by mr fantastic; July 23rd 2009 at 05:46 PM. Reason: Restored question deleted by OP.
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  2. #2
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    Quote Originally Posted by jwbehm View Post
    I think I am going to start helping people in algebra or something to earn my keep here. Anyways, here is the one I am workin on.

    Find where graph of y= cosx/ (1+sinx) has horizontal tangent lines on the interval (0, pi/2)

    I have simplified the derivative down to:

    (-sinx-cos^2x)/ (1+ sinx) and I know that i need to find where x = 0 i am just at a loss as to how to find it. Any help is greatly appreciated, as always.

    Thanks

    Jordan
    To have horizontal tangent lines, the derivative is 0.

    So y = \frac{\cos{x}}{1 + \sin{x}}

    Use the Quotient Rule to take the derivative...

    \frac{dy}{dx} = \frac{(1 + \sin{x})(-\sin{x}) - \cos{x}(\cos{x})}{(1 + \sin{x})^2}

     = \frac{-\sin{x} - \sin^2{x} - \cos^2{x}}{(1 + \sin{x})^2}

     = \frac{-\sin{x} - 1}{(1 + \sin{x})^2}

     = -\frac{1}{1 + \sin{x}}.


    Set the derivative equal to 0 and solve for x...

    0 = -\frac{1}{1 + \sin{x}}.


    I think you can see here that x does not have a solution. So there does not exist any x that will give the original function horizontal tangent lines.
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  3. #3
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    Quote Originally Posted by jwbehm View Post
    I think I am going to start helping people in algebra or something to earn my keep here. Anyways, here is the one I am workin on.

    Find where graph of y= cosx/ (1+sinx) has horizontal tangent lines on the interval (0, pi/2)

    I have simplified the derivative down to:

    (-sinx-cos^2x)/ (1+ sinx) and I know that i need to find where x = 0 i am just at a loss as to how to find it. Any help is greatly appreciated, as always.

    Thanks

    Jordan


    Image taken from http://webgraphing.com/graphing_advanced.jsp



    You can see that nowhere on the graph is there a horizontal tangent.
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