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Math Help - differentiate the trig function

  1. #1
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    differentiate the trig function

    1. sinx/1-cosx.
    (1-cosx)(sinx)'-(sinx)(1-cosx)'
    -----------------------------------
    (1-cosx)^2

    (1-cosx)(cosx)-(sinx)(sinx)
    --------------------------------
    (1-cosx)^2

    (cosx-cosx^2)-sinx^2
    --------------------------------
    (1-cosx)^2

    the answer comes out to be 1/cos(x)-1.

    2. 1+sinx/2-cosx
    (2-cosx)(1+sinx)'-(1+sinx)(2-cosx)'
    ---------------------------------------
    (2-cosx)^2

    (2-cosx)(cosx)-(1+sinx)(sinx)
    ---------------------------------------
    (2-cosx)^2

    2cosx-cosx^2-sinx-sinx^2
    ---------------------------------------
    (2-cosx)^2

    the answer comes out to be (2cosx -sinx -1)/(2-cosx)^2.

    3.sinx+cosx/sinx-cosx

    (sinx-cosx)(sinx+cosx)'-(sinx+cosx)(sinx-cosx)'
    -----------------------------------------------
    (sinx-cosx)^2

    (sinx-cosx)(cos-sinx)-(sinx+cosx)(cosx+sinx)
    -----------------------------------------------
    (sinx-cosx)^2

    and after a whole lot of crossing out, I got (sinxcosx+sinxcosx-sinxcosx+sinxcosx)/(sinx-cosx)^2

    but the answer says -2/(sinx-cosx)^2
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  2. #2
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    See here.
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  3. #3
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    1.) \frac{\sin{x}}{1-\cos{x}}
    =\frac{(1-\cos{x})(\sin{x})' - (1-\cos{x})'(\sin{x})}{(1-\cos{x})^2}
    =\frac{(1-\cos{x})(\cos{x}) - (-(-\sin{x})(\sin{x}))}{(1-\cos{x})^2}
    =\frac{\cos{x} - \cos^2{x} - \sin^2{x}}{(1-\cos{x})^2}
    =\frac{\cos{x} - (\cos^2{x} + \sin^2{x})}{(1-\cos{x})^2}
    =\frac{\cos{x} - 1}{(1-\cos{x})^2}
    =-\frac{1-\cos{x}}{(1-\cos{x})^2}
    cancel 1- cos(x)
    =-\frac{1}{1-\cos{x}}

    yep there's a negetive sign
    Last edited by ^_^Engineer_Adam^_^; July 15th 2007 at 07:16 AM.
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  4. #4
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    Quote Originally Posted by driver327 View Post
    2. 1+sinx/2-cosx
    (2-cosx)(1+sinx)'-(1+sinx)(2-cosx)'
    ---------------------------------------
    (2-cosx)^2

    (2-cosx)(cosx)-(1+sinx)(sinx)
    ---------------------------------------
    (2-cosx)^2

    2cosx-cosx^2-sinx-sinx^2
    ---------------------------------------
    (2-cosx)^2

    the answer comes out to be (2cosx -sinx -1)/(2-cosx)^2.
    you are correct you just need to use some identities:
    \frac{2\cos{x}-\sin{x}-(\cos^2{x}+\sin^2{x})}{(2-\cos{x})^2}

    where \cos^2{x}+\sin^2{x} is 1
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  5. #5
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    Hello, driver327!

    3)\;\;y \:=\:\frac{\sin x + \cos x}{\sin x - \cos x}

    \frac{dy}{dx} \;=\;\frac{(\sin x - \cos x)(\cos x-\sin x) -(\sin x + \cos x)(\cos x +\sin x)}{(\sin x-\cos x)^2} . . . . Keep going

    Multiply: . \frac{(\sin x\cos x - \sin^2x - \cos^2x+ \sin x \cos x) - (\sin x\cos x + \sin^2x + \cos^2x + \sin x\cos x)}{(\sin x-\cos x)^2}

    . . = \;\frac{\left[2\sin x\cos x - \overbrace{(\sin^2x + \cos^2x)}^{\text{This is 1}}\right]    - \left[2\sin x\cos x + \overbrace{(\sin^2x + \cos^2x)}^{\text{This is 1}}\right]} {(\sin x -\cos x)^2}

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

    . . = \;\frac{-2}{(\sin x -\cos x)^2}

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