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Math Help - Trig Derivative - # 2

  1. #1
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    Trig Derivative - # 2

    y = \dfrac{8x}{1 - \cot x}

    y' = \dfrac{[1 - \cot x][8]-[8x][-\csc^{2} x]}{(1 - \cot x)^{2}}

    y' = \dfrac{[8 - 8 \cot x]-[-8 \csc^{3} x]}{(1 - \cot x)^{2}}

    y' = \dfrac{[8(1 - \cot x)]-[-8 \csc^{3} x]}{(1 - \cot x)^{2}}
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  2. #2
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    Re: Trig Derivative - # 2

    Not quite. You made a mistake on line 2, then another one on line 3. Recalculate the derivative of 1-\cot x. Then (8x)(\csc^2 x) \neq 8\csc^3 x
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    Re: Trig Derivative - # 2

    Ok, let's look at his again:

    y = \dfrac{8x}{1 - \cot x}

    \dfrac{[1 - \cot x][8] - [8x][1 - (- \csc^{2} x)]}{(1 - \cot x)^{2}}

    \dfrac{[1 - \cot x][8] - [8x][1 + \csc^{2} x]}{(1 - \cot x)^{2}}

    \dfrac{8 - 8\cot x - 8x - \csc^{2} x}{(1 - \cot x)^{2}} - This answer is still wrong on computer homework.
    Last edited by Jason76; October 14th 2013 at 02:17 PM.
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  4. #4
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    Re: Trig Derivative - # 2

    What is the derivative of 1-\cot x? It is not 1-(-\csc^2 x). Even if it were, (8x)(1+\csc^2 x) \neq 8x + \csc^2x.
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    Re: Trig Derivative - # 2

    I figured out the right answer. The derivative of 1 - \cot x = -(-csc x) or \csc x So you just put that in there, and you get the right answer.
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