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Math Help - Trig Derivative

  1. #1
    Member VitaX's Avatar
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    Trig Derivative

    Find the derivative

    r = \left(\frac{1+Sin \Theta}{1-Cos \Theta}\right)^2

    I believe the chain rule and quotient rule are used here. So here's my first step, if someone can confirm I did this correct than I could probably go on from there.

    r' = 2 \left(\frac{1+Sin \Theta}{1-Cos \Theta}\right) \left(\frac{(1-Cos \Theta)(Cos \Theta) - (1 + Sin \Theta)(Sin \Theta)}{(1-Cos \Theta)^2}\right)
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  2. #2
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    looks good to me, from there on is pure trig.
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  3. #3
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    r' = 2 \left(\frac{1+Sin \Theta}{1-Cos \Theta}\right) \left(\frac{Cos \Theta - Cos^2 \Theta - Sin \Theta + Sin^2 \Theta}{(1-Cos \Theta)^2}\right)

    Would be final answer right?
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  4. #4
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    If you want to complicate it
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  5. #5
    Member VitaX's Avatar
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    Quote Originally Posted by Arturo_026 View Post


    If you want to complicate it
    Nah I'll stick with what I posted xD
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