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  1. #1
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    implicit different.

    coty = x - y

    i worked it to...
    dy/dx = 1/(1-csc球)
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  2. #2
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    Quote Originally Posted by nuclear123 View Post
    coty = x - y

    i worked it to...
    -csc球*(dy/dx) = 1 - (dy/dx)

    unsure of how to finish this haveing dy/dx on both sides of the equation? if i move them both over to one side i get -csc球*(dy/dx) + (dy/dx) = 1 but still unsure on how to finish it? any help -thx
    From your last step you can factor out a dy/dx

    -\csc ^2(y)\, \frac{dy}{dx} + \frac{dy}{dx} = \frac{dy}{dx}\,(1- \csc ^2(y))

    Edit, if it helps: 1- \csc^2(y) = - \cot^2(y)
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  3. #3
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    s

    i got that but over 1....not sure what i did wrong i added the dy/dx over and it left 1 on the right then divided
    Last edited by nuclear123; February 23rd 2010 at 02:33 PM.
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  4. #4
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    bump

    bump*
    anyone explain why i cant work it the way i did? i dont get how u can factor out a dy/dx..
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  5. #5
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    Quote Originally Posted by nuclear123 View Post
    coty = x - y

    i worked it to...
    dy/dx = 1/(1-csc球)
    Your answer is correct
    You could simplify your denominator

    \frac{1}{1-\csc^2y}=\frac{1}{-\cot^2y}=-\tan^2y

    Quote Originally Posted by nuclear123 View Post
    bump*
    anyone explain why i cant work it the way i did? i dont get how u can factor out a dy/dx..
    ab + b = b(a + 1)

    -\csc ^2y\, \frac{dy}{dx} + \frac{dy}{dx} =1

    \frac{dy}{dx}\,(- \csc ^2y+1)=1

    \frac{dy}{dx}=\frac{1}{- \csc ^2y+1}
    Last edited by ione; February 23rd 2010 at 06:05 PM.
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