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Math Help - Differentiating function that contains y and x?

  1. #1
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    Differentiating function that contains y and x?

    hey guys can you help me with this problem.

    Given that x^2 + x – y^2 = 0


    a) find dy/dx.

    is it this:
    \frac{2x+1}{2y} = \frac{dy}{dx}

    b) Find dy/dt at x = 5 if dx/dt = 2/3

    what do I do here?

    is it:
    \frac{dy}{dt} =\frac{dy}{dx} *\frac{dx}{dt}
    \frac{dy}{dt} = \frac{2(5)+1}{2y} = \frac{dy}{dx} * \frac{2}{3}?
    \frac{11}{3y} = \frac{dy}{dt}
    Last edited by mr fantastic; May 8th 2011 at 07:38 PM.
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  2. #2
    Super Member TheChaz's Avatar
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    Quote Originally Posted by RezMan View Post
    hey guys can you help me with this problem.

    Given that x^2 + x – y^2 = 0


    a) find dy/dx.

    is it this:
    \frac{2x+1}{2y} = \frac{dy}{dx}

    b) Find dy/dt at x = 5 if dx/dt = 2/3

    what do I do here?

    is it:
    \frac{dy}{dt} =\frac{dy}{dx} *\frac{dx}{dt}
    \frac{dy}{dt} = \frac{2(5)+1}{2y} = \frac{dy}{dx} * \frac{2}{3}?
    \frac{11}{3y} = \frac{dy}{dt}
    It's called "implicit differentiation".
    Basically, you can navigate through this if you always use the chain rule, and remember that dx/dx = 1 (= x', when differentiating w.r.t. x)

    So for
    x^2 + x – y^2 = 0
    (x^2)' + (x)' - (y^2)' = 0
    2x*x' + 1*x' -2y*y' = 0
    2x + 1 - 2y*y' = 0
    2x + 1 = 2y*y'
    (2x + 1)/2y = y'
    Last edited by mr fantastic; May 8th 2011 at 07:39 PM.
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  3. #3
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    Thankyou.

    so what do I do for the second part?
    Last edited by mr fantastic; May 8th 2011 at 07:39 PM.
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  4. #4
    Super Member TheChaz's Avatar
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    It's like the first way, but x' = dx/dt (NOT 1!) and y' = dy/dt
    (First, you'll want to put x = 5 into the original equation and solve for y)

    x^2 + x – y^2 = 0
    (x^2)' + (x)' - (y^2)' = 0
    2x*x' + 1*x' -2y*y' = 0
    Then use the given values of x' = dx/dt = 2/3, x = 5, y = (whatever you found it to be above) and solve for y' = dy/dt

    And I'm sticking with "whatsoever ye do, whether in word or deed..."

    edit...
    Oh, yeah, I guess you could work from your result in part (a). That would make some sense!
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  5. #5
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    does this frankenstien monster look like it:

    2x(\frac{2}{3})+1(\frac{2}{3})-2\sqrt{30}(\frac{2x+1}{2y})

    yes?
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  6. #6
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    so would you say this is correct 11/3y=dy/dt
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  7. #7
    Super Member TheChaz's Avatar
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    Quote Originally Posted by RezMan View Post
    so would you say this is correct 11/3y=dy/dt
    Where y = +/-√30
    If you're going this route, you can forego the steps I suggested in post #4.
    If you needed to find dy/dt from scratch, you would take the steps in post #4!
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  8. #8
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    okay cool. I think I've been able to put this to rest lol
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