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Math Help - A different implicit differentiation

  1. #1
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    A different implicit differentiation

    Alright, so I'm having some problems. I am supposed to find dy/dx in terms of x and y by implicit differentiation. I wasn't doing too bad until I hit two problems:

    Problem 1) y^2 + ye^x + e^2x=3
    Problem 2) e^y^2 - x^2 - y^2=0

    For the first one, I am not certain how to handle the ones to the power of x and 2x when the first one is y to the power of two. I might be overthinking it though.

    For the second one, I am not certain how to handle the e^y^2, so I am just confused with that one.

    Any help I can get would be much appreciated!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Mike5055 View Post
    Alright, so I'm having some problems. I am supposed to find dy/dx in terms of x and y by implicit differentiation. I wasn't doing too bad until I hit two problems:

    Problem 1) y^2 + ye^x + e^2x=3
    Problem 2) e^y^2 - x^2 - y^2=0

    For the first one, I am not certain how to handle the ones to the power of x and 2x when the first one is y to the power of two. I might be overthinking it though.

    For the second one, I am not certain how to handle the e^y^2, so I am just confused with that one.

    Any help I can get would be much appreciated!
    so how about showing us your solution and we'll check for you..
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  3. #3
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    Alright. I couldn't get to the solution, but for the first one, what I thought is turn it into

    2y + 2e(x') + 2xe^x = 3

    and for the second one, if you derive it to

    (y^2)(e) - 2x - 2y = 0

    I'm not certain if that was the correct step to make with them or not.
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  4. #4
    MHF Contributor kalagota's Avatar
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    ok, here are some hints..

    D_x[y] = \frac{dy}{dx} since y is a variable too.. thus, D_x[y^2] = 2y\cdot D_x[y] = 2y\cdot \frac{dy}{dx} by the power and chain rules..

    now, for ye^x, you may want to use product rule..

    and, e^{2x}, try using chain rule..

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  5. #5
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    Thank you! I'll go through and give that a try!
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  6. #6
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    Verify this please...

    1) y^2 + ye^x + e^(2x) = 3

    2y(dy/dx) + e^x(dy/dx) + ye^x + 2e^(2x) = 0

    2y(dy/dx) + e^x(dy/dx) = -ye^x - 2e^(2x)

    (dy/dx) = [-ye^x - 2e^(2x)] / [ 2y + e^x ]


    2) e^(y^2) - x^2 - y^2 = 0

    2ye^(y^2)(dy/dx) - 2x - 2y(dy/dx) = 0

    (dy/dx)(2ye^(y^2) - 2y) = 2x

    (dy/dx) = x / [ ye^(y^2) - y]


    Edited: Sorry, I didn't see your last post, I agree, give it a try by yourself first.
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