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Math Help - implicit diff

  1. #1
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    implicit diff

    sorry i entered it wrong the first time

    e^(x+y)+x-y^2=3
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  2. #2
    Super Member Aryth's Avatar
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    All we do for this is differentiate implicitly with respect to x, treating y as a separate variable:

    We have:

    e^{x + y} + x - y^2 = 3

    Now we differentiate:

    \frac{d}{dx}[e^{x + y}] + 1 - \frac{d}{dx}[y^2] = 0

    \frac{d}{dx}[e^x]\frac{d}{dx}[e^y] - 2y\frac{dy}{dx} = -1

    When we differentiate y, we differentiate it just like we would x, but we add a \frac{dy}{dx} after it:

    e^x(e^y\frac{dy}{dx}) - 2y\frac{dy}{dx} = -1

    \frac{dy}{dx}(e^{x + y} - 2y) = -1

    \frac{dy}{dx} = -\frac{1}{e^{x + y} - 2y}

    And there you go.
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  3. #3
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    Quote Originally Posted by Aryth View Post
    {\color{red}\frac{d}{dx}[e^{x + y}]} + 1 - \frac{d}{dx}[y^2] = 0

    {\color{red}\frac{d}{dx}[e^x]\frac{d}{dx}[e^y]} - 2y\frac{dy}{dx} = -1
    That is not true. The product rule needs to be used if you were to go about it that way: \frac{d}{dx} e^{x+y} = \frac{d}{dx} \left[e^{x}e^{y}\right] = \left(e^{x}\right)'e^y + e^{x}\left(e^y\right)' = \hdots

    However, it is sufficient to simply know that: \frac{d}{dx}e^{u(x)} = e^{u(x)} \cdot u'(x). Here, we see that u = x + y, so:
    e^{x + y} + x - y^2 = 3
    (x+y)'e^{x+y} + 1 - 2yy' = 0
    (1 + y')e^{x+y} - 2yy' = -1

    etc. etc.
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