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Math Help - Clear this up for me.

  1. #1
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    Clear this up for me.

    e^y+6-e^-1 =3x^2 + 3y^2

    on the first term:

    e^y

    That uses chain rule? or does it remain the same?
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  2. #2
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    Quote Originally Posted by Zanderist View Post
    e^y+6-e^-1 =3x^2 + 3y^2

    on the first term:

    e^y

    That uses chain rule? or does it remain the same?
    so what do you want to do?
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  3. #3
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    Quote Originally Posted by dedust View Post
    so what do you want to do?

    Well right now, I'm trying to figure out how it's derivative is.

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  4. #4
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    Smile

    Quote Originally Posted by Zanderist View Post
    Well right now, I'm trying to figure out how it's derivative is.

    if you differentiate it with respect to y then it remain the same
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  5. #5
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    Quote Originally Posted by Zanderist View Post
    e^y+6-e^-1 =3x^2 + 3y^2

    on the first term:

    e^y

    That uses chain rule? or does it remain the same?
    Using partial derivative

    e^y+6-e^{-1} =3x^2 + 3y^2

    e^y+6-e^{-1} -3x^2 - 3y^2 = 0

    F_x = -6x

    F_y = e^y -6y

    so

    \frac{dy}{dx} = -\frac{F_x}{F_y} = \frac{6x}{e^y -6y}
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  6. #6
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    Quote Originally Posted by Zanderist View Post
    e^y+6-e^-1 =3x^2 + 3y^2

    on the first term:

    e^y

    That uses chain rule? or does it remain the same?
    Quote Originally Posted by dedust View Post
    so what do you want to do?
    Quote Originally Posted by Zanderist View Post
    Well right now, I'm trying to figure out how it's derivative is.

    The point was- the derivative with respect to what variable?

    The derivative of e^y with respect to y is e^y.

    The derivative of e^y with respect to x is e^y \frac{dy}{dx}.
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  7. #7
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    Quote Originally Posted by dedust View Post
    if you differentiate it with respect to y then it remain the same
    Right, this also means then.

    -e^-1 Differentiate to.

    -log(e)
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