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Math Help - Implicit differentiation help

  1. #1
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    Implicit differentiation help

    I need help with yet another implicit differentiation
    e^x/y=x-y
    I've tried

    ((1/y)(e^(x/y))((y-xy')/(y^2)))=1-y'

    (e^(x/y))-1=(-y'y^2)/(y-xy')
    What am I doing wrong? I don't see a way
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  2. #2
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    Quote Originally Posted by ppark View Post
    I need help with yet another implicit differentiation
    e^x/y=x-y
    I've tried

    ((1/y)(e^(x/y))((y-xy')/(y^2)))=1-y'

    (e^(x/y))-1=(-y'y^2)/(y-xy')
    What am I doing wrong? I don't see a way
    I presume this is
    \displaystyle e^{x/y} = x + y

    \displaystyle e^{x/y} \cdot \left ( \frac{1}{y} - \frac{x}{y^2} \cdot y' \right )  = 1 + y'

    \displaystyle e^{x/y} \cdot \frac{1}{y} - e^{x/y} \cdot \frac{x}{y^2} \cdot y' = 1 + y'

    Can you take it from there?

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    I presume this is
    \displaystyle e^{x/y} = x + y

    \displaystyle e^{x/y} \cdot \left ( \frac{1}{y} - \frac{x}{y^2} \cdot y' \right )  = 1 + y'

    \displaystyle e^{x/y} \cdot \frac{1}{y} - e^{x/y} \cdot \frac{x}{y^2} \cdot y' = 1 + y'

    Can you take it from there?

    -Dan
    Does the 1/y on e^(x/y) not need to come down when taking the derivative of e

    EX: d/dx e^(x/y)= 1/y e^(x/y)
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ppark View Post
    Does the 1/y on e^(x/y) not need to come down when taking the derivative of e

    EX: d/dx e^(x/y)= 1/y e^(x/y)
    It's there. Try it using the chain rule and put u = x/y. Then
    \displaystyle \frac{d}{dx} e^{x/y} = \frac{d}{dx}e^u = e^u \cdot \frac{d}{dx}u

    \displaystyle \frac{d}{dx}u = \frac{d}{dx}\left ( \frac{x}{y} \right ) = \frac{1}{y} - \frac{x}{y^2}y' <--- This is correct version of the term you wanted to "bring down."

    So we get
    \displaystyle \frac{d}{dx}e^{x/y} = e^{x/y} \left ( \frac{1}{y} - \frac{x}{y^2}y' \right )
    as before.

    -Dan
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