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Implicit exponential differentiation?

  1. Jan 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Find an equation of the tangent line to the curve xe^y+ye^x=1 at point (0,1).

    2. Relevant equations

    I do not recall seeing the Implicit Function Theroem before, I even went back in my book (Stewart Calculus 6th) to check. I found this post but it does not help me understand the steps involved: https://www.physicsforums.com/showthread.php?t=79275

    3. The attempt at a solution

    [tex]xe^{y}+ye^{x}=1[/tex]
    [tex]\frac{d}{dx}(xe^{y})+\frac{d}{dx}(ye^{x})=\frac{d}{dx}(1)[/tex]
    I am not sure if this requires the product rule but,
    [tex]\frac{d}{dx}(xe^{y})=e^{y}+xe^{y}[/tex]
    Then, I'm not sure on the other part???
    [tex]\frac{d}{dx}(ye^{x})=y'e^{x}+ye^{x}[/tex]
     
  2. jcsd
  3. Jan 13, 2012 #2

    Dick

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    Don't forget the chain rule.
    [tex]\frac{d}{dx}(e^{y})=e^{y} y'[/tex].
     
  4. Jan 13, 2012 #3

    Bacle2

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    Yes, the point is that if fx(x,y) and/or fy(x,y) are not 0 in a 'hood of (0,1), where y=f(x), or x=f(y) , depending on whether fx or fy is non-zero. The standard example ( I know of) , is that of the sphere.
     
  5. Jan 13, 2012 #4
    why can't i use product rule?
     
  6. Jan 13, 2012 #5

    Dick

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    Both parts require the product rule. So, of course you need it. You will need the chain rule too.
     
  7. Jan 13, 2012 #6
    Yes, I now realize I need to use both rules.
     
  8. Jan 13, 2012 #7
    So,
    [tex]\frac{d}{dx}(ye^{x})=ye^{x}+y'e^{x}[/tex]
    giving me,
    [tex]xe^{y}y'+e^y+ye^{x}+e^{x}y'=0[/tex]
    [tex]y'(xe^{y}+e^{x})=-e^{y}-ye^{x}[/tex]
    [tex]y'=\frac{-e^{y}-ye^{x}}{xe^{y}+e^{x}}[/tex]
    at point (0,1).
    slope=(-e-1)
     
  9. Jan 13, 2012 #8

    Dick

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    Seems ok to me.
     
  10. Jan 13, 2012 #9
    Thank you for your help.
     
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