Implicit exponential differentiation?

  • Thread starter jrjack
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  • #1
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Homework Statement



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

Homework 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

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]
 

Answers and Replies

  • #2
Dick
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Don't forget the chain rule.
[tex]\frac{d}{dx}(e^{y})=e^{y} y'[/tex].
 
  • #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.
 
  • #4
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why can't i use product rule?
 
  • #5
Dick
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why can't i use product rule?

Both parts require the product rule. So, of course you need it. You will need the chain rule too.
 
  • #6
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Yes, I now realize I need to use both rules.
 
  • #7
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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)
 
  • #8
Dick
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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)

Seems ok to me.
 
  • #9
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Thank you for your help.
 

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