Implicit exponential differentiation?

[jrjack]

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

xe^{y}+ye^{x}=1
\frac{d}{dx}(xe^{y})+\frac{d}{dx}(ye^{x})=\frac{d}{dx}(1)
I am not sure if this requires the product rule but,
\frac{d}{dx}(xe^{y})=e^{y}+xe^{y}
Then, I'm not sure on the other part?
\frac{d}{dx}(ye^{x})=y'e^{x}+ye^{x}

 

[Dick]

Don't forget the chain rule.
\frac{d}{dx}(e^{y})=e^{y} y'.

 

[Bacle2]

Yes, the point is that if f_x(x,y) and/or f_y(x,y) are not 0 in a 'hood of (0,1), where y=f(x), or x=f(y) , depending on whether f_x or f_y is non-zero. The standard example ( I know of) , is that of the sphere.

 

[timacho]

why can't i use product rule?

 

[Dick]

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.

 

[jrjack]

Yes, I now realize I need to use both rules.

 

[jrjack]

So,
\frac{d}{dx}(ye^{x})=ye^{x}+y'e^{x}
giving me,
xe^{y}y'+e^y+ye^{x}+e^{x}y'=0
y'(xe^{y}+e^{x})=-e^{y}-ye^{x}
y'=\frac{-e^{y}-ye^{x}}{xe^{y}+e^{x}}
at point (0,1).
slope=(-e-1)

 

[Dick]

So,
\frac{d}{dx}(ye^{x})=ye^{x}+y'e^{x}
giving me,
xe^{y}y'+e^y+ye^{x}+e^{x}y'=0
y'(xe^{y}+e^{x})=-e^{y}-ye^{x}
y'=\frac{-e^{y}-ye^{x}}{xe^{y}+e^{x}}
at point (0,1).
slope=(-e-1)

Seems ok to me.

 

[jrjack]

Thank you for your help.