SOLUTION: what is the solution to: implicit differentiation e^x + e^y = e^x+y

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Question 765940: what is the solution to:
implicit differentiation
e^x + e^y = e^x+y
Answer by MaartenRU(13) About Me  (Show Source):
You can put this solution on YOUR website!
I will assume you mean e%5Ex%2Be%5Ey=e%5E%28x%2By%29, and not e%5Ex%2Be%5Ey=e%5Ex%2By, because that would require e%5Ey=y, which has no real solution.

Alright, we now have
e%5Ex%2Be%5Ey=e%5E%28x%2By%29
We will first substitute u=e%5Ey:
e%5Ex%2Bu=u%2Ae%5Ex
Now we want to separate u from the rest of the equation.
e%5Ex%2Bu-u%2Ae%5Ex=0
Factor out u:
e%5Ex%2Bu%281-e%5Ex%29=0
u%28e%5Ex-1%29=e%5Ex
u=%28e%5Ex%29%2F%28e%5Ex-1%29
So putting the substitution back gives us
e%5Ey=%28e%5Ex%29%2F%28e%5Ex-1%29
And so:
y=ln%28%28e%5Ex%29%2F%28e%5Ex-1%29%29

Of course this requires that the denominator is not zero, which translates to x not equal to zero. For any other x, this is the y that satisfies the equation.