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use implicit differentiation to find dy/dx: xe^y-ye^x=1
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yay!!! implicits
Dx(x e^y) = Dx(y e^x) = Dx(1)
I LOVE implicits!
x e^y y' + e^y - (y' e^x + e^x) = 0
lol .... me too :)
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I am your fan
aww shucks....ill fan you just so you can be cool like me :)
x e^y y' + e^y - y' e^x - e^x = 0 if it aint got a y'...move it to the otherside
y' (x e^y) - y'(e^x) = e^x - e^y
factor out you y' and divide the rest...
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y' [(x e^y) - (e^x)] = e^x - e^y y' = e^x - e^y ---------- x e^y - e^x
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