Derivative Problem

[Warr]

e^{xy} = ln (x+y)

I need to find dy/dx...but its difficult to get the answer in the book

I tried this:

ln e^{xy} = ln (ln (x + y))

xy = ln (ln (x+y))

taking the derivitive in terms of x

y + \frac{dy}{dx}x = (\frac {1}{ln (x+y)})(\frac {1}{x+y})(1 + \frac {dy}{dx})

If I were to continue and solve for dy/dx It would not even be close to the books answer. I'm probably doing it wrong anyways...can someone show me the right way?

[AKG]

Try implicitly differentiating right off the bat. I find sometimes with implicit differentiation questions, your answers and the books answers can often seem pretty different (when in fact being the same).

e^{xy} = ln(x + y)
e^{xy}(y + xy') = \frac{1 + y'}{x + y}
y' = \frac{(x + y)(y)(e^{xy}) - 1}{1 - xe^{xy}}

If you can find some way to further simplify, go ahead.

[Warr]

Hey, thanks :)