Find dy/dx of ln(xy) = e^(x+y)

I'm having trouble solving for y' after taking the derivative, could someone please lend me a hand?

Question

Find dy/dx of ln(xy) = e^(x+y)

I'm having trouble solving for y' after taking the derivative, could someone please lend me a hand?

luigi0210 · Answer

What did you get after you took the derivative?

anonymous · Answer

y'/(xy) = e^(x+y)*(1+y').. I've been out of the math game for a while so I'm struggling to set one side = to y'

luigi0210 · Answer

Wouldn't xy be product rule?

anonymous · Answer

Uhh I used the chain rule am I wrong on that part?

luigi0210 · Answer

Actually, I think it would be easier to just make it $\Large ln(xy)=lnx+lny$

anonymous · Answer

What rule allows that? Sorry if it's obvious hahah

luigi0210 · Answer

luigi0210 · Answer

The Product one ^

anonymous · Answer

Okay thanks! Makes sense :) Let me try it from here

anonymous · Answer

Nope hahaha I couldn't figure out how to set y' on one side... It's been a while since I've taken a math class so I'm bad

anonymous · Answer

I think using the log product rule I can get it though this seems easier