Implicit diff. of $(xy)^x = e$ giving two different answers Initial question: take $\dfrac{dy}{dx}$ of $(xy)^x = e$.
Take $\ln$ of both sides:
$$\ln\left((xy)^x\right) = \ln(e) \implies x\ln(xy) = 1$$
I will call this equation S.
From equation S, you could take $\frac{dy}{dx}$ of both sides and you will get:
$$\frac{dy}{dx} = \frac{-y(1 + \ln(xy))}{x}$$
However, you could also start with equation S and divide both sides by $x$ to get $\ln(xy) = \frac 1x$.  When this equation is differentiated implicity, I get:
$$\frac{dy}{dx} = \frac{-y(1 + (1/x))}{x}$$
Im probably missing something really simple but it is really annoying me that I am getting two different answers.  Help!

There are many forms you can have, but normally if we can reduce to just one variable we do that. Then you can argue if you spot any problem. $x$ and $y$ in your equation are not independent.
So:
$$\frac{dy}{dx} = \frac{-y(1 + \frac{1}{x})}{x}$$
$$\frac{dy}{dx} = \frac{-e^{\frac{1}{x}}(1 + \frac{1}{x})}{x^2}=\frac{-e^{\frac{1}{x}}(x + 1)}{x^3}$$
The other way
$$\frac{dy}{dx} = \frac{-y(1 + \ln(xy))}{x}$$
$$\frac{dy}{dx} = \frac{-e^{\frac{1}{x}}(1 + \ln(x\frac{e^{\frac{1}{x}}}{x}))}{x^2}$$
$$\frac{dy}{dx} = \frac{-e^{\frac{1}{x}}(1 + \frac{1}{x}))}{x^2}=\frac{-e^{\frac{1}{x}}(x + 1)}{x^3}$$
So, they are the same.
