# How do you find the derivative of y in the equation ln(xy)=x+y?

You have to remember that in this "implicit" function $y$ is a function of $x$ so deriving you get:
$\frac{1}{x y} \cdot \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 1 + \frac{\mathrm{dy}}{\mathrm{dx}}$ where you used the Chain Rule on the $\ln$
$\frac{1}{x} + \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{\mathrm{dy}}{\mathrm{dx}}$
So you collect $\frac{\mathrm{dy}}{\mathrm{dx}}$ and get:
dy/dx=(1-1/x)/(1/y-1)=((x-1)/x)(y/(1-y))=(y(x-1))/(x(1-y)