Find the partial deriavtive with respect to x and y

Question

$$f(x,y)=\ln \frac{x-y}{(x+y)^2}$$

Use log properties

I started with this

$$\ln(x-y)-2\ln(x+y)$$

I got this for $x$:

$$\frac{1}{x-y}-\frac{2}{x-y}$$

I got this for $y$:

$$-\frac{1}{x-y}-\frac{2}{x+y}$$

would this be the final answers or am I missing something with my work?

Answer 1

No, this is not true.

$$\frac{\partial}{\partial x} f(x,y) = \frac{1}{x-y}-\frac{2}{x+y}$$

$$\frac{\partial}{\partial y} f(x,y) = -\frac{1}{x-y}-\frac{2}{x+y}$$

Small mistake in the derivative with respect to $x$. The derivative with respect to $y$ is okay.

To see this, note that the partial derivative with respect to $x$ of $\ln(x-y)$ is $\frac{1}{x-y}$ by the chain rule, and the partial derivative with respect to $y$ of $\ln(x-y)$ is $-\frac{1}{x-y}$ by the chain rule. For the other two the chain rule can be used as well.