# Partial derivatives of function log(x^2+y^2)

1. Jun 20, 2012

### Chromosom

1. The problem statement, all variables and given/known data
I have got a question concerning the following function:

$$f(x,y)=\log\left(x^2+y^2\right)$$​

Partial derivatives are:

$$\frac{\partial^2f}{\partial x^2}=\frac{y^2-x^2}{\left(x^2+y^2\right)^2}$$​

and

$$\frac{\partial^2f}{\partial y^2}=\frac{x^2-y^2}{\left(x^2+y^2\right)^2}$$​

The conclusion is that the following equation is right:

$$\frac{\partial^2f}{\partial x^2}=-\frac{\partial^2f}{\partial y^2}$$​

But I can not understand, how can it be possible. The role of x and y variables are exactly the same, then why derivatives are not the same?

Sorry for my English - it is my second language. I am from Poland.

2. Jun 20, 2012

### Curious3141

There's a factor of 2 missing in all your second derivatives.

The result is exactly as you'd expect. The variable you're differentiating with respect to, matters. If it's x, then y is treated as a constant, and vice versa. So if the "active" variable is leading in the numerator in one derivative, the same should apply in the other. It's just that the "active" variable is x in one case and y in the other, and the other variable acts like a constant.

3. Jun 21, 2012

### dimension10

How did you get those partial derivatives? They are wrong.

P.S. There's nothing wrong with your English, and even if there were, there is nothing to apologise for.