Partial derivatives of $\ln(x^2+y^2)$

Question

I am new to partial derivatives and they seem pretty easy, but I am having trouble with this one:

$$\frac{\partial}{\partial x} \ln(x^2+y^2)$$ now if this was just $\frac{d}{dx}\ln(x^2)$ we would get $\frac{2x}{x^2}$. So I feel we would get:$$\frac{\partial}{\partial x} \ln(x^2+y^2)=\frac{2x}{x^2+y^2}$$

and with respect to $y$ $$\frac{\partial}{\partial y} \ln(x^2+y^2)=\frac{2y}{x^2+y^2}.$$ Is that right?

Answer 1

It is correct. To justify your feeling, you can apply the chain rule to the maps $g\colon x\mapsto x^2+y^2$ (where $y$ is fixed) and $f\colon x\mapsto \ln x$.