Derivative of $\log(r)$ with respect to $x$

Question

$r$ is a polar radius coordinate, $x$ is a horisontal cartesian coordinate.

Going one way, i get: $\log(r)_x=\log(\sqrt{z\bar{z}})_x$=$\frac{ (\sqrt{z\bar{z}})_x }{ \sqrt{z\bar{z}}}= \frac{0.5((\bar{z})_xz+ \bar{z}z_x) }{ \bar{z}z } = \frac{x}{r^2} $. This is correct.

But going other way, i get: $\log(r)_x=\log(r)_rr_x= \frac{r_x}{r}= \frac{1}{r} $, since $x=re^{i0}=r$ in polar coordinates.

My question is, why is the second way wrong, and are there any easier methods to arrive at $\log(r)_x= \frac{x}{r^2}$, without using complex numbers perhaps?

Answer 1

Your mistake is when you wrote "$x=re^{i0}=r$", this is false. In fact since $r^2=x^2+y^2$, you can take partial derivatives with respect to $x$ on both sides to get $2r\ r_x=2x$, or in other words $r_x=x/r$. Using this corrected formula in place of the incorrect one you wrote ($r_x=1$) in the second method gives the correct answer there as well.

Answer 2

$$\log r= \frac{1}{2}\log(x^2+y^2) \implies \frac{\partial}{\partial x} \log r= \frac{x}{x^2+y^2}=\frac{x}{r^2}. $$