The problem is:
In cylindrical coordinate system, we have a scalar field: p = (r, $\theta$, z) Prove: $$\nabla p = \frac{\partial p}{\partial r}e_r + \frac{1}{r}\frac{\partial p}{\partial \theta}e_\theta + \frac{\partial p}{\partial z}e_z$$
Here is my work:
r = $\sqrt{x^2+y^2}$ ; $\theta = arctan\frac{y}{x}$ ; z = z
Then:
p = (r, $\theta$, z) = p ($\sqrt{x^2+y^2}, acrtan\frac{y}{x},z)= f (x,y,z)$ (convert to cartesian coordinate system)
$$\frac{\partial f}{\partial x}= \frac{\partial p}{\partial r} \frac{\partial r}{\partial x}+\frac{\partial p}{\partial \theta} \frac{\partial \theta}{\partial x}+\frac{\partial p}{\partial z} \frac{\partial z}{\partial x}= \frac{\partial p}{\partial r} \frac{x}{r}-\frac{\partial p}{\partial \theta} \frac{y}{r^2}$$
$$\frac{\partial f}{\partial y}= \frac{\partial p}{\partial r} \frac{\partial r}{\partial y}+\frac{\partial p}{\partial \theta} \frac{\partial \theta}{\partial y}+\frac{\partial p}{\partial z} \frac{\partial z}{\partial y}= \frac{\partial p}{\partial r} \frac{y}{r}+\frac{\partial p}{\partial \theta} \frac{x}{r^2}$$ $$......$$
We must show:$$ \frac{\partial p}{\partial r} = \sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2}$$ (I convert again to cylindrical coordinate system)
but I got: $$ \sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2} = \sqrt{\left(\frac{\partial p}{\partial r}\right)^2+\left(\frac{\partial p}{\partial \theta}\right)^2\frac{1}{r^2}} \neq \frac{\partial p}{\partial r}$$
Where was I wrong?? Are there any other shorter way to prove this ?