Divergence in Polar Coordinates

[neutrino2063]

Why is
$$\nabla\cdot\vec{A}=\frac{1}{r}\frac{\partial}{\partial r}(rA_{r})+\frac{1}{r}\frac{\partial}{\partial \theta}(A_{\theta})$$

Where
$$\vec{A}=A_{r}\hat{r}+A_{\theta}\hat{\theta}$$
And
$$\nabla=\hat{r}\frac{\partial}{\partial r}+\hat{\theta}\frac{1}{r}\frac{\partial}{\partial \theta}$$
Instead of just:

$$\nabla\cdot\vec{A}=\frac{\partial}{\partial r}(A_{r})+\frac{1}{r}\frac{\partial}{\partial \theta}(A_{\theta})$$

 

[nathan12343]

Because the unit vectors are actually functions of position in cylindrical coordinates. This means all the derivative in the gradient operator act not only on the components of a particular vector, but also the unit vectors themselves.