Gradient in polar coordinates I was reading the answer from here, which is showing 
$$\nabla = \partial_r e_r + \frac{1}{r} \partial_\theta e_\theta.$$
The answer actually made the choice 
$$e_r = (\cos\theta, \sin\theta), \quad e_\theta = (-\sin\theta, \cos\theta)$$
I would like to clarify this step more naturally.
More generally, let $g(x,y) = (u,v)$ be the change of variable map. Given a real valued function $f(x,y)$, we know in the new variable $\tilde f (u,v)= f(g^{-1}(u,v))$.
Now given a vector field 
$$v(x,y) = v_1\frac{\partial}{\partial x} +  v_2\frac{\partial}{\partial y}$$ 
we would like to calculate $\tilde v(u,v)$ in the new variable, I think this is given by $dg(v)$ following the change of variable formula
$$dg\left(\frac{\partial}{\partial x}\right) = \frac{\partial u}{\partial x} \frac{\partial}{\partial u}+\frac{\partial v}{\partial x} \frac{\partial}{\partial v}$$
$$dg\left(\frac{\partial}{\partial y}\right) = \frac{\partial u}{\partial y} \frac{\partial}{\partial u}+\frac{\partial v}{\partial y} \frac{\partial}{\partial v}$$
Let us denote the coordinates of $v$ as $[v] = (v_1, v_2)$, we get 
$$[\tilde v(u,v)] = (Jg)(g^{-1}(u,v))[v(g^{-1}(u,v))].$$
I tried the following calculation.
Given $f(x,y)$, $h(r,\theta) = (x,y)$ the change of variable map, and $\tilde f = f \circ h$.
Given the vector field: gradient of $\tilde f$
$$\tilde v(r, \theta) = \frac{\partial \tilde f}{\partial r}\frac{\partial  }{\partial r}+ \frac{1}{r^2}\frac{\partial \tilde f}{\partial \theta}\frac{\partial }{\partial \theta}$$
Then we see that 
$$dh(\tilde v(r, \theta)) = \frac{\partial \tilde f}{\partial r}dh\left(\frac{\partial  }{\partial r}\right)+ \frac{1}{r^2}\frac{\partial \tilde f}{\partial \theta}dh\left(\frac{\partial }{\partial \theta}\right)\\
=  \frac{\partial \tilde f}{\partial r}\left(\cos\theta \frac{\partial  }{\partial x}+ \sin\theta \frac{\partial  }{\partial y}\right)+ \frac{1}{r^2}\frac{\partial \tilde f}{\partial \theta}\left(-{r}\sin\theta \frac{\partial  }{\partial x}+ {r}\cos\theta \frac{\partial  }{\partial y}\right)\\
=  \left(\cos\theta\frac{\partial \tilde f}{\partial r} -\frac{1}{r}\sin\theta \frac{\partial \tilde f}{\partial \theta}\right) \frac{\partial  }{\partial x}+  \left(\sin\theta\frac{\partial \tilde f}{\partial r} +\frac{1}{r}\cos\theta \frac{\partial \tilde f}{\partial \theta}\right) \frac{\partial  }{\partial y}\\
=  \frac{\partial  f }{\partial x}\frac{\partial  }{\partial x}+  \frac{\partial  f }{\partial y}\frac{\partial  }{\partial y}.
$$
Now $e_r = dh\left(\frac{\partial  }{\partial r}\right)$ and $e_\theta = \frac{1}{r} dh\left(\frac{\partial  }{\partial \theta}\right)$
 A: To clarify the confusion, observe that in Cartesian coordinates, we have that
\begin{align}
[\nabla f]_{\text{Cartesian}} =  \frac{\partial f}{\partial x}\mathbf{e}_1+ \frac{\partial f}{\partial y}\mathbf{e}_2
\end{align}
where $\mathbf{e}_i$ are the Cartesian basis vectors. However, we also know that
\begin{align}
\frac{\partial f}{\partial x} =& \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}+\frac{\partial f}{\partial r}\frac{\partial r}{\partial x} = -\frac{\partial f}{\partial \theta}\frac{y}{x^2+y^2}+\frac{\partial f}{\partial r}\frac{x}{\sqrt{x^2+y^2}}\\
\frac{\partial f}{\partial y} =& \frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial y}+\frac{\partial f}{\partial r}\frac{\partial r}{\partial y} = \frac{\partial f}{\partial \theta}\frac{x}{x^2+y^2}+\frac{\partial f}{\partial r}\frac{y}{\sqrt{x^2+y^2}}
\end{align}
which means
\begin{align}
[\nabla f]_{\text{Cartesian}} =&\ \left( -\frac{\partial f}{\partial \theta}\frac{y}{x^2+y^2}+\frac{\partial f}{\partial r}\frac{x}{\sqrt{x^2+y^2}}\right)\mathbf{e}_1+\left( \frac{\partial f}{\partial \theta}\frac{x}{x^2+y^2}+\frac{\partial f}{\partial r}\frac{y}{\sqrt{x^2+y^2}}\right)\mathbf{e}_2\\
=&\ \frac{\partial f}{\partial r}\left( \frac{x}{\sqrt{x^2+y^2}}\mathbf{e}_1+\frac{y}{\sqrt{x^2+y^2}}\mathbf{e}_2\right)+\frac{1}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial \theta}\left( -\frac{y}{\sqrt{x^2+y^2}}\mathbf{e}_1+\frac{x}{\sqrt{x^2+y^2}}\mathbf{e}_2\right)
\end{align}
where $\partial f/\partial r$ and $\partial f/\partial \theta$ are expressed in terms of $(x, y)$. 
However, if you choose to use polar coordinates in space, i.e., replace $x = r\cos\theta$ and $y = r\sin\theta$ then you will see that
\begin{align}
[\nabla f]_{\text{Cartesian}} = \frac{\partial f}{\partial r}\cdot (\cos\theta \ \mathbf{e}_1+ \sin\theta\ \mathbf{e}_2) + \frac{1}{r}\frac{\partial f}{\partial \theta}\cdot (-\sin\theta\ \mathbf{e}_1+\cos\theta \mathbf{e}_2).
\end{align}
Despite the fact that you have changed to polar coordinates, you are still expressing your tangent vectors in terms of the cartesian basis.
