Expressing Basis Vectors in Polar Coordinates Consider the polar coordinate transformation
$$
x = r\cos \theta 
$$
$$
y = r\sin \theta
$$
I am trying to find the most direct way to compute the coordinate basis vectors
$$
\frac{\partial}{\partial x}, \frac{\partial}{\partial y}
$$
The exterior derivatives of $x$ and $y$ are given by
$$
dx = \cos \theta dr - r\sin\theta d\theta
$$
$$
dy = \sin \theta dr + r \cos \theta d\theta
$$
Considering that $dx$ and $dy$ comprise a basis in the cotangent space dual to the tangent space with basis elements $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$, in principle, one ought to be able to compute the latter by using the duality relations
$$
dx \left(\frac{\partial}{\partial x}\right) = 1, dx \left(\frac{\partial}{\partial y}\right) = 0, dy \left(\frac{\partial}{\partial y}\right) = 1, dy \left(\frac{\partial}{\partial x}\right) = 0
$$
Unfortunately, I am unable to make the algebra for this work out as expected. Is this approach promising, and perhaps I'm simply missing some algebraic manipulations or is there a better approach to this problem? 
 A: Yes, it is promising. Write 
$$
  \frac{\partial}{\partial x} = a_{xr} \frac{\partial}{\partial r} + a_{x \theta} \frac{\partial}{\partial \theta} \qquad 
   \frac{\partial}{\partial y} = a_{yr} \frac{\partial}{\partial r} + a_{y \theta} \frac{\partial}{\partial \theta} 
$$
Now
$$
  \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 
   \begin{pmatrix} \langle \mathrm{d} x, \frac{\partial}{\partial x}  \rangle & \langle \mathrm{d} y, \frac{\partial}{\partial x}  \rangle \\ 
   \langle \mathrm{d} x, \frac{\partial}{\partial y}  \rangle & \langle \mathrm{d} y, \frac{\partial}{\partial y}  \rangle \end{pmatrix} = 
\begin{pmatrix} 
    \cos\theta & \sin \theta \\
    -r \sin\theta & r \cos \theta
\end{pmatrix} \cdot 
\begin{pmatrix} \langle \mathrm{d} r, \frac{\partial}{\partial r}  \rangle & \langle \mathrm{d} \theta, \frac{\partial}{\partial r}  \rangle \\ 
   \langle \mathrm{d} r, \frac{\partial}{\partial \theta}  \rangle & \langle \mathrm{d} \theta, \frac{\partial}{\partial \theta}  \rangle \end{pmatrix} \cdot
  \begin{pmatrix} 
       a_{x r} & a_{x \theta} \\ a_{y r} & a_{y \theta}
\end{pmatrix}
$$
Thus
$$
   \begin{pmatrix} 
       a_{x r} & a_{x \theta} \\ a_{y r} & a_{y \theta}
\end{pmatrix} = \begin{pmatrix} 
    \cos\theta & \sin \theta \\
    -r \sin\theta & r \cos \theta
\end{pmatrix}^{-1} = \begin{pmatrix} 
    \cos\theta & -\frac{1}{r}\sin \theta \\
    \sin\theta & \frac{1}{r}  \cos \theta
\end{pmatrix} 
$$
Thus
$$
   \frac{\partial}{\partial x} = \cos(\theta) \frac{\partial}{\partial r} - \sin(\theta) \frac{1}{r} \frac{\partial}{\partial \theta} \qquad
   \frac{\partial}{\partial y} = \sin(\theta) \frac{\partial}{\partial r} + \cos(\theta) \frac{1}{r} \frac{\partial}{\partial \theta} 
$$
A: One way is to use the chain rule:
$$\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\frac{\partial}{\partial\theta}$$
and similarly for $\frac{\partial}{\partial y}$.
Your method should also work, however; can you show us what went wrong?
A: using $x=r\cos\theta, y=r\sin\theta, r=\sqrt{x^2+y^2}, \theta=\arctan(y/x)$ we have
$$
\frac{\partial r}{\partial x}=\cos\theta, \frac{\partial r}{\partial y}=\sin\theta, \frac{\partial \theta}{\partial x}=-\frac{\sin\theta}{r},\frac{\partial \theta}{\partial y}=\frac{\cos\theta}{r}
$$
then using the chain rule
$$
\frac{\partial}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\frac{\partial}{\partial\theta}=\cos\theta\frac{\partial}{\partial r}-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}
$$
$$
\frac{\partial}{\partial y}=\frac{\partial r}{\partial y}\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial y}\frac{\partial}{\partial\theta}=\sin\theta\frac{\partial}{\partial r}+\frac{\cos\theta}{r}\frac{\partial}{\partial \theta}
$$
