A question from Lee's Introduction to Smooth Manifolds asks to convert the vector field $X = 2x \frac{\partial}{\partial x}$ from standard coordinates to polar coordinates (on some open subset on which they are defined) and my solution is at odds with that of a provided solution. So I am really looking for some proof verification/explanation of what I may have done incorrectly.
So in polar coordinates $(r , \theta)$, we know that there exists unique $v_{r}, v_{\theta} \in \mathbb{R}$ such that $$\frac{\partial}{\partial x} = v_{r} \frac{\partial}{\partial r} + v_{\theta} \frac{\partial}{\partial \theta}$$ and $v_{r},v_{\theta}$ can be found the action of $\partial/\partial x$ on the component functions $r$ and $\theta$: \begin{eqnarray*} v_{r} & = & \frac{\partial}{\partial x} \left( r \right) = \frac{\partial}{\partial x} \left( \sqrt{x^2 + y^2} \right) = \frac{x}{\sqrt{x^2 + y^2}} = \frac{ r \cos \theta}{r} = \cos \theta \end{eqnarray*} \begin{eqnarray*} v_{\theta} & = & \frac{\partial}{\partial x}\left( \theta \right) = \frac{\partial}{\partial x}\left( \tan^{-1}\left( \frac{y}{x} \right) \right) = \frac{-yx^{-2}}{1 + (y/x)^2} = \frac{-y}{x^2 + y^2} = \frac{-r\sin\theta}{r^2} = - \frac{\sin \theta}{r}. \end{eqnarray*} It then follows that $$ X = 2 r \cos \theta \left( \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r}\frac{\partial}{\partial \theta} \right).$$
Another approach yields the same answer: The Jacobian of the transformation $(r , \theta) \mapsto (r \cos \theta , r \sin \theta)$ is given by the matrix
$$
J =
\begin{pmatrix}
\cos \theta & - r \sin \theta \newline
\sin \theta & r \cos \theta
\end{pmatrix},
$$
which implies that the Jacobian of the transformation $(x , y) \mapsto (\sqrt{x^2 + y^2}, \tan^{-1}(y/x))$ is
$$
J^{-1} =
\begin{pmatrix}
\cos \theta & \sin \theta \newline
-\frac{\sin \theta}{r} & \frac{\cos \theta}{r}
\end{pmatrix}.
$$
Then $\partial/\partial x \equiv (1 , 0)^t$ and so we have
$$
J^{-1}\begin{pmatrix} 1 \newline 0 \end{pmatrix} =
\begin{pmatrix}
\cos \theta & \sin \theta \newline
-\frac{\sin \theta}{r} & \frac{\cos \theta}{r}
\end{pmatrix}\begin{pmatrix} 1 \newline 0 \end{pmatrix} = \begin{pmatrix} \cos \theta \newline - \frac{\sin \theta}{r} \end{pmatrix} \Longrightarrow \frac{\partial}{\partial x} = \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta}.
$$
Hence, using this second method, we still get
$$
X = 2 r \cos \theta \left(\cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r}\frac{\partial}{\partial \theta} \right).
$$
So my main questions are:
Are either of these solutions correct?
If it is the case that neither are correct, then what is the correct answer?