Doubt about differentia operatorl in polar coordinates

Question

The differential operator $d$ can be written as $$d=\partial_i dx^i$$ where $\partial_i\equiv\frac{\partial}{\partial x_i}$. If I have $d$ expressed in cartesian coordinate and I want to obtain the operator in terms of a new set of coordinate, polar for example ($x=r\cos\theta$ $y=r\sin\theta$), I was thinking to rewrite $$d=\partial_x dx+\partial_y dy$$ using the fact that $$\partial_x=\partial_xr\partial_r+\partial_x\theta\partial_\theta,\quad \partial_y=\partial_yr\partial_r+\partial_y\theta\partial_\theta$$ and that $$ dx=\partial_rxdr+\partial_\theta xd\theta,\quad dy=\partial_rydr+\partial_\theta yd\theta. $$ The substitution gives \begin{align} d&=A\partial_r dr+B\partial_\theta d\theta+C\partial_rd\theta+D\partial_\theta dr\\ &=(\partial_xr\partial_rx+\partial_yr\partial_ry)\partial_rdr+(\partial_x\theta\partial_\theta x+\partial_y\theta\partial\theta y)\partial_\theta d\theta+(\partial_xr\partial_\theta x+\partial_yr\partial_\theta y)\partial_rd\theta+(\partial_x\theta\partial_rx+ \partial_y\theta\partial_ry)\partial_\theta dr \end{align} and while I found correctly ($A=1$, $C=D=0$), my $B$ is different from the correct one $B=1\neq\frac{1}{r}$. I am using the following Jacobian \begin{bmatrix}\partial_rx=\cos\theta&\partial_\theta x=-r\sin\theta\\\partial_r y=\sin\theta&\partial_\theta y=r\cos\theta\end{bmatrix} and its inverse \begin{bmatrix}\partial_xr=\cos\theta&\partial_yr=\sin\theta\\\partial_x \theta=\frac{-\sin\theta}{r}&\partial_y\theta=\frac{\cos\theta}{r}\end{bmatrix}Where is my flaw?