According to the wikipedia definition of Jacobian, if $f:\mathbb{R}^{n}\to \mathbb{R}$, then $Jf=\nabla f^{T}$, where $Jf$ is the jacobian of $f$ and $\nabla f$ is its gradient. Note: I simply took the special case $f:\mathbb{R}^{n}\to \mathbb{R}^{m}, m=1$
However, in polar coordinates, we know the gradient of a function $f(r,\theta)$ is: $$ \begin{pmatrix} \frac{ \partial f }{ \partial r } (r,\theta) \\\\ \frac{1}{r}\frac{ \partial f }{ \partial \theta }(r,\theta) \end{pmatrix} $$ while the Jacobian, which does not depend on the coordinate system, should still be: $$ \begin{pmatrix} \frac{ \partial f }{ \partial r } (r,\theta) , \quad \frac{ \partial f }{ \partial \theta }(r,\theta) \end{pmatrix} $$
So its off by the factor $\frac{1}{r}$. Am I wrong somewhere, if not how can we explain this?
