Jacobian Matrix in 'Polar Coordinates'..? I'm pretty familiar with the del operator and I've just become familiar with the Jacobian Matrix. From my basic understanding of what the Jacobian Matrix is, the gradient is a subset of the Jacobian Matrix. 
So, how would you get a gradient from the Jacobian Matrix? Also, since there is a such thing as "the gradient in polar coordinates", is there a such thing as "the jacobian matrix in polar coordinates" as well?
Clearly I'm going about this all wrong. Any clarification would be much appreciated.
 A: I believe all the basic Del formulas you seek can be found here.
In general, you can use parameters for your Cartesian coordinates and then calculate the determinant of the Jacobian derivative matrix.
In case of polar coordinates: $(x=rcos\theta,y=rsin\theta )$ and then calculate the Jacobian matrix the same way you usually do. 1st column is r, 2nd column is $\theta$. For further reading you can read this.
Regarding the gradient: each row $i$ in the Jacobian Matrix is the gradient of $f_i$, this is the case in general when we speak of a vector-field. 
further explanation:
In polar coordinates - your Jacobian Matrix will be $n\times 2$. The first column being $\frac{\partial f_1}{\partial r}$ until $\frac{\partial f_n}{\partial r}$ whilst the 2nd column is  $\frac{\partial f_1}{\partial \theta}$ until $\frac{\partial f_n}{\partial \theta}$.
In physics, by using this Jacobian Matrix you can calculate "the Jacobian" $r$ [meters] (for polar) which fills the role of filling the "unit" gap when you switch between coordinates in order to use parameters and solve integrals. $\iint dxdy=\iint rdrd\theta$.
Also in spherical coordinates you get $r^2\sin\phi drd\theta d\phi$ for that same reason.
Mathematics behind the grad formula in polar coordinates:
$$
\frac{\partial{f}}{\partial r}=\frac{\partial{f}}{\partial x}\frac{\partial{x}}{\partial r}+\frac{\partial{f}}{\partial y}\frac{\partial{y}}{\partial r}=\frac{\partial{f}}{\partial x}\cos\theta+\frac{\partial{f}}{\partial y}\sin\theta\\
\frac{\partial{f}}{\partial \theta}=\frac{\partial{f}}{\partial x}\frac{\partial{x}}{\partial \theta}+\frac{\partial{f}}{\partial y}\frac{\partial{y}}{\partial \theta}=-\frac{\partial{f}}{\partial x}r\sin\theta+\frac{\partial{f}}{\partial y}r\cos\theta\\
$$
We can use those two equations to get:
$$
\frac{\partial{f}}{\partial r}=\cos\theta\frac{\partial f}{\partial x}+\sin\theta\frac{\partial f}{\partial y}\\
\frac{1}{r}\frac{\partial f}{\partial \theta}=-\sin\theta\frac{\partial f}{\partial x}+\cos\theta\frac{\partial f}{\partial y}
$$
if you write those two linear equations in a matrix you get the definition of gradient in polar coordinates.
$$
\pmatrix{\frac{\partial{f}}{\partial r}\\\frac{1}{r}\frac{\partial f}{\partial \theta}}=\pmatrix{\cos\theta&sin\theta\\-\sin\theta&\cos\theta}\pmatrix{\frac{\partial{f}}{\partial x}\\\frac{\partial{f}}{\partial y}}
$$
Since this is an orthogonal matrix we know that the $A^T=A^{-1}$ so we get:
$$
\pmatrix{\frac{\partial{f}}{\partial x}\\\frac{\partial{f}}{\partial y}}=\pmatrix{\cos\theta&-sin\theta\\\sin\theta&\cos\theta}\pmatrix{\frac{\partial{f}}{\partial r}\\\frac{1}{r}\frac{\partial f}{\partial \theta}}=\pmatrix{\cos\theta\\\sin\theta}\frac{\partial f}{\partial r}+\pmatrix{-\sin\theta\\\cos\theta}\frac{1}{r}\frac{\partial f}{\partial \theta}\\
\nabla = \hat r \frac{\partial f}{\partial r} + \hat \theta \frac{1}{r}\frac{\partial f}{\partial \theta}
$$
