I am reading Marsden and Tromba's Vector Calculus book, it says that the derivative for $f \circ g$ is $D(f(g)) D(g(x)$, where $D$ is the matrix of partial derivatives.
They have an example in which they try to calculate given $f(x,y)$, and $x = r \cos \theta$, $y = r \sin \theta$, a formula for $\partial f / \partial \theta$.
I tried obtaining the result they have through matrix multiplication using a function composition: where $f = f(x,y)$ and $g = g(r \cos \theta, r \sin \theta)$. Then I tried to compute the matrix derivatives of $f \circ g$, where $Df$ is a 2x2 identity matrix, and Dg is the 2x2 matrix of partials of the components of $g$ and $r$ and $\theta$. I get a transformation matrix: $\cos \theta, -r\sin \theta, \sin \theta, r \cos \theta$ (reading from (1,1) to (2,2)), but I don't know how to get $\partial f / \partial r$ and $\partial f / \partial \theta$ from this.
What you want is for $g(r,\theta) = \begin{bmatrix}r\cos\theta & r\sin\theta\end{bmatrix}^\intercal$
So the relevant composition is: $(f\circ g) (r,\theta) = f(r\cos\theta, r\sin\theta)$.
Then the Jacobian chain rule gives you
$$\begin{align}\mathrm D(f\circ g)(r,\theta) &= \mathrm D f\circ g(r,\theta)~\mathrm Dg(r,\theta) \\[1ex]&= \begin{bmatrix}\tfrac{\partial f}{\partial x}\circ g(r,\theta)&\tfrac{\partial f}{\partial y}\circ g(r,\theta)\end{bmatrix}\begin{bmatrix}\tfrac{\partial g_1}{\partial r}(r,\theta) & \tfrac{\partial g_1}{\partial\theta}(r,\theta)\\ \tfrac{\partial g_2}{\partial r}(r,\theta) & \tfrac{\partial g_2}{\partial\theta}(r,\theta)\end{bmatrix}\\[2ex] &=\begin{bmatrix}\tfrac{\partial f}{\partial x}(r\cos\theta, r\sin\theta) & \tfrac{\partial f}{\partial y}(r\cos\theta, r\sin\theta)\end{bmatrix}\begin{bmatrix}\cos\theta &-r\sin\theta\\\sin\theta & r\cos\theta\end{bmatrix}\end{align}$$
Hence $\tfrac{\partial (f\circ g)}{\partial r}(r,\theta) = \cos(\theta) \,\tfrac{\partial f}{\partial x}(r\cos\theta,r\sin\theta) + \sin(\theta)\,\tfrac{\partial f}{\partial y}(r\cos\theta,r\sin\theta)$
And $\tfrac{\partial (f\circ g)}{\partial\theta}(r,\theta) = {-}\sin(\theta)\,\tfrac{\partial f}{\partial x}(r\cos\theta,r\sin\theta)+r\cos(\theta)\,\tfrac{\partial f}{\partial y}(r\cos\theta,r\sin\theta)$
$~$
Often this is abbreviated to $\tfrac{\partial (f\circ g)}{\partial r} = \cos(\theta)\,\tfrac{\partial f}{\partial x}+\sin(\theta)\,\tfrac{\partial f}{\partial y}~$
and $~\tfrac{\partial(f\circ g)}{\partial\theta} = -r\sin(\theta)\,\tfrac{\partial f}{\partial x}+r\cos(\theta)\,\tfrac{\partial f}{\partial y}$