A question about notation. $\frac{\partial g}{\partial\theta}(r,\theta)\neq\frac{\partial f}{\partial\theta}(r\cos\theta,r\sin\theta)$? Let $f(r\cos\theta,r\sin\theta)=:g(r,\theta)$.
We often write $\frac{\partial f}{\partial\theta}$ by which we mean $\frac{\partial g}{\partial\theta}$.
I am confused about this notation.
Is the following right?
$g(r,\theta)=f(r\cos\theta,r\sin\theta).$
$\frac{\partial g}{\partial\theta}(r,\theta)\neq\frac{\partial f}{\partial\theta}(r\cos\theta,r\sin\theta)$.
$\frac{\partial g}{\partial\theta}(r,\theta)=\frac{\partial f}{\partial\theta}(r,\theta)$.
 A: What you have written is correct.
Think of $f(r\cos \theta ,r\sin \theta)$ as a compound function, $f(x,y)$ where $x(r,\theta) = r\cos\theta$ and $y(r,\theta)=r\sin\theta$. When you take the derivative with respect to $\theta$, you need to use the chain rule. This gives $\frac{\partial f}{\partial \theta} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta}$, which is indeed the same as $\frac{\partial g}{\partial \theta}$.
If we add into our notation the inputs, then we have
\begin{align}
\frac{\partial f}{\partial \theta}(r,\theta) &= \frac{\partial f}{\partial x}(x,y)\frac{\partial x}{\partial \theta}(r,\theta) + \frac{\partial f}{\partial y}(x,y)\frac{\partial y}{\partial \theta}(r,\theta) \\
  &= \frac{\partial f}{\partial x}(r\cos\theta,r\sin\theta)\frac{\partial x}{\partial \theta}(r,\theta) + \frac{\partial f}{\partial y}(r\cos\theta,r\sin\theta)\frac{\partial y}{\partial \theta}(r,\theta)
\end{align}
If we want to be absolutely precise, in the first line above we are treating $\frac{\partial f}{\partial \theta}$ as a function of $x$, $y$, $r$, and $\theta$, but since $x$ and $y$ are themselves functions of $r$ and $\theta$, we can make the obvious substitution and then treat $\frac{\partial f}{\partial \theta}$ as a function of just $r$ and $\theta$.
Note the inputs on the right hand side are all in terms of $r$ and $\theta$, but are not all of the form $(r,\theta)$. If we were to take $\frac{\partial f}{\partial \theta}(r\cos\theta,r\sin\theta)$, we would have to replace each $r$ and $\theta$ on the RHS with $r\cos\theta$, and $r\sin\theta$, giving
$$
\frac{\partial f}{\partial x}\big((r\cos\theta)\cos(r\sin\theta),(r\cos\theta)\sin(r\sin\theta)\big)\frac{\partial x}{\partial \theta}(r\cos\theta, r\sin\theta) +\\
\qquad \frac{\partial f}{\partial y}\big((r\cos\theta)\cos(r\sin\theta),(r\cos\theta)\sin(r\sin\theta)\big)\frac{\partial y}{\partial \theta}(r\cos\theta, r\sin\theta)
$$
Basically, think of $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ as functions of $x$ and $y$, but think of $\frac{\partial f}{\partial \theta}$ as a function of $r$ and $\theta$.
