Implicit use of the Implicit Function Theorem when finding tangent lines to polar curves. Recently I found myself having to teach students how to find the slope of a tangent line to a curve in $\mathbb R^2$ given in polar coordinates by the equation $r = f(\theta)$. The students' calculus book instructs them that surely the slope of the tangent line must be given by $\frac{dy}{dx}$ and uses the chain rule to calculate $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$ Then, using the fact that $x = r\cos\theta$ and $y = r\sin\theta$ they are therefore able to find a formula for $\frac{dy}{dx}$ in terms of $r$ and $\theta$. There is however a problem with this line of reasoning, namely that there is no compelling reason for $y$ to be defined even locally as a function of $x$. Knowledge of the implicit function theorem allows one to formulate the hypothesis necessary to make the above use of the chain rule correct. However it would be nice to justify the use of the chain rule here using only methods available to a first year calculus student. My question is therefore
Is there a nice way (intuitive or rigorous) of explaining when and why $y$ is a differentiable function of $x$ in this case, using only methods available to a first year calculus student?
Yes, it is possible to look through the proof of the implicit function theorem and simplify it in this case, but I am especially interested in hearing the thoughts of experienced teachers, so I hope I am justified in asking the question.
 A: I can think of two ways to do it. Whether they are adequate for first year calculus students depends on the syllabus.
The first is to solve for $\theta$ as a function of $x$. If you write $x-f(\theta)\cos\theta=0$, you use the implicit function theorem to do it. But if you write it $x=f(\theta)\cos\theta:=g(\theta)$, it is just the inverse function theorem for real functions of one real variable, which is included in he syllabus of many first year calculus courses. If it is not included, you can justify it as follows: if $g'(\theta_0)\ne0$, then $g$ is strictly monotone on a neighborhood around $\theta_0$, there is an inverse, the inverse is also strictly monotone and is differentiable.
The second is writing the curve in parametric form
$$\begin{align*}
x&=f(\theta)\cos\theta,\\
y&=f(\theta)\sin\theta.
\end{align*}$$
Then $(x',y')$ is a vector in the direction of the tangent (if it is not $(0,0)$). From it you can compute the slope of the tangent, and find when that tangent is vertical.
