Why is the derivative of a polar function $dy/dx$ and not $dr/d\theta$? I don't understand. 
If $r = 2\cos(\theta)$ then why is the derivative: 
$dy/dx$? I have a "hypothesis," 
By the polar equation are you really describing a curve in the cartesian plane? So is that why?
 A: $\dfrac {\mathrm dr}{\mathrm d\theta}$ is a perfectly kosher derivative. If you have a set of parametric equations in $x$ and $y$, you can also consider other derivatives.
"Derivative" is ambiguous. There's the derivative of "something" with respect to "something", and often there's more than one useful derivative to consider for a particular problem.
edit: the OP says he or she's looking for the slope of the tangent line. I looked up finding the tangent line of a curve written in polar coordinates in "Calculus" by Larson. 
He says, if you have parametric equations at the point you're interested in, and you want to find the slope of the tangent lines, then use:
$\displaystyle \frac {\mathrm dy}{\mathrm dx}= \frac{\frac {\mathrm dy}{\mathrm d\theta}}{\frac {\mathrm dx}{\mathrm d\theta}}$
He actually says more than that, but you can definitely derive everything he says. The moral of the story is that if the book wants $\mathrm dy/\mathrm dx$, you can find it without having $y$ as an explicit function of $x$. 
A: Your polar equation is equivalent to some cartesian equation. Are you able to find it?
