Computing slopes of tangent lines

To compute slopes of tangent lines to a polar curve $r=f(\theta)$, we treat it as a parametrized curve with $\theta=t$ and $r=f(t)$. (Equivalently, we can use $\theta$ as our parameter). This means that $$x= r \cos(\theta) = f(t) \cos(t); \qquad y = r \sin(\theta) = f(t) \sin(t).$$ Taking derivatives we get $$dx/dt = -f(t) \sin(t) + f'(t) \cos(t); \qquad dy/dt = f(t)\cos(t) + f'(t) \sin(t).$$ That's enough information for us to compute $dy/dx$: $$\begin{eqnarray*}\frac{dy}{dx} &=& \frac{dy/dt}{dx/dt} \cr &=& \frac{f(t)\cos(t) + f'(t) \sin(t)}{-f(t) \sin(t) + f'(t) \cos(t)} \cr &=& \frac{r \cos(\theta) + r' \sin(\theta)}{-r\sin(\theta) + r' \cos(\theta)}, \end{eqnarray*}$$ where $r'$ means $dr/d\theta$.