Derivative of polar equations

Consider an arbitrary polar equation defined as: \[ r(\theta), \ \ \ 0 \le \theta < 2 \pi \] The polar curve can be converted into Cartesian coordinates using parametric equations. The relationship between polar equations and their parametric equivalent is neatly shown below:

So we can define the polar curve parametrically using $\theta$ as the parameter. This is: \[\begin{align} y &= r(\theta) \sin \theta \\ x &= r(\theta) \cos \theta \end{align}\]

The derivative of the two parameters can be found using the product rule, this is a trivial step so I have not included the method: \[\begin{align} \frac{\mathrm{d}y}{\mathrm{d} \theta} = r(\theta) \cos \theta + r'(\theta) \sin \theta \\ \\ \frac{\mathrm{d} x}{\mathrm{d} \theta} = r'(\theta) \cos \theta - r(\theta) \sin \theta \end{align} \]

We can apply the chain rule to find the derivative of this set of parametric equations: \[\begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}y}{\mathrm{d} \theta} \times \frac{\mathrm{d} \theta}{\mathrm{d}x} \\ \\ &= \frac{\mathrm{d}y}{\mathrm{d} \theta} \times \frac{1}{\frac{\mathrm{d} x}{\mathrm{d} \theta}} \end{align}\] So substituting in the two derivatives: \[ \frac{\mathrm{d}y}{\mathrm{d}x} = r(\theta) \cos \theta + r'(\theta) \sin \theta \times \frac{1}{r'(\theta) \cos \theta - r(\theta) \sin \theta } \] Hence the derivative of an arbitrary polar equation is: \[\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ r(\theta) \cos \theta + r'(\theta) \sin \theta }{r'(\theta) \cos \theta - r(\theta) \sin \theta} \]