Polar equations can seem tricky, especially when it comes to taking derivatives. However, the derivative of a polar equation can be taken directly in polar coordinates without needing to convert to rectangular form. This article will guide you through the process with helpful examples.

### Which Derivative Are You Computing?

In polar coordinates, you may want to find $\dfrac{dr}{d\theta}$, which represents the rate of change of the radial distance $r$ with respect to the angle $\theta$. This is the most common derivative in polar equations and can be computed directly. However, if you’re looking for $\dfrac{dy}{dx}$, the slope of the curve in rectangular coordinates, you will need to use the relationships between polar and rectangular coordinates, which we’ll discuss later.

### The Process for Finding the Derivative

For many problems, you are interested in $\dfrac{dr}{d\theta}$. This can be computed directly by differentiating the polar equation $r = f(\theta)$. If you want the slope of the curve, $\dfrac{dy}{dx}$, you will need to use the conversion formulas

\[ x = r\cos\theta \quad \text{and} \quad y = r\sin\theta. \]

By the Chain Rule,

\[ \frac{dy}{dx} = \frac{dy}{d\theta}\cdot\frac{d\theta}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}. \]

Applying the Product Rule to the conversion formulas yields

\[ \begin{align*} &\frac{dy}{d\theta} = \frac{d}{d\theta} (r\sin\theta), &\frac{dx}{d\theta} = \frac{d}{d\theta} (r\cos\theta). \end{align*} \]

### Examples

**Example 1:** Find $\dfrac{dr}{d\theta}$ for $r = \theta\sin\theta$.

Answer: Since we only need $\dfrac{dr}{d\theta}$, we differentiate by using the Product Rule:

\[ \dfrac{dr}{d\theta} = \boxed{\sin\theta + \theta\cos\theta}. \]

**Example 2:** Find the slope of the curve $r = 2 + \cos\theta$ at $\theta = \dfrac{\pi}{3}$.

Answer: Note that we wish to find $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d\theta}}{\dfrac{dx}{d\theta}}$. We first, find $\dfrac{dy}{d\theta}$ and $\dfrac{dx}{d\theta}$:

\[ \begin{align*} &\dfrac{dy}{d\theta} = \frac{d}{d\theta}(r\sin\theta)=\frac{d}{d\theta}((2+\cos\theta)\sin\theta) = \sin\theta\cos\theta +\cos\theta(2 + \cos\theta),\\ &\dfrac{dx}{d\theta} = \frac{d}{d\theta}(r\cos\theta)=\frac{d}{d\theta}((2+\cos\theta)\cos\theta) =-\sin\theta(2 + \cos\theta) – \cos\theta\sin\theta. \end{align*} \]

Thus,

\[ \begin{align*} \dfrac{dy}{dx}\bigg|_{\textstyle\theta= {\frac{\pi}{3}}} &= \frac{\sin\left({\dfrac{\pi}3}\right)\cos\left({\dfrac{\pi}3}\right)+\cos\left({\dfrac{\pi}3}\right)\left(2 + \cos\left({\dfrac{\pi}3}\right)\right)}{-\sin\left({\dfrac{\pi}3}\right)\left(2 + \cos\left({\dfrac{\pi}3}\right)\right) – \cos\left({\dfrac{\pi}3}\right)\sin\left({\dfrac{\pi}3}\right)}\\ &=\frac{\dfrac{\sqrt3}2\cdot\dfrac12+\dfrac12\left(2 + \dfrac12\right)}{-\dfrac{\sqrt3}2\left(2 + \dfrac12\right) – \dfrac12\cdot\dfrac{\sqrt3}2}\\ &=\frac{\sqrt3 + 5}{-\sqrt3 \cdot 5 – \sqrt3}\\ &=-\frac{(\sqrt3+5)(5\sqrt3-\sqrt3)}{(5\sqrt3+\sqrt3)(5\sqrt3 – \sqrt3)}\\ &=\boxed{-\frac{3+5\sqrt3}{18}}. \end{align*} \]

### Conclusion

Now you know how to differentiate polar equations both in polar form and to find the slope in rectangular coordinates. With practice, these techniques will become more intuitive. Good luck with your future math adventures!