Derivative of the sine function when the argument is measured in degrees

Yes, it is correct, but keep in mind that what you are calculating is:

$$f(x)=\sin(πx/180)$$

$$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=\frac{\pi}{180}\cos(\pi x/180),$$ where $x$ is expressed in degrees.

This annoyed me when I revisited this material years later and had to teach this material to someone else, so I'm posting an answer here. (The other answer is perfectly fine, but I wanted to give a more complete exposition.)

Now here's the thing: you're told to find the derivative of $\sin(\theta)$ when $\theta$ is in degrees. At a first glance, this seems simple: it should just be $\cos(\theta)$. However, this answer is wrong, because you found that $\sin(\theta)$ has derivative $\cos(\theta)$ under the assumption that $\theta$ is measured in radians, and not in degrees.

Here's how you should approach the problem.

Notice that $\sin(\theta)$, when $\theta$ is in degrees or when $\theta$ is in radians, gives two different values. So, in fact, for this problem, writing $\sin(\theta)$ is itself ambiguous, because it isn't clear if $\theta$ is in degrees or radians. (However, for the rest of your studies, you'll likely assume the radian form.)

For the moment, suppose that $\sin_d(\theta)$ denotes $\sin(\theta)$ when $\theta$ is measured in degrees, and $\sin_r(\theta)$ denotes $\sin(\theta)$ when $\theta$ is measured in radians. We apply similar meaning for $\cos_d$ and $\cos_r$.

The motivation for this seems confusing at a first glance: why do we need to do this? It's because when we're calculating $\sin(\theta)$ depending on whether or not $\theta$ is in degrees or radians, we are fundamentally working with two different functions. Assuming we only care about the standard unit circle, when $\theta$ is in degrees, the domain is $[0, 360)$. When $\theta$ is in radians, the domain is $[0, 2\pi)$. Hence why we use $\sin_d$ and $\sin_r$ for these two different cases.

Now, back to the problem: what you are asked to find is $$\dfrac{\text{d}}{\text{d}\theta}\sin_d(\theta)\text{.}$$ You know for a fact that $$\dfrac{\text{d}}{\text{d}\theta}\sin_r(\theta) = \cos_r(\theta)\text{.}$$ So now, the problem boils down to this: how can we write $\sin_d$ in terms of $\sin_r$ so that we can apply the chain rule to find $\dfrac{\text{d}}{\text{d}\theta}\sin_d(\theta)$?

Recall that $\pi$ radians is $180$ degrees. So, if we have an angle which is $\theta$ degrees, it follows that the equivalent radian angle is $\dfrac{\theta}{180}\pi$. It follows that $$\sin_d(\theta)=\sin_r\left(\dfrac{\theta}{180}\pi \right)\text{.}$$

Hence,

$$\dfrac{\text{d}}{\text{d}\theta}\sin_d(\theta) = \dfrac{\text{d}}{\text{d}\theta}\sin_r\left(\dfrac{\theta}{180}\pi \right) = \dfrac{\pi}{180}\cos_r\left(\dfrac{\pi}{180}\theta \right)$$ from an application of the chain rule.

Lastly, note that the angle $\dfrac{\pi}{180}\theta$ is in radians. The equivalent degree measure would be $$\dfrac{\pi}{180}\theta \cdot \dfrac{180}{\pi} = \theta$$ hence, $$\dfrac{\pi}{180}\cos_r\left(\dfrac{\pi}{180}\theta \right) = \dfrac{\pi}{180}\cos_d(\theta)$$ and then we obtain $$\dfrac{\text{d}}{\text{d}\theta}\sin_d(\theta) = \dfrac{\pi}{180}\cos_d(\theta)$$ as desired.

Remark: This is an exercise in Stewart's text. I wouldn't expect a typical Calculus I student to be able to do this exercise, given that very little time (in my experience) is spent on functions and given the strange notation (this would look strange to a Calc. I student - remember, most of these people haven't even read proofs) involved. As you might guess, although this problem is interesting, given the way this exercise is usually written and implemented in calculus courses, I'm not a fan. More time should be dedicated to thinking about functions than usually done for a Calc. I course should this problem be assigned as an exercise.