I Should Divergent Integrals Justify Ignoring Derivatives?

[Tio Barnabe]

I wanted to argue that $$\frac{d}{d \cos \theta} \sin \theta, \ \theta \in [0, \pi]$$ should be ignored in the given interval, because integration of it leads to a divergent integral. Is this acceptable as a reason?

 

[Mark44]

I wanted to argue that $$\frac{d}{d \cos \theta} \sin \theta, \ \theta \in [0, \pi]$$ should be ignored in the given interval, because integration of it leads to a divergent integral. Is this acceptable as a reason?

First off, $\frac{d}{d \cos \theta} \sin \theta$ is pretty unwieldy, as $\sin(\theta)$ isn't a function of $\cos(\theta)$ at first glance. However, you could write the derivative as $\frac d {d(\cos(\theta))} (\pm \sqrt{1 - \cos^2{\theta}})$, and then use the chain rule. So, no, I don't see that it's reasonable to ignore it.

Second, integration and differentiation are different operations, so the fact that the integral of some function on some interval is divergent doesn't have any bearing here.

 

[FactChecker]

In addition to what @Mark44 said, the idea that something can be ignored because it is divergent is wrong. If it was very small compared to other terms, that would be different, but being too large makes it impossible to ignore.

 

[NFuller]

Just to further elaborate @Mark44's point, it isn't very hard to actually calculate the derivative and show that it does exist. So to drive home the point, here is the result.

Using the transformation $\xi(\theta)=\text{cos}(\theta)$
$$\text{sin}(\theta)=\pm\sqrt{1-\xi^{2}}$$
so
$$\frac{d}{d\text{cos}(\theta)}\text{sin}(\theta)=\pm\frac{d}{d\xi}\sqrt{1-\xi^{2}}=\pm\frac{\xi}{\sqrt{1-\xi^{2}}}=\pm\frac{\text{cos}(\theta)}{\text{sin}(\theta)}=\pm\text{cot}(\theta)$$

 

[Tio Barnabe]

Thank you to everyone