How to find the derivative of $\frac{d\theta}{d\cos \theta}$? How can I find the derivative of $\frac{d\theta}{d\cos \theta}$? Would it be just the inverse of $\frac{d\cos\theta}{d\theta}$? thanks.
 A: You're going to want to write $\theta$ as a function of $\cos(\theta)$. Observe that you can do this by writing $\theta = \cos^{-1}(\cos(\theta))$ for $0 \le \theta \le \pi$. It should be more clear now how you'd find $\dfrac{d}{d\cos(\theta)} \cos^{-1}(\cos(\theta))$. You can generalize this to all values of $\theta$ by defining different "branches" of $\cos^{-1}(\theta)$ to output values of $\theta$ in the ranges of $n\pi \le \theta \le (n+1)\pi, n \in \mathbb{Z}$.
EDIT: There is indeed an easier way to do this using the inverse function theorem... Again, we would need to restrict $\theta$ to $0 \le \theta \le \pi$, then we could write $\cos^{-1}(\cos(\theta))=\theta$, from which we can differentiate implicitly with respect to $\theta$ to write:
$$\frac{d}{d\theta} \cos^{-1}(\cos(\theta))=\frac{d}{d\theta}\theta$$
$$\Rightarrow \dfrac{d\cos^{-1}(\cos(\theta))}{d\cos(\theta)} \cdot \dfrac{d\cos(\theta)}{d\theta}=1$$
$$\Rightarrow \dfrac{d\theta}{d\cos(\theta)} \cdot \dfrac{d\cos(\theta)}{d\theta}=1$$
$$\Rightarrow \dfrac{d\theta}{d\cos(\theta)} = -\dfrac{1}{\sin(\theta)}$$
Which is perhaps a preferred way of finding this derivative since the first method will have you messing around with triangles in order to simplify it to this form.
A: By the chain rule:
$$\frac{\mathrm{d}\theta}{\mathrm{d}\cos{\theta}}\frac{\mathrm{d}\cos{\theta}}{\mathrm{d}\theta}=\frac{\mathrm{d}\theta}{\mathrm{d}\theta} = 1$$
So
$$\frac{\mathrm{d}\theta}{\mathrm{d}\cos{\theta}} = \frac{1}{\left(\frac{\mathrm{d}\cos{\theta}}{\mathrm{d}\theta}\right)}=-\frac{1}{\sin\theta}$$
So yes; it is the multiplicative inverse of $\frac{\mathrm{d}\cos{\theta}}{\mathrm{d}\theta}$.

See also the more general case.
