Solve differential in respect to a function rather than a variable In an exercise I have to solve this differential. Now what confuses me is that we take the derivative in respect to a function rather then a variable. Now I am not sure how to proceed?
$$\frac{d}{d(\cos (\theta))} \cos(n\theta)$$
 A: Note: I'd rather call this a 'derivative' as opposed to 'differential'; in my experience, a differential refers to something like '$dy$' or '$dx$' considered on its own.
Hint: to solve this problem, use the chain rule:
$$
\frac{d\cos n\theta}{d\cos\theta}=\frac{d\cos n\theta}{d\theta} \cdot \frac{d\theta}{d\cos\theta} \, .
$$
As for your more general question about what
$$
\frac{d\cos n\theta}{d\cos\theta}
$$
actually means, this is a prime example of implicit differentiation. The equation
$$
x^2+y^2=1
$$
gives us a simpler example of this. There is no single way to write $y$ as a 'function of $x$'. Either
$$
y(x) = \sqrt{1-x^2}
$$
or
$$
y(x)=-\sqrt{1-x^2} \, .
$$
Implicit differentiation tells us that regardless of which of these two functions we are considering, the following relationship holds:
$$
\frac{dy(x)}{dx} = -\frac{x}{y(x)} \, .
$$
In your example, it might be difficult or impossible to find an explicit formula connecting $\cos n\theta$ and $\cos \theta$; that is, there is no single way to write $\cos n\theta$ as a 'function of $\cos\theta$'. However, we can still find an 'implicit' equation that tells us how $\cos n\theta$ changes for infinitesimal changes in $\cos\theta$, treating $\cos\theta$ as a variable just like any other.

In Lagrange notation, these manipulations would read differently. Let $f$ be an arbitrary function satisfying
$$
f(\cos\theta) = \cos n\theta
$$
for all $\cos\theta$. To make this more readable, let $g(\theta)=\cos \theta$. Then,
$$
\cos n \theta = f(g(\theta)) \, .
$$
By the chain rule,
\begin{align}
-n\sin n \theta &= f'(g(\theta))g'(\theta) \\
&= -f'(\cos\theta)\sin\theta
\end{align}
which implies
$$
f'(\cos\theta) = \frac{n\sin n \theta}{\sin\theta} \, .
$$

Finally, another way to solve this equation is to parameterise $\cos n\theta$ and $\cos \theta$ with respect to $\theta$. In general, parameterisation is where you express two variables $x$ and $y$ in terms of a third variable $t$. In general, if we find two functions $f$ and $g$ such that $x=f(t)$ and $y=g(t)$, then
$$
\frac{dx}{dt}=f'(t) \text{ and } \frac{dy}{dt}=g'(t)
$$
which implies that
$$
\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{g'(t)}{f'(t)} \, .
$$
In this case, if we let $y=\cos n\theta$ and $x=\cos \theta$, then
$$
\frac{dy}{d\theta}=-n\sin n\theta \text{ and } \frac{dx}{d\theta} = -\sin\theta \, .
$$
Hence,
$$
\frac{dy}{dx} = \frac{-n\sin(n\theta)}{-\sin\theta}=\frac{n\sin n\theta}{\sin\theta} \, .
$$
This method is in essence the same as the first one I showed you, but it's helpful to get another perspective.
