I am working through the following solved problem which uses separation of variables to get two ODEs. The problem is to show that
$$\frac{1}{\sin\theta P}\frac{\mathrm{d}}{\mathrm{d}\theta}\left(\sin\theta\frac{\mathrm{d}P}{\mathrm{d}\theta}\right)-\frac{m^2}{\sin^2\theta}=-\lambda,$$
can be expressed as
$$\left(1-x^2\right)\frac{\mathrm{d}^2P}{\mathrm{d}x^2}-2x\frac{\mathrm{d}P}{\mathrm{d}x}+\left[l\left(l+1\right)-\frac{m^2}{1-x^2}\right]P=0,$$
where $\lambda=l\left(l+1\right)$ and $x=\cos\theta$.
I can solve the problem fine but I am uncomfortable with some of the notation that I used and want to know / understand better if it is correct.
Essentially the answer involves stating that
$$\frac{\mathrm{d}}{\mathrm{d}\theta}=\frac{\mathrm{d}x}{\mathrm{d}\theta}\frac{\mathrm{d}}{\mathrm{d}x}=-\sin\theta\frac{\mathrm{d}}{\mathrm{d}x},$$
and substituting it in.
I can see that this is an application of the chain rule, but I don't feel comfortable in it being expressed this way. I think what bothers me is the first part it is not specified what function we are taking the derivative of, which in this case is $x$ but then that would leave you with
$$\frac{\mathrm{d}x}{\mathrm{d}\theta}=\frac{\mathrm{d}x}{\mathrm{d}\theta}=-\sin\theta,$$
which doesn't give you the manipulation that is needed to re-express the question. This sort of manipulation gets used a lot and I would like to become more comfortable with understanding why it is valid to state such a thing as to me it just doesn't feel 100% correct.
Assume you have two differentiable functions $f:I\rightarrow \mathbb{R} $ and $g:J\rightarrow \mathbb{R} $ such that $f(I)\subset J$, where $I$ and $J$ are two intervals $\varnothing \neq I\subset \mathbb{R} $ and $\varnothing \neq J\subset \mathbb{R} $. Then the chain rule can be expressed as \begin{equation*} \left( g\circ f\right) ^{\prime }(a)=g^{\prime }(b)f^{\prime }(a),\qquad a\in I,b=f(a)\text{.} \end{equation*}
If $y=g(x)$ and $x=f(\theta )$ this corresponds to the following Leibnitz notation for the derivative of the composite function $y=\left( g\circ f\right) \left( \theta \right) =g\left( f(\theta )\right) $ \begin{equation*} \frac{dy}{d\theta }=\frac{dy}{dx}\frac{dx}{d\theta }, \end{equation*} where $dy/d\theta $, $dy/dx$ and $dx/d\theta $ are evaluated, respectively, at $\theta =a\in I$, $b=f(a)$ and $\theta =a$.