I have a series which involves an expression of the form
$\displaystyle \frac{\partial^{n} a(x)}{\partial x^n} + c(x)\frac{\partial^{n+1} b(x)}{\partial x^{n+1}}$
Is there a way to factor out the $\frac{\partial^n }{\partial x^n}$ operator? I believe I can write the expression above as
$\displaystyle \frac{\partial^{n} a(x)}{\partial x^n} + c(x)\frac{\partial^n }{\partial x^n}\frac{\partial b(x)}{\partial x}$
Since the coefficient $c$ is a function of $x$, I'm pretty sure factoring out the operator any further is not an option (at least not in a nice way). For clarity, I'm wondering if the following can be achieved
$\displaystyle \frac{\partial^{n} }{\partial x^n}\left(a(x) + c(x)\frac{\partial b(x)}{\partial x}\right) $
The motivation of this question is to better understand how powers work with operators, and not just when an operator coefficient is constant. This brings me to the second part of my question. If I define
$\displaystyle f^n = c(x)^n\frac{\partial^n}{\partial x^n}$
then based on that definition,
$\displaystyle f^{n+1} = c(x)^{n+1}\frac{\partial^{n+1}}{\partial x^{n+1}}$
however, the rules of powers (e.g. $g^{n+1} = g^ng$) don't seem to apply when $f$ is defined with an operator? Specifically, if $f^{n}f$ should be interpreted as a composition , then the differential operator in $f^n$ would have to be applied to $c(x)$ I believe. However, I'm not sure how to proceed, because I feel like some different rules are in play here? Specifically, the term $f$ is not commutative, so I'm wondering if I'm entering a different area of mathematics?
You are correct that you cannot factor the $\frac{\partial^n}{\partial x^n}$ operator further specifically because the following are not equal:
$$c(x) \frac{\partial^n}{\partial x^n} [\text{something}] \neq \frac{\partial^n}{\partial x^n}c(x) [\text{something}]$$
Looking at the right hand side it's clear you'd need some multiple-derivative product rule (https://en.wikipedia.org/wiki/General_Leibniz_rule)
On the other hand,
$$\frac{\partial^n}{\partial x^n} a(x)+\frac{\partial^n}{\partial x^n}c(x) [\text{something}]$$ $$=\frac{\partial^n}{\partial x^n}(a(x)+c(x)[\text{something}]).$$ Due to the sum rule.
Your definition of $f^n$ isn't just $n$ applications of some operator $f^1$. If it were, the rules of powers would apply. For instance you could define $$f^n = \underbrace{c(x) \frac{\partial}{\partial x} c(x) \frac{\partial}{\partial x} \cdots c(x) \frac{\partial}{\partial x}}_{n \text{ times}},$$ but this is different from your definition specifically because multiplying by $c(x)$ and taking the partial derivative don't commute.