I have a quantum mechanics problem asking me to prove that the commutator of $x$ and $p$ is equal to some value. In my computation I get something along the lines of
$$ (a*x) \frac{d}{dx} - (b)\frac{d(x)}{dx} $$
Is this expression equivalent to the expression
$$ (a*x) \frac{d(\textbf{1})}{dx} - (b)\frac{d(x)}{dx} $$ Where $\textbf{1}$ is the one function, not the number one.
On
Though this isn't what you asked, what you actually want to do is write
$$(x \circ \frac{d}{dx}-\frac{d}{dx} \circ x)f=x\frac{df}{dx}-\frac{d(xf)}{dx}$$
and then simplify the result. You should find that it is just $-f$. Thus $x \circ \frac{d}{dx}-\frac{d}{dx} \circ x$ corresponds to multiplication by $-1$.
The point here is that $x \circ \frac{d}{dx}-\frac{d}{dx} \circ x$ is not a number, it is an operator, so it only does something when applied to an appropriate function.
$\frac{d}{dx}:C^1 \to C$ is an operator, while $\frac{d1}{dx}$ is a function of $x$.
Here $C^1$ denotes the space of first order differentiable functions, and $C$ denotes continuous functions.