The function
$f(x)=x+\sin(x)$
is easily checked to be a bijection from the reals to itself, and so it has a unique inverse $y\mapsto g(y)$ such that $f\circ g=g\circ f$ are both the identity map.
Now $g$ will almost certainly be a function which is not expressible using "the functions in a high-schooler's toolkit" (by which I guess I mean $\exp$, $\log$, and, if you like, the usual trigonometric functions and their friends like $\sinh$, although of course these can all be of course built from exponentials anyway). For purely recreational reasons (stemming from conversations I've had whilst teaching undergraduates) I'm interested in how one proves this sort of thing.
A few years ago I was interested in a related question, and took the trouble to learn some differential Galois theory. My motivation at the time was learning how to prove things like why $h(t):=\int_0^t e^{x^2} dx$ is not expressible in terms of these calculator-button functions (I'm sure there's a better name for them but I'm afraid I don't know it). I've realised that since then I've forgotten most of what I knew, but furthermore I am also unclear about whether this is the way one is supposed to proceed. Is the idea that I come up with some linear differential equation satisfied by $g$ and then apply some differential Galois theory technique? In fact, one of the many things that I have forgotten is the following: if $F$ is a field equipped with a differential operator $D$, and $E/F$ is the field extension obtained by adding a non-zero root of $Dh=ch$, with $c\in F$, then the Galois group of $E/F$ is solvable, whereas the equation itself might not be, in terms of calculator-button functions, if I can't integrate $c$.
Can a more enlightened soul explain to me how one is supposed to proceed? I wonder whether I am somehow conflating two ideas and the differential Galois theory business is a red herring, but it seemed simpler to ask rather than continuing to flounder around.