TITLE: Why 1/x is elementary function?
QUESTION [6 upvotes]: This is so obvious, that $\frac{1}{x}$ is elementary function. But how this can be proven? I've been searching for information, and have found a whole list of elementary functions, and $\frac{1}{x}$ is one of them. It is like an axiom, which is always true. There are many ways to prove that different compound functions are elementary, but $\frac{1}{x}$ is always considered to be elementary.
We were give a list of properties, based on which we should prove that $\frac{1}{x}$ is also elementary. I'm quite confused, because I have no idea how to start proving. Would appreciate any kind of help.

Base Cases.

Identity function, $id(x) = x$ is in EF.
Any constant function is in EF.
The sine function $sin(x)$ is in EF

Constructor Cases. If $f,g \in EF$, then so are

$f+g$, $fg$, $2^g$
The inverse function $f^{-1}$;
The composition $f \circ g$.


Original: Given properties

REPLY [2 votes]: Recall that $\cot x=\tan(\pi/2-x)$, so
$$
\frac{1}{x}=\cot\arctan x
$$
More precisely, for $x>0$ you have
$$
\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}
$$
so
$$
\frac{1}{x}=\tan\left(\frac{\pi}{2}-\arctan x\right)=\cot\arctan x
$$
and, for $x<0$,
$$
\arctan x+\arctan\frac{1}{x}=-\frac{\pi}{2}
$$
so
$$
\frac{1}{x}=\tan\left(-\frac{\pi}{2}-\arctan x\right)=
-\cot(-\arctan x)=\cot\arctan x
$$
The tangent and the cotangent are elementary, because so are the sine and the cosine.
The exponential function is elementary, because $e^x=2^{x/\!\log 2}$. Therefore also the natural logarithm is elementary. Thus
$$
\frac{1}{\cos^2x}=\exp(-\log(\cos^2x))
$$
is elementary and
$$
\tan x=\sin x\cos x\frac{1}{\cos^2x}
$$
Similarly for the cotangent.