Why 1/x is elementary function? 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
 A: 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.
A: Since $2^x$ is elementary and elementary functions are closed under inverses, $\log_2(x)$ is elementary. Then: $$\frac{1}{x} = x^{-1} = 2^{\log_2(x^{-1})} = 2^{(-1)\log_2(x)}.$$
Edit: As noted in the comments, this expression is only defined for $x>0$. Following egreg's suggestion:
$$\frac{1}{x} = x(x^{-2}) = x(2^{\log_2(x^{-2})}) = x(2^{(-1)\log_2(x^2)}).$$
This is defined for all $x\neq 0$, since then $x^2>0$.
A: Wikipedia says that

In mathematics, an elementary function is a function of one variable which is the composition of a finite number of arithmetic operations (+ – × ÷), exponentials, logarithms, constants, and solutions of algebraic equations (a generalization of nth roots).

Clearly, going by the above definition, $\frac{1}{x}$ is elementary.
That pretty much answers your question, right?
