Prove $\tan{(x)} \in \left(\frac{-\pi}{2},\frac{\pi}{2}\right)$ has a differentiable inverse defined on $\mathbb{R}$. By the Inverse Function Theorem, this seems obviously, where I would first prove $\tan(x)$ is continuously differentiable on this interval. $\tan(x)$ is continuous and differentiable by it being composed of $2$ continuous/ differentiable functions. Would this suffice? It seems too simple.
Definitions and theorems I can use are: Inverse Function Theorem, Global Inverse Function Theorem, injective, pretty much everything with differentiability and continuity.
I CANNOT use anything dealing with integration. Thank you!
 A: If you believe that $\sin(x)$ and $\cos(x)$ are differentiable, then the quotient rule will get you that $\frac{\sin(x)}{\cos(x)}$ is differentiable whenever $\cos(x)\neq 0$. Doing the quotient rule, you get the usual $\sec(x)^2$. Why is $\sec(x)^2\neq 0$ on $(-\pi/2,\pi/2)$?
A: The function $\tan$ restricted to $(-\pi/2,\pi/2)$ is increasing and continuous.
The first property can be proved without differentiation: if $-\pi/2<x<y<\pi/2$, then we have
$$
\tan y-\tan x=\frac{\sin y\cos x-\cos y\sin x}{\cos x\cos y}=
\frac{\sin(y-x)}{\cos x\cos y}>0
$$
because $0<y-x<\pi$. Thus it is invertible.
It is continuous because it is differentiable, the derivative being $1+\tan^2x$. Note that $1+\tan^2x>0$, so this also proves the function is strictly increasing, hence invertible. By the inverse function theorem, its inverse is everywhere differentiable as well, because the derivative of $\tan$ never vanishes.
The domain of the inverse is $\mathbb{R}$, because of
$$
\lim_{x\to-\pi/2^+}\tan x=-\infty
\qquad
\lim_{x\to\pi/2^-}\tan x=\infty
$$
and the intermediate value theorem.
The derivative of the inverse, usually called $\arctan$ can be found from considering
$$
\tan\arctan y=y
$$
so differentiating both sides with respect to $y$ yields
$$
(1+\tan^2\arctan y)\arctan'y=1
$$
and therefore
$$
\arctan'y=\frac{1}{1+\tan^2\arctan y}=\frac{1}{1+y^2}
$$
