Infinitely differentiable How can one find if a function $f$ is infinitely differentiable?
 A: There is no magic recipe to tell whether a function is infinitely differentiable, or even once differentiable, or even continuous indeed. It really depends on the function itself. Of course there are really general and trivial statements like all polynomials are infinitely differentiable and so on.
A: Depending on how your function is given, that can be very easy or very hard. For functions given by expressions in closed form, you can try deriving a general formula for the $n$-th derivative. E.g. this way you can prove that all polynomials are smooth (=infinitely differentiable), so are sin, cos and exponential functions and any concatenation, product and sum of smooth functions is smooth.
Often, you will have to work with the actual derivative as a limit and differentiate until you find either a non-smooth point or some closed form that you recognise, you maybe some periodicity (i.e. so that some $n$-derivative is equal to some $m$-the derivative, like with trigonometric functions) or you have to get really clever.
A: As Willie Wong points out, that's the only way to do it. By the way $\sin{x},\cos{x}$ are examples of infinitely differentiable functions, and most famous example is the function $f(x)=e^{x}$. This function is infinitely differentiable and it's derivative, is $e^{x}$. To see this:
\begin{align*}
e^{x}  = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots \\ \frac{\textrm{d}} {\textrm{dx}}e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \cdots 
\end{align*}
A: By differentiating it an infinite number of times? 
But seriously, that's what you do. Just that usually you can infer what the higher order derivatives will be, so you don't have to compute it one by one. 
Example To see that $\sin(x)$ is infinitely differentiable, you realize the following: $\frac{d}{dx}\sin(x) = \cos(x)$, and $\frac{d}{dx} \cos(x) = -\sin(x)$. So you see that $(\frac{d}{dx})^4\sin(x) = \sin(x)$, and so the derivatives are periodic. Therefore by continuity of $\sin(x)$ and its first three derivatives, $\sin(x)$ must be infinitely differentiable. 
Example To see that $(1 + x^2)^{-1}$ is infinitely differentiable, you realize that $\frac{d}{dx}(1+x^2)^{-n} = -2n x (1+x^2)^{-n-1}$. So therefore by induction you have the following statement: all derivatives of $(1+x^2)^{-1}$ can be written as a polynomial in $x$ multplied by $(1 +x^2)^{-1}$ to some power. Then you can use the fact that (a) polynomial functions are continuous and (b) quotients of polynomial functions are continuous away from where the denominator vanishes to conclude that all derivatives are continuous. 
The general philosophy at work is that in order to show all derivatives are bounded and continuous, you can take advantage of some sort of recursive relationship between the various derivatives to inductively give a general form of the derivatives. Then you reduce the problem to showing that all functions of that general form are continuous. 
A: for functions:
Each analytic function is infinitely differentiable.
Each polynomial function is analytic.
Each Elementary function is analytic almost everywhere. I assume this is valid also for the Liouvillian functions.
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for function terms:
The set of the function terms of the Elementary functions is closed regarding differentiation.
The set of the function terms of the Liouvillian functions is closed regarding differentiation.
If $n_1,n_2\in\mathbb{N}_0$, $n_1\ne n_2$, $f\colon z\mapsto f(z)$, and
$$\frac{d^{n_1}}{dz^{n_1}}f(z)=\frac{d^{n_2}}{dz^{n_2}}f(z),$$
then $f(z)$ is infinitely often differentiable.
There are some general differentiation rules for calculating $n$-th derivatives, e.g. higher factor rule, higher sum rule, higher product rule, higher chain rule.
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A function with an infinitely differentiable function term is infinitely differentiable if each of its $n$-th derivatives is differentiable.
A: If a function is infinitely differentiable, then one of 3 things holds true, concerning the sequence $S$, that consists of 1. $f$, 2. the derivative of $f$, 3. the derivative of the derivative of $f$, and so on. (If f is infinitely differentiable, S is infinite, and vice versa.)


*

*There is no repetition. 

*The sequence repeats - one of its elements is a function G, that is its own derivative. (I might call this convergence.) Thus, S ends up just being G over and over again forever.

*The sequence repeats - it contains a sequence of consecutive elements C, such that the derivative of the last element of C is the first element of C. Thus, S ends up just being C over and over again forever. (3. is a generalization of 2.)


$2.$ or $3.$ are somewhat useful properties - if we prove the sequence $S$ has them, $S$ is infinite, and $f$ is infinitely differentiable.
2.
Let $f$ be $x$. Then $S$ is $x$, $1$, $0$, $0$, $...$
Since the derivative of $0$ is $0$, $S$ is infinite, and $f$ is infinitely differentiable.
3.
Let $f$ be $sin(x)$. Then $S$ is $sin(x), cos(x), -sin(x), -cos(x), sin(x), cos(x)$$, -sin(x), -cos(x), ...$ 
Since the derivative of -cos(x) is sin(x), S is just the same 4 functions over and over again, forever. Since $S$ is infinite $f$ is infinitely differentiable.
