How To Prove That This Particular Single Variable Function Is Infinitely Differentiable? I'm a physics student and this weird looking function came up in my doctoral research $\frac{\cos ^{-1}\left(e^{3 B}\right)}{\sqrt{1-e^{6 B}}}$ and I would like to be able to prove that it is infinitely differentiable in B. I know using L'Hospital's rule that in the limit when B approaches zero that it converges. I can also check using Mathematica that for small values of n that the nth derivative with respect to B of this function also converges in the limit of B approaching zero. However for the work that I will be doing I really need to a rigorous proof that this function is infinitely differentiable. Can anyone offer any suggestions.
Thanks
 A: Short answer: think about power series representations.
Long answer: you should first specify the interval of $B$-values you are about, e.g., the function isn't defined if $B = 1$ since $e^{3} = 20.08\ldots$ isn't in the domain of $\arccos x$. And since $B$ only occurs in the context of a power of $e^{3B}$, focus on the function $$f(x) = \frac{\arccos(x)}{\sqrt{1-x^2}}$$ for $x \in (-1,1)$. Your function is $f(e^{3B})$, defined for $B < 0$, and a composition of infinitely differentiable functions is infinitely differentiable, so it suffices to show $f$ is infinitely differentiable.  We'll do that by using power series, which are automatically infinitely differentiable on the interval of convergence.
Look at power series expansions of the numerator and denominator near $0$: if $x \in (-1,1)$ then 
$$
\arccos x = \frac{\pi}{2} - \sum_{n \geq 0} \frac{(2n)!}{2^{2n}(n!)^2}x^{2n+1} = \frac{\pi}{2} - x - \frac{1}{6}x^3 - \cdots
$$
and
$$
\sqrt{1 - x^2} = \sum_{n \geq 0} \binom{1/2}{n}(-1)^nx^{2n} = 1 - \frac{1}{2}x^2 - \frac{1}{8}x^4 - \cdots
$$
The ratio of these two power series that each converge for $-1 < x < 1$ also converges for $-1 < x < 1$, although the simplest proof of this comes from thinking of these power series as defined on the unit disc $|z| < 1$ in $\mathbf C$ so you can appeal to properties of analytic functions (where having a first derivative on an open disc implies having a power series representation from the center that converges on the whole disc). A ratio of analytic functions on $|z| < 1$ with a denominator that is nonvanishing there is analytic on $|z| < 1$, so $f(z)$ is analytic for $|z| < 1$.   Thus $f$ is infinitely differentiable on $|z| < 1$, so the composition $f(e^{3B})$ is infinitely differentiable for $B < 0$.
You could use a computer algebra package to find the first few power series coefficients for $f(z)$ at $z = 0$.
For your function $f(e^{3B})$, which is initially defined for $B < 0$, as $B \rightarrow 0^{-}$ we have $e^{3B} \rightarrow 1^{-}$ and as $x \rightarrow 1^{-}$ we have $\arccos(x) \rightarrow \arccos(1) = 0^{+}$ and $\sqrt{1-x^2} \rightarrow 0^{+}$. Perhaps your actual interest all along was the behavior of $f(e^{3B})$ around $B = 0$, where most of what I wrote above is not directly relevant.
A: Here's a different approach. 
First, replace $x = e^{3B}$; since $\exp$ is $C^\infty$ it suffices to look at the function 
$$ x \mapsto \frac{\cos^{-1}(x)}{\sqrt{1 - x^2}}$$
Note that for $x \in [0,1]$, we have that 
$$ \cos^{-1}(x) = \sin^{-1} (\sqrt{1-x^2}) $$
Consider the function $f(y) = \sin^{-1}(y) / y$. We have that $\sin^{-1}(y)$ has a power series (around zero) with radius convergence 1, and is an odd function, and hence the function $\sin^{-1}(y) / y$ is an even function with radius convergence 1. 
This implies that $f(y)$ has Maclaurin series that only contains even powers in $y$, and hence there is a power series with radius of convergence 1, converging to some function $g(y)$, such that $f(y) = g(y^2)$. The positive radius of convergence implies that $g(y)$ is smooth. 
Finally, our original function is $f(\sqrt{1-x^2}) = g(1 - x^2)$ now can be seen as a composition of smooth functions (when $1-x^2$ is near 0), and therefore is smooth near $x = 1$. 
