How smooth is the generalized power mean function? Consider the function of power mean of $1$ and $e$ (so that the computation is simpler). That is, define $\displaystyle f(x)=(\frac{1+e^x}{2})^{1/x}$ when $x\neq 0$ and $f(0)=\sqrt{e}$.
It's not difficult to use L'Hopital rule with std. exponential trick to verify the function is continuous.
The question is, do we know how smooth (in which $C^k$) is this function (and a closed form formula of $f^{(n)}(0)$ if possible)?
The only way I can think of is trying to compute $\displaystyle \frac{f^{(n)}(h)-f^{(n)}(0)}{h}$ to verify if $f^{(n+1)}(0)$ exists, but the formula becomes extremely complicated and it's practically impossible to compute that limit even for $n=1$.

I will stick to $x\in \mathbb R.$ We can write
$$\tag 1 f(x) = \exp [(1/x)\ln ((e^x+1)/2)]$$
for $x\ne 0.$ Now $\ln ((e^x+1)/2)$ is real analytic on $\mathbb R.$ Why? Because $(e^x+1)/2$ is real analytic and positive on $\mathbb R,$ $\ln x$ is real analytic on $(0,\infty),$ and compositions of real analytic functions are real analytic. Furthermore, $\ln ((e^x+1)/2)$ equals $0$ when $x=0.$ It follows that $(1/x)\ln ((e^x+1)/2)$ is real analytic on $\mathbb R,$ once we define $f(0)$ to be the value you found, namely $e^{1/2}.$ Again using "compositions of real analytic functions are real analytic", we see from $(1)$ that $f$ is real analytic on $\mathbb R.$ (Since real analytic functions are $C^\infty,$ we have $f\in C^\infty(\mathbb R).$)
Sorry, I don't have any insight towards a nice formula for $f^{(n)}(0).$
