Edit note: I mis-wrote the expression originally, resulting in helpful-but-incorrect feedback. The function is now correctly expressed.
I suspect that the answer to this is a resounding 'no', but I'm curious...
Is there a way to differentiate Riemann's approximation to the prime counting function
$$R(x)=\sum_{n=0}^{\infty} \frac{\mu(n)}{n}\text{li}(x^{1/n})$$
where $\mu(n)$ is the Moebius function, and $\text{li}(x)$ is the logarithmic integral?
Note that $R(x)$ can also be expressed as
$$R(x)=1+\sum_{k=1}^{\infty} \frac{\bigr(\log(x)\bigl)^k}{k! k\zeta(k+1)}$$
I am aware that there are differentiable approximations to $R(x)$. My question is whether the function itself can be differentiated.
With the last formula, you can easily prove that $R$ is infinitely differentiable. since its is normaly convergent for the $\mathscr C^p$ norm on every compact $[a,b]\subset ]0,+\infty[$: $$ \left\| \frac{\log^k(x)}{k! k \zeta(k+1)} \right\|_{\mathscr C^p} \leq \frac{\max(|\log(a)|,|\log(b)|)^k}{k!}$$ and $$\sum_{k=1}^{+\infty} \frac{\max(|\log(a)|,|\log(b)|)^k}{k!} = \exp(\max(|\log(a)|,|\log(b)|))-1 < +\infty $$