Is the given expression, monotonically increasing or decreasing with increasing x?
$\frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$
EDIT: This is the derivative of the prime counting function $\pi(x)$ w.r.t. x, ie., $\frac{d\pi(x)}{dx} \sim \frac{1}{x \log(x)} \sum_{i=1}^{\infty} \frac{\mu(i)}{i} x^{1/i}$ Also, you might check Bruce C. Berndt's "Ramanujan's NoteBooks Part IV" page 123, equation 10.16, which gives = sign instead of $\sim$
I'm not sure I should bother answering this question, because it seems like the original poster may not have asked the right question. However, it is a nice exercise in basic asymptotics.
For $x$ sufficiently large, the sum in question is decreasing.
First, note that this sum is equal to $$\sum_{k \geq 1} \frac{(\log x)^{k-1}}{x k! \zeta(k+1)}.$$ (See here for a very similar series; we are using the highly nontrivial identity $\sum \mu(i)/i=0$ to get rid of the "$k=0$" term.)
Substituting $x=e^u$, we want to know whether or not $$e^{-u} \sum_{k \geq 1} \frac{u^{k-1}}{k! \zeta(k+1)}$$ is increasing or decreasing in $u$. One can justify taking term by term derivatives, so we want to know whether $$e^{-u} \left( \sum_{k \geq 2} \frac{(k-1) u^{k-2}}{k! \zeta(k+1)} - \sum_{k \geq 1} \frac{u^{k-1}}{k! \zeta(k+1)} \right)$$ is positive or negative.
Rearranging terms, we are interested in the sign of $$e^{-u} \sum_{\ell \geq 0} \frac{u^{\ell}}{\ell!} \left( \frac{ 1}{(\ell+2) \zeta(\ell+3)} - \frac{1}{(\ell+1) \zeta(\ell+2)} \right).$$
The quantity in parenthesis is $-1/(\ell+1)(\ell+2) + O(2^{- \ell})$. So we are interested in the sign of $$e^{-u} \left( - \sum_{\ell \geq 0} \frac{u^{\ell}}{(\ell+2)!} + \sum O(\frac{ u^{\ell} 2^{- \ell}}{\ell !}) \right) =$$ $$e^{-u} \left( - \frac{e^u -1-u}{u^2} + O(e^{u/2}) \right)=$$ $$-1/u^2 + O(e^{-u/2}).$$
This is negative for $u$ sufficiently large.
The expression $\pi(x)\approx\sum_{n>0}\frac{\mu(n)}{n}Li(x^{1/n})$ is certainly not an exact equality of any kind, and must be interpreted very carefully. This infinite series is sometimes called the Gram series; the real interest in this expression is that numerically it seems to have a much better error than the simple approximation $\pi(x)=Li(x)+error$. Ramanujan was misled by its remarkable accuracy (for smallish values of $x$) and made some very strong conjectures on the relation between $\pi(x)$ and the Gram series, conjectures which are demonstrably false. In fact Littlewood proved that the Gram series is a worse approximation to $\pi(x)$ than $Li(x)$ infinitely often.
Besides, $\pi(x)$ is a step function; it doesn't make any sense to differentiate a continuous approximation and expect to retain any meaning. Consider, for example, what would happen if you differentiated the expression $\left\lfloor x \right\rfloor\approx x+\sin(10^{10}x)$...
In fact, if you plot the Ramanujan's series in Mathematica, up to i=1000, or so, what you get is a monotonically decreasing function. This function is indistinguishable of the derivative of the Riemann's prime counting function, R(x).
In other words, if there exists a function that match the prime counting function Pi(x), for all x (not for the steps, but for the mean behaviour), its derivative should be the Ramanujan's series.
The Ramanujan's series for the derivative of Pi(x) is a continuous function.
