The Standard Deviation of The Derivative of a Signal
As pointed out in pichenettes' answer, you need to know more about the original signal $x(t)$. If the signal's power spectrum $S_x(\omega)$ is known then the power spectrum of its $n^{th}$ derivative is

$$S_y(\omega)=\omega^{2n}S_x(\omega)$$

The power of the $n^{th}$ derivative can then be computed as

$$E\{|y(t)|^2\}=\frac{1}{2\pi}\int_{-\infty}^{\infty}\omega^{2n}S_x(\omega)d\omega\tag{1}$$

(assuming convergence of the integral, e.g. by assuming that $x(t)$ is band-limited). Since the mean of the derivative is zero (even if the mean of the original signal isn't), the variance of $y(t)$ equals its power, and so the standard deviation is the square root of the expression in (1).