You can, but have you to take a small detour. Remember that the standard deviation is the square root of the variance, which is a lot easier to work with:
$$
s^2 = \frac{1}{N-1} \sum_{i=1}^N (x_i - \bar{x}_N)^2
$$

Now suppose we have computed the variance $s_{N-1}^2$ and mean $\bar{x}_{N-1}$ based on N-1 samples, and a new sample $x_N$ arrives and you want to compute $s_N^2$. If you look at the difference in the sums of the squared differences used in the formulas for variance you get something surprisingly simple:

$$
(N-1)s_N^2 - (N-2)s_{N-1}^2 = (x_N-\bar{x}_N)(x_N - \bar{x}_{N-1}) \\
$$

(See the full derivation)

Therefore the updated variance is
$$
s_N^2 = \frac{(N-2)s_{N-1}^2 + (x_N-\bar{x}_N)(x_N - \bar{x}_{N-1})}{N-1}
$$

and the updated standard deviation would be just the square root. If you want the population variance or standard deviation replace N-1 with N and N-2 with N-1.