Errors and differentials In the Heisenberg Uncertainty Principle
$$ \Delta p \Delta x \geq \hbar/2$$
we write the inherent errors in measurement as $ \Delta p $ and $ \Delta x $ .
Does these two errors behave the same as the differentials $ dp $ or $ dx $ or some other form of differentials?
 A: The Heisenberg inequality deals with a highly specific measure of "uncertainty", i.e. spread in the probability distributions that govern observed measurement outcomes applied to the same quantum state. Here, the $\Delta$s are the standard deviations of the relevant probability distributions, and have nothing to do with differentials.
More specifically, two quantum observables $\hat{x}$ and $\hat{p}$ are conjugate, i.e. fulfill the canonical commutation relationship $[\hat{x},\,\hat{p}]=i\,\hbar\,\mathrm{id}$, then if we apply $\hat{x}$ and $\hat{p}$ to the same quantum state, the measurements we glean from them have standard deviations related by the Heisenberg inequality. The inequality is saturated (becomes equality) if and only if the probability distributions are both Gaussian.

Furthermore, it can be shown that (1) $\hat{x}$ and $\hat{p}$ have continuous spectrums, and that measurements represented by them can take on any real value and (2) if one writes the quantum state in $\hat{x}$-co-ordinates (i.e. wherein $\hat{x}$ is a simple multiplication operator) so we represent the quantum state by the $\mathbf{L}^2$ function $\psi(x)$, then $\hat{p}=-i\,\hbar\,\mathrm{d}_x$ and so (3) the Fourier transform (a unitary mapping) maps the quantum state and co-ordinates into the Hilbert space wherein the $\hat{p}$ operator becomes the simple multiplication operator.
Accordingly, the Heisenberg inequality is a statement about what the spread of function tells us about the spread of its Fourier transform. If a function is confined to be nonzero in a finite, compact region, then it is a fundamental fact that the Fourier transform must have nonzero values over an infinite subset of the real line ("a function and its Fourier transform cannot both have compact support"). The Heisenberg inequality is simply a less coarse, more quantitative version of this theorem, as I discuss in my answer here.
