should I apply low-pass filter when calculating central derivative?
No, you are not increasing the sampling time.  The sample rate stays the same.  The Central difference derivative filter is simply longer than the First difference derivative filter.  A longer filter just means more delay before you get your result.

No, you should not low pass filter without a clear reason for why it is needed to meet your signal processing objectives.  Your reasoning is flawed in this case, so don't apply a LPF.

Note that the Ideal derivative filter multiplies a signal's spectrum by $j\omega \cdot e^{j\omega \mu}$ in the frequency domain.  In the time domain, this ideal filter is an infinitely long filter with taps:

$$h[n] = \dfrac{1}{n+\mu}\left[\cos(\pi \left[n+\mu\right])-\mathrm{sinc}(n+\mu)\right]$$

Where

$n$ is the filter tap index in $(-\infty, \infty).$

$\mu$ corresponds to an inter-sample shift of the filter taps in the range $[0.0, 1.0]$ samples.

Given the above expression for $h[n]$, note that:

The Central difference is $h[n]$ truncated to 3 taps with $\mu = 0$.

The First difference is a scaled version of $h[n]$ truncated to 2 taps with $\mu = 0.5$.

Update to add filter magnitude responses plot