Backward finite difference differentiation filter frequency response
In discrete-time signal processing terminology a backward (one sample) difference filter is
$$y[n] = x[n] - x[n-1]$$
which is a causal, LTI system with an FIR impulse response of
$$h[n] = \delta[n] -\delta[n-1]$$
The associated frequency response is then the DTFT of the impulse response as
$$\begin{align}
H(e^{j\omega}) = \mathscr{F}\{ h[n] \} &amp;= 1 - e^{-j\omega} \\
&amp;= e^{-j\omega/2}(e^{-j\omega/2} - e^{-j\omega/2}) \\
&amp;= 2j e^{-j\omega/2} \ \tfrac1{2j}(e^{-j\omega/2} - e^{-j\omega/2}) \\
&amp;= e^{-j\omega/2} 2j  \ \sin(\omega/2) \\
&amp;\approx e^{-j\omega/2} j  \ \omega \qquad \text{for } |\omega| \ll \pi \\
\end{align}$$
So you have a delay of $\frac12$ sample, an additional phase shift of +90° (due to the $j$ factor), and a gain proportional to frequency (at least for low frequencies, relative to Nyquist).  Other than the half-sample delay, this is what you expect from a differentiator.
If you happen to use an $N$-sample backward difference such as
$$y[n] = x[n] - x[n-N]$$
then the associated frequency response will be
$$H_N(e^{j\omega}) = 1 - e^{-jN\omega} $$
This is similar to the first orde case, but the delay is $\frac{N}2$ samples, and the gain is $N \omega / 2$ .