So in continuous time, the relationship is:
$$\mathcal{F}(\frac{dx(t)}{dt}) \leftrightarrow j\omega X(e^{j\omega})$$ And from: $$\mathcal{F}(x(t)) = X(e^{j\omega})$$ We can recover $F(x(t))$ from the relationship by simply dividing to get: $$\frac{\mathcal{F}(\frac{dx(t)}{dt})}{j\omega} = \mathcal{F}(x(t))$$
But I wanted to see if I could make the relationship work in the case of Discrete Time to recover $X(e^{j\omega})$, so I started by breaking everything down into finite terms: $$given: \ \ \ \ n = [0,N]; x[n] \begin{cases} x[n] = 0; n>N_1 \\ x[n] = 0;n<N_0 \\ x[n] \ne 0;|N_0| \le n \le |N_1| \end{cases}; \frac{\Delta f[x]}{\Delta x} = \frac{f(x+a)-f(x-a)}{2(x-a)};\\ and: \ \mathcal{F}(x[n_i]) = X(e^{j\omega})_i = \sum_{n=-\infty}^{+\infty}x[n]e^{-j\omega n}$$
$$Then:\mathcal{F}(\frac{\Delta x[n_i]}{\Delta n}) = F\bigg(\frac{x[n_{i+1}]-x[n_{i-1}]}{2(\frac{n_N-n_0}{N-1}=\Delta n)}\bigg)_i = \frac{1}{2\Delta n}\bigg(\sum_{n=N_0+1}^{N_1-1} (x[n_{i+1}] - x[n_{i-1}])e^{-1j\omega_i n}\bigg) = \frac{1}{2\Delta n}\bigg(X(e^{j\omega})_{i+1}-X(e^{j\omega})_{i-1}\bigg)= \frac{1}{2\Delta n}\bigg(\Delta X(e^{j\omega})_{i}\bigg)\\$$
Or would it be: $$\frac{1}{2\Delta n}\bigg(\sum_{n = N_0+1}^{N_1-1}(x[\color{red}{n}]e^{j\omega_{\color{red}{i+1}}n}-x[\color{red}{n}]e^{j\omega_{\color{red}{i-1}}n})\bigg) = \frac{1}{2\Delta n}\bigg(X(e^{j\omega_{\color{red}{i+1}}})-X(e^{j\omega_{\color{red}{i-1}}})\bigg)= \frac{1}{2\Delta n}\bigg(\Delta X(e^{j\omega})\bigg)_i$$
Regardless, how would you recover $X(e^{j\omega n})_i$ if all we can recover instead is $2\Delta n \mathcal{F}(\frac{\Delta x[n_i]}{\Delta n}) = \Delta X(e^{j\omega})_i$? What am I supposed to do with $\Delta X(e^{j\omega})_i$ to recover $X(e^{j\omega})_i$? Or is my formulation entirely wrong?
And are there any examples I can attempt to perform this on?
I had the formulation wrong, the solution is worked out as follows:
$$\mathcal{F\bigg(\frac{x[n]-x[n-1]}{\Delta n}}\bigg) = \frac{1}{\Delta n}\bigg(\sum_{n = -\infty}^{+\infty}x[n]e^{-1j\omega} +\sum_{n = -\infty}^{+\infty}x[n-1]e^{-1j\omega} \bigg)$$ and we substitute m = n-1 and n = m+1 gives: $$\frac{1}{\Delta n}\bigg(\sum_{n = -\infty}^{+\infty}x[n]e^{-1j\omega n} -\sum_{m = -\infty-1=-\infty}^{\infty-1=\infty}x[m]e^{-1j\omega(m+1)} \bigg) \\ = \frac{1}{\Delta n}\bigg(\sum_{n = -\infty}^{+\infty}x[n]e^{-1j\omega n} -e^{-j\omega}\sum_{m = -\infty-1=-\infty}^{\infty-1=\infty}x[m]e^{-1j\omega(m)} \bigg) \\$$ $$\frac{1-e^{-j\omega}}{\Delta n}\bigg(X(e^{j\omega}) \bigg)$$