Calculate the Inverse DTFT of the DTFT Derivative in Terms of $ x \left[ n \right] $
This is pretty straight forward using the definition of the Discrete Time Fourier Transform (DTFT).

The definition of the DTFT:

$$ X \left( {e}^{j \omega} \right) = \sum_{m = -\infty}^{\infty} x \left[ m \right] {e}^{-j \omega m} $$

Differentiating with respect to $\omega$:

$$\begin{align*}
\frac{d}{d \omega} X \left( {e}^{j \omega} \right) &amp; =  \sum_{m = -\infty}^{\infty} \frac{d}{d \omega} x \left[ m \right] {e}^{-j \omega m} \\
&amp; = \sum_{m = -\infty}^{\infty} \left( -j m \right) x \left[ m \right] {e}^{-j \omega m} \\
&amp; = \frac{1}{j} \sum_{m = -\infty}^{\infty} m x \left[ m \right] {e}^{-j \omega m} \\
\end{align*}$$

Now, just plug it into the Inverse DTFT:

$$\begin{align*}
\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{1}{j} \sum_{m = -\infty}^{\infty} m x \left[ m \right] {e}^{-j \omega m} {e}^{j \omega n} d \omega &amp; = \frac{1}{j} \sum_{m = -\infty}^{\infty} m x \left[ m \right] \frac{1}{2 \pi} \int_{0}^{2 \pi} {e}^{-j \omega m} {e}^{j \omega n} d \omega \\
&amp; = \frac{1}{j} \sum_{m = -\infty}^{\infty} m x \left[ m \right] \delta \left[ n - m \right] \\
&amp; = \frac{n}{j} x \left[ n \right]
\end{align*}$$

This implies:

$$ j \frac{d}{d \omega} X \left( {e}^{j \omega} \right) = j \frac{n}{j} x \left[ n \right] = n x \left[ n \right] $$