How to prove: high-slope/more change is high freq part of signal
The Fourier transform of the derivative of a signal $x(t)$ is given by

$$\dot{x}(t)\Longleftrightarrow j\omega X(\omega)\tag{1}$$

where $\dot{x}(t)$ denotes the derivative of $x(t)$, and $X(\omega)$ is the Fourier transform of $x(t)$. From (1) we have

$$\dot{x}(t)=\frac{j}{2\pi}\int_{-\infty}^{\infty}\omega X(\omega)e^{j\omega t}\,d\omega\tag{2}$$

If we now assume that $x(t)$ is band-limited with cut-off frequency $\omega_c$ (in radians), then (2) becomes

$$\dot{x}(t)=\frac{j}{2\pi}\int_{-\omega_c}^{\omega_c}\omega X(\omega)e^{j\omega t}\,d\omega\tag{3}$$

From (3) we can determine an upper bound for $|\dot{x}(t)|$:

$$|\dot{x}(t)|\le \frac{1}{2\pi}\int_{-\omega_c}^{\omega_c}|\omega| |X(\omega)|\,d\omega
\le\frac{\omega_c}{2\pi}\int_{-\omega_c}^{\omega_c}|X(\omega)|\,d\omega\tag{4}$$

The upper bound (4) on the magnitude of the derivative of $x(t)$ increases with increasing $\omega_c$, so you can expect faster changes of $x(t)$ with increasing values of the cut-off frequency $\omega_c$.