What is the rate of change of amplitude? Frequency?
Fourier transforms and differentiation are indeed very different, but your intuition is still not entirely wrong. Here's how to make sense of it.

Let's assume you have a signal that is essentially sine-shaped locally, i.e. we can write it as $s(t) = a \sin(2 \pi f t)$ for a times $t$ close to some time $t_0$. In other words: The signal oscillates like a sine and has a sufficiently slowly changing envelope and instantaneous frequency.

The time derivative of this signal is $s'(t) = 2 a \pi f \cos(2 \pi f t)$. This signal is also oscillating with the same frequency and its amplitude is scaled by $2\pi f$ compared to the original signal. So the frequency is encoded in the derivative. 

There are a few different methods for decoding the frequency from this derivative, and I'll describe the one that is probably closest to your intuition. We have to get rid of the oscillating part. So let's take first the magnitude of both signals and afterwards apply a low pass filter $H_l$ that removes the residual oscillation and leaves you with just the average value:

$$H_l(|s(t)|) = a \frac{2}{\pi}$$
$$H_l(|s'(t)|) = 2 a \pi f \frac{2}{\pi}$$

( The 2/$\pi$  comes from averaging over $|\sin(t)|$. ) With this, we can find the quotient of the two results and have

$$\frac{H_l(|s'(t)|)}{H_l(|s(t)|)}=2\pi f$$

which demonstrates that the derivative of the signal in fact yields a very accurate approximation of the frequency under the assumptions we made.

For more general signals this method calculates not the instantaneous frequency but the so called spectral centroid frequency.