Why is the time-domain derivative equivalent to multiplication by frequency ($s$) in the Laplace transform?
Why is the time-domain integral equivalent to division by frequency ($\frac{1}{s}$) in the Laplace transform?
Intuitively, I thought the reason was that the frequency was a sort of "rate of change", so it was somehow equivalent to $\frac{df}{dt}$. The Laplace transform turns rate of change into a variable ($s$), and holds that rate of change constant throughout the problem, which is why algebraic manipulations in the frequency domain are possible.
Am I on the right track?