TITLE: Why is the derivative multiplication by frequency in Laplace transform?
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: I had a similar question here and came to the following conclusion:
The Laplace transform involves the exponential function: e^n*x.
Differentiating this means ne^nx - which is simply multiplication by n.
Integrating this means 1/ne^nx - which is simply division by n.
This holds true for power series which are the discrete form of integral transforms (sort of) and originally stems from the power rule (for differentiation) - although you would have a pesky division by the base using "ordinary" power terms. This is prevented by using the exponential function.