Let us suppose that we are modeling a particle traversing some distance in space & that f(x)=x². Let the x-axis be time (hrs) & the y-axis be position in space (miles). What is the instantaneous velocity of said particle at t=4 hours? Let's assign variables to each coordinate in our x, y table. So A(2,4), B(3,9), C(4,16), etc. & see what we can figure out. From A to B, the particle traversed a distance of 5 miles, hence 9-4=5. From B to C, the particle traversed a distance of 7 miles. C to D, 9 miles, & so on. What's the particle doing? The particle is traversing a larger & larger distance over each time interval. It's accelerating. Well at what rate? At a rate of 2mph/hr. In terms of variables, the y variable is accelerating twice as fast with respect to the x variable, hence f'(x)=2x. Using this argument, I can solve the velocity (the slope) at any point on the curve x², including x=4, which is 8mph. The wonderful thing here is that you could reason this problem out without using any calculus. No need for limits here. You can also use this argument to find the total distance traversed over any time interval. My younger sibling told me I wasn't smart enough to reason out a problem involving a differential equation & so farshe's right! I can find h(t) of a falling body using f=ma, but I can't reason it out like I can do for the differential & the integral calculus. I'm sure there are many individuals on this forum that are much brighter than I am. Maybe give it a go? & don't feel limited to falling bodies either... Thanks! *I am aware that acceleration is typically used to describe the 2nd derivative, especially in physics problems like the one I listed. I have used acceleration to describe the 1st derivative ...even though that's exactly what acceleration is ...the rate at which something is changing with respect to the other. Anyways!