merc, 1 year ago There’s no reason it shouldn’t work. dy/dx is the same as (y1 - y2) / (x1 - x2) as the distance between the two points approaches zero. “dx” and “dy” aren’t very useful measurements on their own though.
There’s no reason it shouldn’t work.
dy/dx is the same as (y1 - y2) / (x1 - x2) as the distance between the two points approaches zero. “dx” and “dy” aren’t very useful measurements on their own though.