You are here:



When calculating the "rate of change"  increase in x (dx) will suddenly change to "0" why? Besides, why only "dx" change to  0. And increase in y(dy) is not. Please sir, i'm curious to know!!!

Hi Johnny,
To understand differentiation from first principles, it is important that you get familiar with the idea of limits.
y = f(x)
y + Δy = f(x + Δx)
Subtracting the two equations and then dividing by Δx leads to
Δy/Δx = [f(x + Δx) - f(x)] / Δx
Now, Δy/Δx is the ratio of the changes in y and x. When Δx becomes very small, the ratio approaches the slope of the function f(x). And so we define dy/dx (which is not exactly a ratio but a notation) as the limit of Δy/Δx (i.e what it approaches) as Δx goes to zero. This now represents the slope function of f(x).
You should note that dx is not the same thing as Δx and in the expressions it is the latter that approaches 0, written as Δx → 0. It doesn't suddenly change to zero, we just try to find what happens to the ratio of changes when the changes are very very small. Since y is a function of x, the change in y, Δy, depends on the change in x, Δx, and so Δy doesn't exclusively need to go to zero in the considerations.

I hope it is clearer now.



All Answers

Answers by Expert:

Ask Experts


Ahmed Salami


I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I believe i would be very helpful in calculus and can as well help a good deal in Physics with most emphasis directed towards mechanics.


Aspiring theoretical physicist. I have been doing maths and physics all my life.

I teach mathematics and engineering physics.

©2016 All rights reserved.