# Derivative and difference (intervals) between values

Having only a high-school math background (~30 years ago), I'm having issues with the idea of derivative.

I understand that it's to be considered as a sort of "acceleration" of the values of a function, i.e. the rate at which such values change. Intuitively, this explains why a straight line has a derivative of 0.

Does this mean that it's possible to calculate one value from the previous one by using the increment specified in the derivative?

For example, let's say we have a simple function, maybe

$$f(x)=x^2$$

The values of f(x) for x = 1, 2, 3, 4, 5, ... are 1, 4, 9, 16, 25, .... The increments between each value are 3, 5, 7, 9, ..., but the derivative is 2x, which means that the increments should be 2, 4, 6, 8, ... ?

I know that I'm missing something, I just can't see what.

• Functions have derivatives. A straight line is not a function, so it doesn't have a derivative. A straight line is the graph of a function, and that function has a derivative, but that derivative isn't zero unless the straight line is horizontal. – Gerry Myerson Oct 7 at 12:11
• Let's write $\delta$ for an increment. Then $f(x+\delta)$ is approximately $f(x)+\delta f'(x)$, and the approximation gets better as the increment gets smaller. In your example the increment is one, which isn't very small, so the approximation isn't very good. Try it for $x=1$ and $x=1.01$, you should see the approximation using the derivative is very good. – Gerry Myerson Oct 7 at 12:14

$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$
which is essentially $$\frac{\text{rise}}{\text{run}}$$ when the points are very, very close together.
Bonus question: given your function $$f(x) = x^2$$, can you use the limit definition of the derivative to prove that $$f'(x) = 2x$$?