# Derivative: Definition

How to find the derivative of a function at the point on the line by using the definition of the derivative: definition, 1 example, and its solution.

## Definition

### Definition

Think of two points on the graph y = f(x):

(a, f(a)) and (a + h, f(a + h)).

Draw a line that passes through these two points.

The slope of the line is

[f(a + h) - f(a)]/[(a + h) - a]

= [f(a + h) - f(a)]/h.

The change of x is h.

Then, as h → 0,

(a + h, f(a + h)) goes to (a, f(a)).

Then the line becomes the tangent of y = f(x) at x = a.

Limit of a Sequence

f'(a), f prime a, is the slope of y = f(x) at x = a.

So f'(a) = lim_{h → 0} [f(a + h) - f(a)]/h.

This is the definition of the derivative f'(a).

## Example

### Example

### Solution

f'(2) = lim_{h → 0} [f(2 + h) - f(2)]/h.

f(x) = x^{2} - 3x + 1

Then f(2 + h) = (2 + h)^{2} - 3(2 + h) + 1.

And f(2) = 2^{2} - 3⋅2 + 1.

(2 + h)^{2} = 2^{2} + 2⋅2⋅h + h^{2} = 4 + 4h + h^{2}

Square of a Sum

-3(2 + h) = -6 - 3h

Multiply a Monomial and a Polynomial

-[2^{2} - 3⋅2 + 1] = -[4 - 6 + 1] = -4 + 6 - 1

Cancel 4 and -4.

Cancel -6 and +6.

And cancel +1 and -1.

+4h - 3h = +h

Then lim_{h → 0} [h^{2} + h]/h.

Divide both of the numerator and the denominator by h.

Then lim_{h → 0} [h + 1]/1.

Then lim_{h → 0} [h + 1]/1 = [0 + 1]/1 = 1.

So f'(2) = 1.

### Graph

This curve is the graph of y = f(x):

y = x^{2} - 3x + 1.

f'(2) = 1 means

the slope of y = f(x) at x = 2 is 1.