1. Definition of the Derivative Using Average Rate (Page 129 - 133 and 160 in the book)
f(a+h)
Slope of the line =
Average Rate of Change =
f(a+h) – f(a) h
f(a+h) – f(a)
f(a) h h a
a+h

2. Now, Watch what happens when: Point a is fixed and the size of the interval h shrinks
a+h h a
a+h h a
a+h h a a h

3. h f(a+h) f(a) a+h a As h shrinks and approaches zero (but not = 0),
the line becomes a Tangent Line h
f(a+h)
f(a)
a+h a
Slope of the line =
Average Rate of Change =
f(a+h) – f(a) h
Slope of the Tangent line =
f(a+h) – f(a) h
As h approaches zero

4. The slope of the Tangent Line at a is the Derivative, f ' (a)
As h approaches zero,
or:
f(a+h) – f(a) h =
f(a) a
f(a+h) – f(a) h
= lim h
lim: Limit, as h approaches zero
The slope of the Tangent Line at a is the Derivative, f ' (a)
f(a+h) – f(a) h
lim h
f ' (a) =

5. If f(x) = -2x + 3, then f ' (x) = -2
Example: Use the definition of the derivative to obtain the following result:
If f(x) = -2x + 3, then f ' (x) = -2
f(x+h) – f(x) h
f ' (x) =
lim
h 0
Solution: Using the definition
f (x + h) = -2(x + h) + 3
= (-2x - 2h + 3)
f (x + h) – f (x) h
f ' (x) =
lim
h 0 =
(-2x - 2h + 3) – (-2x + 3) h
lim
h 0 =
(-2h) h
lim
h 0
= -2

6. If f(x) = x2 - 8x + 9, then f ' (x) = 2x - 8
Example: Use the definition of the derivative to obtain the following result:
If f(x) = x2 - 8x + 9, then f ' (x) = 2x - 8
f(x+h) – f(x) h
f ' (x) =
lim
h 0
Solution: Using the definition
f (x + h) = (x + h)2 - 8(x + h) + 9
= (x2 + 2xh + h2 - 8x -8h + 9)
f (x + h) – f (x) h
f ' (x) =
lim
h 0 =
(x2 + 2xh + h2 - 8x - 8h + 9) – ( x2 - 8x + 9) h
lim
h 0 =
(2xh + h2 - 8h) h
lim
h 0 =
h (2x + h - 8) h
lim
h 0 =
(2x + h - 8)
lim
h 0
= 2x - 8