Name:
Mods:
Date:
2.2a Notes on Definition of a Derivative
Yesterday we looked at what happened when you found the average rate of change (slope between any
two points on a function). Then you estimated the instantaneous rate of change (the slope of a function
as we approach a certain point). Today we will be finding these algebraically.
Average vs. Instantaneous Rate of Change
Example 1:
220
h200
The position of an object in free fall is given by ℎ(𝑡) = −16𝑡 2 + 200,
(1,184)
fx = -16 x2+200
180
160
where h is measured in feet and t is measured in seconds.
Find the average rate of change (velocity) from t = 1 to t = 3.
140
120
y
100
80
(3,56)
60
x
40
20
1
2
t
-20
Example 2:
(3,h(3))
Find the velocity at t = 3 (instantaneous rate of change).
To find a slope, we choose a point just beyond 3, we call this 3 + h.
y=h(3+h)-h(3)
Then we take the limit of the function we create as h approaches 0.
𝛥𝑦 ℎ(3 + ℎ) − ℎ(3)
=
(3 + ℎ) − 3
𝛥𝑥
(3+h,h(3+h))
x= 3+h -3
3
Example 3:
Find the slope of f(x) = 𝑥 2 at 𝑥 = 4. This is the derivative at 4 or f’(4)
Δ𝑦
=
Δ𝑥
Example 4:
Find the slope of g(𝑥) = 𝑥 2 − 4𝑥 + 1 at x = 3.
The slope of the curve y = f(x) at x=a is:
𝑚=
Homework: Worksheet 2.2a
Δ𝑦
𝑓(𝑎 + ℎ) − 𝑓(𝑎)
= 𝑓 ′ (𝑎) = lim
ℎ→0
Δ𝑥
ℎ