Lesson Plan: Introduction to Derivatives
Objective
Students will learn both the Geometric and Physical interpretations of derivatives. They will also learn to
compute derivatives using limit definitions along with fundamental rules for solving them. Lastly, they
will learn about real life applications of derivatives through examples and simulations.
Outline
1. Geometric interpretation of derivatives (10 minutes)
lim
∆𝑓
∆𝑥→0 ∆𝑥
= lim
∆𝑥→0
𝑓(𝑥0 + ∆𝑥) − 𝑓(𝑥0 )
= 𝑓′(𝑥𝑜 )
∆𝑥
Illustrate how to apply this definition to find any derivative by working through multiple
examples in class.
a. 𝑓(𝑥) = 𝑥 2
(𝑥 + ℎ)2 − 𝑥 2
ℎ→0
ℎ
𝑓 ′ (𝑥) = lim
2𝑥ℎ + ℎ2
ℎ→0
ℎ
𝑓 ′ (𝑥) = lim
𝑓 ′ (𝑥) = lim 2𝑥 + ℎ
ℎ→0
𝑓 ′ (𝑥) = 2𝑥
1
b. 𝑓(𝑥) = 3𝑥 + 2
3(𝑥 + ℎ) + 2 − (3𝑥 + 2)
ℎ→0
ℎ
𝑓 ′ (𝑥) = lim
𝑓 ′ (𝑥) = 3
c. 𝑓(𝑥) = sin (𝑥)
𝑓 ′ (𝑥) = lim
ℎ→0
sin(𝑥 + ℎ) − sin (𝑥)
ℎ
use identity: sin(𝑥 + ℎ) − sin(𝑥) = 2 cos (
2𝑥 + ℎ
ℎ
) sin ( )
2
2
𝑓 ′ (𝑥) = cos (𝑥)
1
d. 𝑓(𝑥) = 𝑥
𝑓
′ (𝑥)
1
1
−𝑥
𝑥
+
ℎ
= lim
ℎ→0
ℎ
𝑓 ′ (𝑥) = lim −
ℎ→0
𝑓 ′ (𝑥) = −
1
𝑥(𝑥 + ℎ)
1
𝑥2
2. Differentiation Rules (25 minutes)
a. Power Rule:
𝑑(𝑥 𝑛 )
= 𝑛𝑥 𝑛−1 , 𝑓𝑜𝑟 𝑛 ≠ 0
𝑑𝑥
𝑑
𝑑
(𝑘 ∗ 𝑓(𝑥)) = 𝑘 ∗ 𝑑𝑥 𝑓(𝑥)
𝑑𝑥
b. Scalar Multiplication Rule:
c. Constant Rule:
𝑑
[𝑐] = 0
𝑑𝑥
d. Additive Rule and Subtraction Rules:
e. Product Rule:
𝑑
𝑑
𝑑
(𝑓(𝑥) ± 𝑔(𝑥) = 𝑓(𝑥) ± 𝑔(𝑥)
𝑑𝑥
𝑑𝑥
𝑑𝑥
If 𝑦 = 𝑢 ∗ 𝑣, 𝑤ℎ𝑒𝑟𝑒 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑥
𝑑𝑦
𝑑𝑣
𝑑𝑢
then 𝑑𝑥 = 𝑢 ∗ 𝑑𝑥 + 𝑣 ∗ 𝑑𝑥
2
f.
Quotient Rule:
𝑢
𝑣
If 𝑦 = , 𝑤ℎ𝑒𝑟𝑒 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑥
𝑑𝑦
then 𝑑𝑥 =
g. Chain Rule:
𝑑𝑢
𝑑𝑣
−𝑢∗
𝑑𝑥
𝑑𝑥
𝑣2
𝑣∗
If y is a function of v, and v is a function of x, then y is a
function of x and
h. Inverse function Rule
𝑑𝑦
𝑑𝑦 𝑑𝑣
= ∗
𝑑𝑥
𝑑𝑣 𝑑𝑥
𝑑𝑦
1
If 𝑥 = 𝑓(𝑦) then 𝑑𝑥 = 𝑑𝑥
𝑑𝑦
Illustrate the application of above rules through examples in class.
a. 𝑦 = 4𝑥 3
𝑑𝑦
= (4) ∗ (3) ∗ 𝑥 3−1 = 12𝑥 2 [Power Rule]
𝑑𝑥
b. 𝑦 = 4𝑥 2 − 𝑥 3 − 4𝑥
𝑑𝑦
= (4)(2) ∗ 𝑥 2−1 − 3𝑥 3−1 − 4𝑥 1−1 [Additive Subtraction Rule]
𝑑𝑥
𝑑𝑦
= 8𝑥 − 3𝑥 2 − 4
𝑑𝑥
c. 𝑦 = (4𝑥 3 − 3𝑥 + 2)(2𝑥 2 + 4𝑥)
𝑑𝑦
= (4𝑥 3 − 3𝑥 + 2)(4𝑥 + 4) + (2𝑥 2 + 4𝑥)(12𝑥 2 − 3) [Product Rule]
𝑑𝑥
3𝑥+2
d. 𝑦 = 𝑥 2 +4
(𝑥 2 +4)(3)−(3𝑥+2)(2𝑥)
𝑑𝑦
=
[Quotient Rule]
(𝑥 2 +4)2
𝑑𝑥
𝑑𝑦 −3𝑥 2 − 4𝑥 + 12
=
(𝑥 2 + 4)2
𝑑𝑥
e. 𝑦 = (4𝑥 3 + 3𝑥 − 7)4
Let 𝑣 = 4𝑥 3 + 3𝑥 − 7, then 𝑦 = 𝑣 4 [Chain Rule]
𝑑𝑦
= 4(4𝑥 3 + 3𝑥 − 7)3 ∗ (12𝑥 2 + 3)
𝑑𝑥
3
3. Real life applications of derivatives (10 minutes)
Work through examples with students in class for each real life applications of derivatives.
a. Marginal Cost in Economics
Marginal cost (MC) represents the change in total cost that arises when the quantity
produced changes by one unit.
Let’s consider a company that produces chocolates. The total cost C(q) of producing q
chocolates is given by:
𝑞2
𝐶(𝑞) = 1000 + 5𝑞 +
10
Here, 1000 is the fixed cost, 5q is the variable cost per unit, and
𝑞2
is the additional costs
10
that increase with the square of the quantity produced.
The Marginal Cost MC(q) is the derivative of the total cost function C(q).
𝑑
𝑑
𝑞2
1
𝑀𝐶(𝑞) =
𝐶(𝑞) =
(1000 + 5𝑞 + ) = 5 + 𝑞
𝑑𝑞
𝑑𝑞
10
5
b. Linear Approximations
Tangent line approximation or linearization, involves using the derivative of a function
to approximate its value near a specific point.
Consider the function 𝑓(𝑥) = √𝑥
In order to approximate f(x) near x=9, we first find its derivative.
𝑓 ′ (𝑥) =
𝑑
1
√𝑥 =
𝑑𝑥
2√𝑥
Linear approximation, L(x), is given by:
𝐿(𝑥) = 𝑓(9) + 𝑓 ′ (9) ∗ (𝑥 − 9)
𝐿(𝑥) = √9 + (
1
2√9
) (𝑥 − 9)
c. Instantaneous rate of change: velocity, acceleration
Consider an object moving along a straight line. The position s(t) of the object at time t
4
seconds is given by:
𝑠(𝑡) = 3𝑡 2 − 2𝑡 + 5
Its velocity, v(t) in m/s, is then given by:
𝑣(𝑡) =
𝑑𝑠
𝑑
= (3𝑡 2 − 2𝑡 + 5) = 6𝑡 − 2
𝑑𝑡 𝑑𝑡
Its acceleration, a(t) in m/s2 is given by:
𝑎(𝑡) =
𝑑𝑣
𝑑
= (6𝑡 − 2) = 6
𝑑𝑡 𝑑𝑡
4. Summarize everything that was covered in class and assign homework either from textbook or
handouts. Reinforce the importance of learning derivatives as they form the fundamental building
blocks of most physical sciences and play a critical role in most industry applications. (5 minutes)
5