Basic Calculus (Module 9) - Intermediate Value Theorem

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ACTS COMPUTER COLLEGE
Sta. Cruz, Laguna
Basic Calculus
Prepared by:
MS. CHRISHELYN DACSIL

5/16/2022 1:48 PM
ACTS Computer College | Sta. Cruz, Laguna 6
In this lesson you will:
• apply the definition of the derivative of a
function at a given number.
OBJECTIVES:

Definition of Derivative
The derivative of a function 𝑓(𝑥) denoted 𝑓 ′(𝑥) at any 𝑥
in the domain of the given function is defined as
𝑓′
𝑥 = lim
∆𝑥→0
∆𝑥
∆𝑦
= lim
ℎ→0
𝑓 𝑥 + ℎ − 𝑓(𝑥)
ℎ

Example 1:
𝑓 𝑥 = 𝑥2
+ 3𝑥
𝑓′
𝑥 = lim
ℎ→0
𝑓 𝑥+ℎ −𝑓(𝑥)
ℎ
= lim
ℎ→0
(𝑥+ℎ)2+3(𝑥+ℎ)−(𝑥2+3𝑥)
ℎ
= lim
ℎ→0
𝑥2+2ℎ𝑥+ℎ2+3𝑥+3ℎ−𝑥2−3𝑥
ℎ
= lim
ℎ→0
2ℎ𝑥+ℎ2+3ℎ
ℎ
= lim
ℎ→0
ℎ(2𝑥+ℎ+3)
ℎ
= lim
ℎ→0
2𝑥 + ℎ + 3
= lim
ℎ→0
2𝑥 + 0 + 3
𝒇′
𝒙 = 𝟐𝐱 + 𝟑

Example 2:
𝑓 𝑥 = 5𝑥 − 1, 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑓′(1)
𝑓′
𝑥 = lim
𝑥→𝑥0
𝑓 𝑥 −𝑓(𝑥0)
𝑥−𝑥0
𝑓′
1 = lim
𝑥→1
5𝑥 − 1 − [5 1 − 1]
𝑥 − 1
𝑓′
1 = lim
𝑥→1
5𝑥 − 1 − 4
𝑥 − 1
𝑓′
1 = lim
𝑥→1
5𝑥 − 5
𝑥 − 1
𝑓′
1 = lim
𝑥→1
5(𝑥 − 1)
𝑥 − 1
𝑓′
1 = lim
𝑥→1
5
𝒇′
𝟏 = 𝟓

I. DERIVATIVE OF A CONSTANT FUNCTION
● The graph of a horizontal function is a horizontal
line, and a horizontal line has zero slope. Recall
that the derivative measures the slope of the
tangent line, and so the derivative of a constant
term is zero.
● Example : 𝑦 = 300,000 then the 1st derivative is
𝑦’ = 0

I. DERIVATIVE OF A POWER FUNCTION
A function of the form : 𝑦 = 𝑥𝑛
where n is a real number,
is called a power function. In general, this is called the
POWER RULE
If 𝒚 = 𝒙𝒏
then, 𝒚′
= 𝒏𝒙𝒏−𝟏

POLYNOMIAL FUNCTION
1. 𝑦 = 𝑥
𝑦′
= 1 𝑥1−1
= 𝟏
2. If 𝑓 𝑥 = 𝑥3
Then 𝑓′
𝑥 = 3𝑥3−1
;
𝒇′
𝒙 = 𝟑𝒙𝟐

RATIONAL FUNCTION
Find 𝑔′(𝑥), where 𝑔 𝑥 =
1
𝑥2
𝑔 𝑥 =
1
𝑥2 can be written as: 𝑔 𝑥 = 𝑥−2
𝑔′
𝑥 = −2𝑥−2−1
𝑔′
𝑥 = −2𝑥−3
= −
𝟐
𝒙𝟑

RADICAL FUNCTION
Find the derivative of 𝑦 =
4
𝑥3
SOLUTIONS:
𝑦 =
4
𝑥3
𝑦 = 𝑥
3
4 rewrite the expression to exponential form
𝑦′ =
3
4
𝑥
3
4
−1
apply the power rule
=
3
4
𝑥−
1
4 subtract the exponents
𝑦′ =
3
4𝑥 ൗ
1
4
=
𝟑
𝟒𝟒
𝒙
simplify to radical form

RADICAL FUNCTION
Given ℎ 𝑥 =
1
3
𝑥
; find ℎ′(𝑥)
SOLUTIONS:
ℎ 𝑥 =
1
3
𝑥
ℎ 𝑥 =
1
𝑥 ൗ
1
3
= 𝑥 Τ
−1
3 rewrite to exponential form
ℎ′
𝑥 = −
1
3
𝑥 Τ
−1
3−1
apply to power rule
= −
1
3
𝑥 Τ
−4
= −
1
3𝑥 ൗ
4
3
simplified radical expression
= −
𝟏
𝟑
𝟑
𝒙𝟒

CONSTANT MULTIPLE RULE
States that the derivative of a constant times a
differentiable function is the constant times the
derivative of the function.
If 𝑦 = 𝑘𝑓(𝑥) where k is constant ( k is the numerical
coefficient of the function of x ) ; then :
𝑦 ‘ = 𝑘 • 𝑓 ‘ ( 𝑥 )

Given: ℎ 𝑥 = 5𝑥 Τ
3
4; 𝑓𝑖𝑛𝑑
𝑑𝑦
𝑑𝑥
SOLUTION : Let 𝑦 = ℎ ( 𝑥 )
𝑦 = 5𝑥 Τ
3
4
𝑦 = 5 ∙ 𝑥 Τ
3
4 rewrite in the form
𝑘 ∙ 𝑥𝑛
𝑦′
= 5 ∙
3
4
𝑥
3
4
−1
apply constant
multiple
and power rule
=
15
4
𝑥−
1
4 subtract exponents
and combine similar terms
=
15
4𝑥
1
4
expressed in positive
exponent
𝑦′ =
15
44
𝑥
simplest
radical form

Given: 𝑔 𝑥 =
3
𝑥
3
SOLUTION:
𝑔 𝑥 =
3
𝑥
3
; 𝑔 𝑥 =
1
3
𝑥
1
3 rewrite in the form 𝑘 ∙ 𝑥𝑛
𝑔′ 𝑥 =
1
3
∙
1
3
𝑥
1
3
−1
apply constant multiple and power rle
=
𝑥
−
2
3
9
subtract exponents and combine similar
terms
𝑔′ 𝑥 =
1
9𝑥
2
3
=
𝟏
𝟗
𝟑
𝒙𝟐
in positive exponent to simplest radical
form

SUM / DIFFERENCE RULE
Given two differentiable functions g and h, if 𝑦 =
𝑔 𝑥  ℎ 𝑥 , 𝑡ℎ𝑒𝑛 ∶ 𝑦 ′
= 𝑔′
𝑥 ℎ′ ( 𝑥 )
EXAMPLES:
Given: 𝑓 𝑥 = 5𝑥 Τ
3
4 ; 𝑔 𝑥 =
1
3
3
𝑥 𝑎𝑛𝑑 ℎ 𝑥 = − 3𝑥
From the given functions above find the following :
1) 𝑓 ‘ ( 𝑥 ) + 𝑔 ‘ ( 𝑥 )
2) 𝑔 ‘ ( 𝑥 ) − ℎ ‘ ( 𝑥 )

𝑓 𝑥 + 𝑔 𝑥 = 5𝑥
3
4 +
1
3
3
𝑥 combine the given functions by addition as indicated
= 5 ∙ 𝑥
3
4
−1
+
1
3
∙ 𝑥
1
3 apply the constant multiple rule for each function
𝑓′
𝑥 + 𝑔′
𝑥 = 5 ∙
3
4
𝑥
3
4
−1
+
1
3
∙
1
3
𝑥
1
3
−1
use the power
=
15
4
∙ 𝑥−
1
4 +
1
9
∙ 𝑥−
2
=
15
4𝑥
1
4
+
1
9
3
𝑥2
expressed the terms with positive exponents
𝑓′
𝑥 + 𝑔′
𝑥 =
15
4𝑥
1
4
+
1
9
3
𝑥2
simplified radical expression

𝑔 𝑥 − ℎ 𝑥 =
1
3
3
𝑥 − (− 3𝑥) combine the functions by addition as indicated
=
1
3
∙ 𝑥
1
3 + 3 ∙ 𝑥 apply the constant multiple rule for each function
𝑔′
𝑥 − ℎ′
𝑥 =
1
3
∙
1
3
𝑥
1
3
−1
+ 3 1 𝑥1−1
use the power rule
=
1
9
∙ 𝑥−
2
3 + 3 ∙ 𝑥0 subtract the exponents
=
1
9𝑥
2
3
+ 3 ∙ 1 expressed the terms with positive exponents
𝑔′ 𝑥 − ℎ′ 𝑥 =
1
9
3
𝑥2
+ 3 simplest radical form

DERIVATIVE OF EXPONENTIAL AND LOGARITHMIC
FUNCTIONS
● In the previous study of the different function graphs you have learned that
exponential functions play an important role in modeling population
growth and the decay of radioactive materials. Logarithmic functions can
help rescale large quantities and are particularly helpful for rewriting
complicated expressions.
● Just as when we found the derivatives of algebraic functions, we can also
find the derivatives of exponential and logarithmic functions using
formulas.

DIFFERENTIATION FORMULAS : EXPONENTIAL AND
LOGARITHMIC FUNCTIONS
Derivative of Exponential Functions :
If 𝑓 𝑥 = 𝑒𝑥
, then 𝑓′
𝑥 = 𝑒𝑥
If 𝑓 𝑥 = 𝑏𝑥, then 𝑓′ 𝑥 = 𝑏𝑥 ln 𝑏
Derivative of Logarithmic Functions :
If 𝑓 𝑥 = 𝑙𝑜𝑔𝑏𝑥, then 𝑓′ 𝑥 =
1
𝑥 ln 𝑏
Derivative of Natural Logarithm :
If 𝑓 𝑥 = ln 𝑥, then 𝑓′ 𝑥 =
1
𝑥

EXAMPLES : Determine the derivative of the following functions :
1. 𝑦 = 𝑒−𝑥
SOLUTIONS:
𝑦 =
1
𝑒𝑥 expressed with positive exponent
𝑑𝑦
𝑑𝑥
=
𝑒𝑥∙𝑑 1 ∙[1∙𝑑(𝑒𝑥)
(𝑒𝑥)2 apply the quotient rul
=
𝑒𝑥(0)∙(1)(𝑒𝑥)
(𝑒𝑥)2 simplify the terms
𝑑𝑦
𝑑𝑥
=
−𝑒𝑥
(𝑒𝑥)(𝑒𝑥)
= −
1
𝑒𝑥 derivative in simplest form

2. 𝑓 𝑥 = 2𝑥 ∙ 𝑒𝑥
SOLUTIONS: Let 𝑢 = 2𝑥 and 𝑣 = 𝑒𝑥
𝑓 𝑥 = 2𝑥 ∙ 𝑒𝑥 given
𝑓′ 𝑥 = 2𝑥 ∙ 𝑑 𝑒𝑥 + 𝑒𝑥 ∙ 𝑑(2𝑥) apply the product formula
= 2𝑥𝑒𝑥 + 𝑒𝑥 ∙ 2𝑥 ln 2 combine similar terms
𝑓′
𝑥 = 2𝑥
𝑒𝑥
(1 + ln 2) factor and simplify

3. 𝑔 𝑥 = 𝑥2 + (−2 ln 𝑥)
SOLUTIONS:
𝑔 𝑥 = 𝑥2 − 2 ln 𝑥 clear off parenthesis
𝑔′
𝑥 = 2𝑥 − 2 ∙
2
𝑥
apply the differentiation formula
= 2𝑥 −
2
𝑥
combine similar terms
𝑔′
𝑥 =
2𝑥2−2
𝑥
simplified into a single
expression

Basic Calculus (Module 9) - Intermediate Value Theorem

Basic Calculus (Module 9) - Intermediate Value Theorem

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