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UNIT I-Module 7 Differentiation of Logarithmic and exponential Functions

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UNIT I - THE DERIVATIVES
Module 7: Differentiation of Logarithmic and Exponential Functions
Introduction
In this topic, we will discuss differentiation of the other set of transcendental functions, which are
logarithmic and exponential functions
Learning Objectives
At the end of this topic, you should be able to demonstrate the following:
a. Determine the derivative of logarithmic functions;
b. Find the derivative of exponential functions; and
c. Differentiate a function raised to a function by logarithmic differentiation.
Presentation of Contents
Differentiation of Logarithmic Functions
Let u be a differentiable function of x, that is, u = g(x). The following formulas are used in differentiating
logarithmic functions.
๐’…
(๐’๐’๐’ˆ๐’ƒ ๐’–)
๐’…๐’™
1.
๐Ÿ
๐’–
= ๐’๐’๐’ˆ๐’ƒ ๐’† .
๐’…๐’–
๐’…๐’™
If b = 10, ๐‘กโ„Ž๐‘’๐‘› ๐‘™๐‘œ๐‘”๐‘ ๐‘ข = ๐‘™๐‘œ๐‘”๐‘ข called the common logarithm. The formula is reduced to
๐’…
(๐’๐’๐’ˆ๐’–)
๐’…๐’™
๐Ÿ
๐’–
= log๐’† .
๐’…๐’–
๐’…๐’™
If b = e, ๐‘กโ„Ž๐‘’๐‘› ๐‘™๐‘œ๐‘”๐‘’ ๐‘ข = ๐‘™๐‘›๐‘ข called the natural logarithm. The formula is reduced to
๐’…
๐Ÿ
2. ๐’…๐’™ (๐ฅ๐ง ๐’–) = ๐’– .
๐’…๐’–
๐’…๐’™
EXAMPLE 1
๐‘‘๐‘ฆ
a. Find ๐‘‘๐‘ฅ if y = ๐‘™๐‘œ๐‘”5 (3๐‘ฅ 2 − 5).
๐‘‘๐‘ฆ
y = ๐‘™๐‘œ๐‘”5 (3๐‘ฅ 2 − 5) takes the form of y = ๐‘™๐‘œ๐‘”๐‘ ๐‘ข with b = 5 and u = 3๐‘ฅ 2 − 5. To find ๐‘‘๐‘ฅ , we use the
formula
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
=
=
=
=
๐‘‘
๐‘‘๐‘ฅ
1
๐‘‘๐‘ข
(๐‘™๐‘œ๐‘”๐‘ ๐‘ข) = ๐‘ข ๐‘™๐‘œ๐‘”๐‘ ๐‘’ . ๐‘‘๐‘ฅ
1
3๐‘ฅ2 −5
1
3๐‘ฅ2 −5
๐‘‘
2
๐‘™๐‘œ๐‘”5 ๐‘’ . ๐‘‘๐‘ฅ (3๐‘ฅ − 5)
๐‘™๐‘œ๐‘”5 ๐‘’ . (6x)
6๐‘ฅ๐‘™๐‘œ๐‘”5 ๐‘’
3๐‘ฅ2 −5
by substitution
b. Find
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
if y = log ๐‘ ๐‘–๐‘›2 4๐‘ฅ.
๐‘‘๐‘ฆ
Again, y = log ๐‘ ๐‘–๐‘›2 4๐‘ฅ takes the form of y = log ๐‘ข with u =๐‘ ๐‘–๐‘›2 4๐‘ฅ. To find ๐‘‘๐‘ฅ , we use the formula
๐‘‘
(๐‘™๐‘œ๐‘”๐‘ข)
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
1
๐‘ข
= log๐‘’ .
1
๐‘‘๐‘ข
๐‘‘๐‘ฅ
by substitution
๐‘‘
= ๐‘ ๐‘–๐‘›2 4๐‘ฅ log๐‘’ . ๐‘‘๐‘ฅ (๐‘ ๐‘–๐‘›2 4๐‘ฅ)
log๐‘’
๐‘‘
= ๐‘ ๐‘–๐‘›2 4๐‘ฅ . ๐‘‘๐‘ฅ [(sin 4๐‘ฅ)2 ]
log๐‘’
by GPR
๐‘‘
= ๐‘ ๐‘–๐‘›2 4๐‘ฅ . 2 (sin 4๐‘ฅ)2−1 ๐‘‘๐‘ฅ (sin 4๐‘ฅ)
๐‘‘๐‘ฆ
log๐‘’
๐‘‘
= ๐‘ ๐‘–๐‘›2 4๐‘ฅ . 2 sin 4x [cos 4๐‘ฅ . ๐‘‘๐‘ฅ (4๐‘ฅ) ]
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
log๐‘’
= ๐‘ ๐‘–๐‘›2 4๐‘ฅ . 2 sin 4x [cos 4๐‘ฅ . (4)]
=
=
๐‘‘๐‘ฅ
8log๐‘’ sin 4๐‘ฅ cos 4๐‘ฅ
๐‘ ๐‘–๐‘›2 4๐‘ฅ
8log๐‘’๐‘๐‘œ๐‘  4๐‘ฅ
= 8log๐‘’ cot 4๐‘ฅ
sin 4๐‘ฅ
๐‘‘๐‘ฆ
c. Find ๐‘‘๐‘ฅ if y = ln( 4x + 3).
๐‘‘๐‘ฆ
y = ln( 4x + 3) takes the form of y = ln u with u = 4x + 3. To find ๐‘‘๐‘ฅ , we use the formula
๐‘‘
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
1
๐‘ข
(ln ๐‘ข) = .
1
= 4x + 3.
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ข
๐‘‘๐‘ฅ
๐‘‘
๐‘‘๐‘ฅ
(4๐‘ฅ + 3)
1
= 4x + 3. (4)
4
= 4x + 3
๐‘‘๐‘ฅ
d. Find f ‘(x) if f(x) = ๐‘™๐‘›4 (๐‘ฅ + 3)
f(x) = ๐‘™๐‘›4 (๐‘ฅ + 3) can be expressed in power form as f(x) = [ln(๐‘ฅ + 3)]4
Note: ๐’๐’๐Ÿ’ (๐’™ + ๐Ÿ‘) ≠ ln(๐’™ + ๐Ÿ‘)๐Ÿ’
f(x) = [ln(๐‘ฅ + 3)]4 takes the form of f(x) = ๐‘ข๐‘› where u = ln(๐‘ฅ + 3) and x = 4. To find f ‘(x), we
use the GPR.
๐‘‘๐‘ข
f’(x) = n๐‘ข๐‘›−1 . ๐‘‘๐‘ฅ
u = ln(๐‘ฅ + 3)
๐‘‘๐‘ข
๐‘‘๐‘ฅ
1
๐‘‘
= ๐‘ฅ+3 ๐‘‘๐‘ฅ (๐‘ฅ + 3)
n=4
๐‘‘๐‘ข
1
= ๐‘ฅ+3 (1)
๐‘‘๐‘ฅ
๐‘‘๐‘ข
๐‘‘๐‘ฅ
1
= ๐‘ฅ+3
1
f ‘(x) = 4 [ln(๐‘ฅ + 3)]3 ๐‘ฅ+3
f ‘(x) =
4๐‘™๐‘›3 (x+3)
๐‘ฅ+3
๐‘ฅ2
๐‘‘๐‘ฆ
e. Find ๐‘‘๐‘ฅ given that y = ๐‘™๐‘›๐‘ฅ
๐‘ฅ2
๐‘‘๐‘ฆ
y = ๐‘™๐‘›๐‘ฅ takes the form of a quotient of two functions, so to find for ๐‘‘๐‘ฅ , we use the quotient rule.
๐‘‘๐‘ฆ
=
๐‘‘๐‘ฅ
๐‘‘(๐‘ฅ)๐‘›′ (๐‘ฅ)−๐‘›(๐‘ฅ)๐‘‘′(๐‘ฅ)
[๐‘‘(๐‘ฅ)]2
๐‘ฅ2
y = ๐‘™๐‘›๐‘ฅ
n(x) = ๐‘ฅ 2
d(x) = lnx
n’(x) = 2x
d’(x) = ๐‘ฅ ๐‘‘๐‘ฅ (๐‘ฅ)
1 ๐‘‘
1
d’(x) = ๐‘ฅ
๐‘‘๐‘ฆ
=
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
=
=
1
๐‘ฅ
๐‘™๐‘›๐‘ฅ(2๐‘ฅ)−๐‘ฅ 2 ( )
(๐‘™๐‘›๐‘ฅ)2
2๐‘ฅ๐‘™๐‘›๐‘ฅ−๐‘ฅ
๐‘™๐‘›2 ๐‘ฅ
or
๐‘ฅ(2๐‘™๐‘›๐‘ฅ−1)
๐‘™๐‘›2 ๐‘ฅ
Application
Activity 1
๐‘‘๐‘ฆ
Find the ๐‘‘๐‘ฅ of each function below.
3
1. y = log √√12๐‘ฅ
2. y = ๐‘™๐‘›2 ๐‘ฅ − ln(๐‘™๐‘›๐‘ฅ 2 )
3. y = ln(sec ๐‘ฅ + tan ๐‘ฅ)
4. y = ln ๐‘ฅ arctan x
5. xlny = x + ylnx
Differentiation of Exponential Functions
Let u be a differentiable function of x, that is, u = g(x). The following formulas are used in differentiating
exponential functions.
1.
2.
๐’…
๐’…๐’™
๐’…
๐’…๐’™
(๐’‚๐’– ) = ๐’‚๐’– ln a .
(๐’†๐’– ) = ๐’†๐’–
๐’…๐’–
๐’…๐’™
๐’…๐’–
๐’…๐’™
Where: a = constant and a > 0 and a ≠1
e = 2.718281828… ( e is an irrational number)
EXAMPLE 2
๐‘‘๐‘ฆ
a. Find the ๐‘‘๐‘ฅ if y = 34๐‘ฅ
๐‘‘๐‘ฆ
y = 34๐‘ฅ is in the form of y = ๐‘Ž๐‘ข . To find ๐‘‘๐‘ฅ , we use the formula
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐’…
๐’…๐’–
= ๐’…๐’™ (๐’‚๐’– ) = ๐’‚๐’– ln a . ๐’…๐’™
Where a = 3,
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
u = 4x
๐’…
= 34๐‘ฅ ln 3 . ๐’…๐’™ (4x)
= 34๐‘ฅ ln 3 . 4
= 4(34๐‘ฅ )ln3
๐‘‘๐‘ฆ
b. Given y = ๐‘’ 2๐‘ฅ+3 , find ๐‘‘๐‘ฅ .
๐‘‘๐‘ฆ
Again, y = ๐‘’ 2๐‘ฅ+3 is in the form of y = ๐‘’ ๐‘ข . To find ๐‘‘๐‘ฅ , we use the formula
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐’…
= ๐’…๐’™ (๐’†๐’– ) = ๐’†๐’–
๐’…๐’–
๐’…๐’™
u = 2x + 3
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘
= ๐‘’ 2๐‘ฅ+3 ๐‘‘๐‘ฅ (2x + 3)
= ๐‘’ 2๐‘ฅ+3 (2)
= 2 ๐‘’ 2๐‘ฅ+3
c. Find f ‘(x) if f(x) = ๐‘’ ๐‘ฅ ( ๐‘’ 2๐‘ฅ - ๐‘’ ๐‘ฅ )
f(x) = ๐‘’ ๐‘ฅ ( ๐‘’ 2๐‘ฅ - ๐‘’ ๐‘ฅ ) is a product of two functions. Now, instead of using the product rule,
simplify the function using DPMA. That is,
f(x) = ๐‘’ ๐‘ฅ ( ๐‘’ 2๐‘ฅ - ๐‘’ ๐‘ฅ )
using the DPMA
f(x) = ๐‘’ ๐‘ฅ ๐‘’ 2๐‘ฅ - ๐‘’ ๐‘ฅ ๐‘’ ๐‘ฅ
using the product law of exponents
f(x) = ๐‘’ ๐‘ฅ+ 2๐‘ฅ - ๐‘’ ๐‘ฅ+๐‘ฅ
simplifying the exponents
f(x) = ๐‘’ 3๐‘ฅ - ๐‘’ 2๐‘ฅ
Find f ‘(x) using the sum rule
๐‘‘
๐‘‘
f ’(x) = ๐‘‘๐‘ฅ (๐‘’ 3๐‘ฅ ) - ๐‘‘๐‘ฅ (๐‘’ 2๐‘ฅ )
๐‘‘
๐‘‘
f ‘(x) = ๐‘’ 3๐‘ฅ . ๐‘‘๐‘ฅ (3x) - ๐‘’ 2๐‘ฅ . ๐‘‘๐‘ฅ (2x)
f ‘(x) = ๐‘’ 3๐‘ฅ (3) - ๐‘’ 2๐‘ฅ (2)
f ‘(x) = 3๐‘’ 3๐‘ฅ - 2๐‘’ 2๐‘ฅ
๐‘‘๐‘ฆ
d. Given y = √1 + ๐‘’ ๐‘ฅ , find ๐‘‘๐‘ฅ .
1
y = √1 + ๐‘’ ๐‘ฅ in power form is ๐‘ฆ = (1 + ๐‘’ ๐‘ฅ )2 . Since the function is in power form y = ๐‘ข๐‘› ,
๐‘‘๐‘ฆ
we use the GPR to find the ๐‘‘๐‘ฅ .
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ข
= n๐‘ข๐‘›−1 . ๐‘‘๐‘ฅ
1
๐‘ฆ = (1 + ๐‘’ ๐‘ฅ )2
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
u = 1 + ๐‘’๐‘ฅ
1
n=2
๐‘‘๐‘ข
= n๐‘ข๐‘›−1 . ๐‘‘๐‘ฅ
=
=
=
=
1
−1
1
−1
1
−1
๐‘‘
(1 + ๐‘’ ๐‘ฅ ) 2 . (1 + ๐‘’ ๐‘ฅ )
2
๐‘‘๐‘ฅ
๐‘‘
(1 + ๐‘’ ๐‘ฅ ) 2 . (0 + ๐‘’ ๐‘ฅ (๐‘ฅ))
2
๐‘‘๐‘ฅ
(1 + ๐‘’ ๐‘ฅ ) 2 . ๐‘’ ๐‘ฅ
2
๐‘’๐‘ฅ
2 √1+ ๐‘’ ๐‘ฅ
๐‘’ 3๐‘ฅ +1
e. Find f ‘(x) if f(x) = ๐‘’ 3๐‘ฅ −1
๐‘’ 3๐‘ฅ +1
f(x) = ๐‘’ 3๐‘ฅ −1 is in the form of a quotient, so we use the quotient rule.
f ‘(x) =
๐‘‘(๐‘ฅ)๐‘›′ (๐‘ฅ)−๐‘›(๐‘ฅ)๐‘‘′(๐‘ฅ)
[๐‘‘(๐‘ฅ)]2
n(x) = ๐‘’ 3๐‘ฅ + 1
n’(x) = ๐‘’ 3๐‘ฅ
f ‘(x) =
๐‘‘
๐‘‘๐‘ฅ
(3๐‘ฅ) + 0
d(x) = ๐‘’ 3๐‘ฅ − 1
d’(x) = ๐‘’ 3๐‘ฅ
๐‘‘
๐‘‘๐‘ฅ
(3๐‘ฅ) – 0
n’(x) = ๐‘’ 3๐‘ฅ (3)
d’(x) = ๐‘’ 3๐‘ฅ (3)
n’(x) = 3๐‘’ 3๐‘ฅ
d’(x) = 3๐‘’ 3๐‘ฅ
(๐‘’ 3๐‘ฅ −1)3๐‘’ 3๐‘ฅ −(๐‘’ 3๐‘ฅ +1)3๐‘’ 3๐‘ฅ
[๐‘’3๐‘ฅ −1]
2
Using DPMA in the numerator
f ‘(x) =
f ‘(x) =
3๐‘’ 6๐‘ฅ − 3๐‘’ 3๐‘ฅ −3๐‘’ 6๐‘ฅ − 3๐‘’ 3๐‘ฅ
[๐‘’3๐‘ฅ −1]
2
combining similar terms in the numerator
− 6๐‘’ 3๐‘ฅ
[๐‘’3๐‘ฅ −1]
2
Application
Activity 2
๐‘‘๐‘ฆ
Find the ๐‘‘๐‘ฅ for each function below.
3
1. y = √(๐‘’ −2๐‘ฅ + 2๐‘’ ๐‘ฅ )2
2. y = 4๐‘ฅ ln 4๐‘ฅ
3. y = (1 − ๐‘’ −๐‘ฅ )2 ( 1 + ๐‘’ ๐‘ฅ )
4. y = arctan√๐‘’ ๐‘ฅ − 1
๐‘’ 2๐‘ฅ −1
5. y = ln (๐‘’ 2๐‘ฅ +1)
Before we move to the last differentiation technique, observe these three groups of functions.
y = ๐‘ฅ4
y = 4๐‘ฅ
y = ๐‘ฅ๐‘ฅ
y = (2๐‘ฅ + 1)๐‘’
y = ๐‘’ (2๐‘ฅ+1)
๐‘ฆ = (2๐‘ฅ + 1)๐‘’
y = ๐‘ฅ2
y = 2๐‘ฅ
y = ๐‘ฅ2
๐‘ฆ = (ln ๐‘ฅ)2
๐‘ฆ = 2๐‘™๐‘›๐‘ฅ
y = (ln ๐‘ฅ)2๐‘ฅ
Power Functionsthe bases are
functions of x and
the exponents are
constants
Exponential
Functions – The
bases are
constants and the
exponents are
functions of x
Differentiate using
power rule or GPR
Differentiate using
Differentiation of
Exponential
functions
๐‘ฅ
๐‘ฅ
Power Functions
whose base and
exponents are
both differentiable
functions
Differentiate using
Logarithmic
differentiation
Note:
Be able to recognize and differentiate these three types of functions and apply
appropriate differentiation rules or techniques.
Logarithmic Differentiation
Consider a function y = ๐‘ข๐‘ฃ where u = g(x) and v = h(x). Then y = ๐‘”(๐‘ฅ)โ„Ž(๐‘ฅ) is a power function whose base
and exponent are both differentiable functions of x. To find the derivative of this kind of function, we may use the
method of logarithmic differentiation defined by:
If y = y = ๐‘ข๐‘ฃ where u = g(x) and v = h(x), then
๐’…๐’š
๐’…๐’™
๐’…
= ๐’–๐’— ๐’…๐’™ (๐’—๐’๐’๐’–)
EXAMPLE 2
๐’…๐’š
a. Find ๐’…๐’™ if y = ๐’™๐’™
y = ๐’™๐’™ is in the form of y = ๐‘ข๐‘ฃ where u = x and v = x. To find
๐’…๐’š
,
๐’…๐’™
we use the logarithmic differentiation.
That is,
๐’…๐’š
๐’…๐’™
= ๐’–๐’—
๐’…
(๐’—๐’๐’๐’–)
๐’…๐’™
๐’…๐’š
๐’…๐’™
= ๐’™๐’™ ๐’…๐’™ (๐’™๐’๐’๐’™)
dy
dx
= x x [x (x . 1) + ln x (1)]
dy
dx
= x x [1 + ln x ]
๐’…
๐‘‘
using product rule to evaluate ๐‘‘๐‘ฅ (๐‘ฅ๐‘™๐‘›๐‘ฅ)
1
๐’…๐’š
b. Find ๐’…๐’™ if ๐‘ฆ = (2๐‘ฅ + 1)๐‘’
๐‘ฅ
๐‘ฅ
Again, ๐‘ฆ = (2๐‘ฅ + 1)๐‘’ takes the form of y = ๐‘ข๐‘ฃ where u = 2x + 1 and v = ๐‘’ ๐‘ฅ .
๐’…๐’š
๐’…๐’™
= ๐’–๐’— ๐’…๐’™ (๐’—๐’๐’๐’–)
๐’…
dy
dx
= (2๐‘ฅ + 1)๐‘’
dy
dx
= (2๐‘ฅ + 1)๐‘’ [๐‘’ ๐‘ฅ (
dy
dx
= (2๐‘ฅ + 1)๐‘’ [
๐‘ฅ ๐’…
๐’…๐’™
(๐‘’ ๐‘ฅ ln(2x + 1))
1
๐‘ฅ
2๐‘ฅ+1
2๐‘’ ๐‘ฅ
๐‘ฅ
2๐‘ฅ+1
using product rule to evaluate
๐‘‘
(๐‘’๐‘ฅ ln(2x + 1))
๐‘‘๐‘ฅ
. 2) + ln(2x + 1)(๐‘’ ๐‘ฅ . 1) ]
+ ๐‘’ ๐‘ฅ ln(2x + 1) ]
๐‘‘๐‘ฆ
c. Find ๐‘‘๐‘ฅ if ๐‘ฅ ๐‘ฆ + ๐‘’ ๐‘ฅ = 8
๐‘‘๐‘ฆ
๐‘ฅ ๐‘ฆ + ๐‘’ ๐‘ฅ = 8 is an implicit function, so we use implicit differentiation to find for ๐‘‘๐‘ฅ .
Differentiating both sides with respect to x, we have
๐‘‘
๐‘‘๐‘ฅ
๐‘‘
๐‘‘๐‘ฅ
๐‘‘
(๐‘ฅ ๐‘ฆ + ๐‘’ ๐‘ฅ ) = ๐‘‘๐‘ฅ(8)
๐‘‘
(๐‘ฅ ๐‘ฆ ) + ๐‘‘๐‘ฅ (๐‘’ ๐‘ฅ ) = 0
๐‘‘
๐‘ฅ ๐‘ฆ ๐‘‘๐‘ฅ(y ln x) + ๐‘’ ๐‘ฅ (1) = 0
1
๐‘‘
๐‘ฅ ๐‘ฆ [๐‘ฆ (๐‘ฅ . 1) + ln ๐‘ฅ . ๐‘‘๐‘ฅ (๐‘ฆ)] + ๐‘’ ๐‘ฅ = 0
๐‘ฆ
๐‘ฅ ๐‘ฆ (๐‘ฅ + ๐‘™๐‘› ๐‘ฅ
๐‘ฆ(๐‘ฅ ๐‘ฆ )
๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
)+ ๐‘’ ๐‘ฅ = 0
๐‘‘๐‘ฆ
+ ๐‘ฅ ๐‘ฆ ln x ๐‘‘๐‘ฅ + ๐‘’ ๐‘ฅ = 0
๐‘‘๐‘ฆ
๐‘ฅ ๐‘ฆ ln x ๐‘‘๐‘ฅ = -๐‘’ ๐‘ฅ ๐‘‘๐‘ฆ
=
๐‘‘๐‘ฅ
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
=
−๐‘’ ๐‘ฅ −
๐‘ฅ
๐‘ฅ ๐‘ฆ ln x
−๐‘ฅ๐‘’๐‘ฅ − ๐‘ฆ(๐‘ฅ๐‘ฆ )
๐‘ฅ
๐‘ฅ ๐‘ฆ ln x
−๐‘ฅ๐‘’ ๐‘ฅ − ๐‘ฆ(๐‘ฅ ๐‘ฆ )
(๐‘ฅ)๐‘ฅ ๐‘ฆ ln x
๐‘‘๐‘ฆ
=
๐‘‘๐‘ฅ
−๐‘ฅ๐‘’ ๐‘ฅ − ๐‘ฆ(๐‘ฅ ๐‘ฆ )
(๐‘ฅ)๐‘ฅ ๐‘ฆ ln x
๐‘‘๐‘ฆ
−๐‘ฅ๐‘’ ๐‘ฅ − ๐‘ฆ(๐‘ฅ ๐‘ฆ )
=
๐‘‘๐‘ฅ
๐‘ฅ
๐‘ฆ(๐‘ฅ๐‘ฆ )
๐‘‘๐‘ฆ
=
๐‘‘๐‘ฅ
๐‘ฆ(๐‘ฅ ๐‘ฆ )
๐‘ฅ ๐‘ฆ+1 ln x
Application
Activity 3
Find
๐‘‘๐‘ฆ
๐‘‘๐‘ฅ
for each function below.
1. y = (ln ๐‘ฅ)2๐‘ฅ
2. y = (sin ๐‘ฅ)๐‘ฅ
3. y = ๐‘ฅ 2
๐‘ฅ
4. y = ๐‘ฅ ln( 1+ ๐‘’
5. y = ๐‘ฅ √๐‘ฅ
๐‘ฅ)
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