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.
𝒅
(𝒍𝒐𝒈𝒃 𝒖)
𝒅𝒙
1.
𝟏
𝒖
= 𝒍𝒐𝒈𝒃 𝒆 .
𝒅𝒖
𝒅𝒙
If b = 10, 𝑡ℎ𝑒𝑛 𝑙𝑜𝑔𝑏 𝑢 = 𝑙𝑜𝑔𝑢 called the common logarithm. The formula is reduced to
𝒅
(𝒍𝒐𝒈𝒖)
𝒅𝒙
𝟏
𝒖
= log𝒆 .
𝒅𝒖
𝒅𝒙
If b = e, 𝑡ℎ𝑒𝑛 𝑙𝑜𝑔𝑒 𝑢 = 𝑙𝑛𝑢 called the natural logarithm. The formula is reduced to
𝒅
𝟏
2. 𝒅𝒙 (𝐥𝐧 𝒖) = 𝒖 .
𝒅𝒖
𝒅𝒙
EXAMPLE 1
𝑑𝑦
a. Find 𝑑𝑥 if y = 𝑙𝑜𝑔5 (3𝑥 2 − 5).
𝑑𝑦
y = 𝑙𝑜𝑔5 (3𝑥 2 − 5) takes the form of y = 𝑙𝑜𝑔𝑏 𝑢 with b = 5 and u = 3𝑥 2 − 5. To find 𝑑𝑥 , we use the
formula
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
=
=
𝑑
𝑑𝑥
1
𝑑𝑢
(𝑙𝑜𝑔𝑏 𝑢) = 𝑢 𝑙𝑜𝑔𝑏 𝑒 . 𝑑𝑥
1
3𝑥2 −5
1
3𝑥2 −5
𝑑
2
𝑙𝑜𝑔5 𝑒 . 𝑑𝑥 (3𝑥 − 5)
𝑙𝑜𝑔5 𝑒 . (6x)
6𝑥𝑙𝑜𝑔5 𝑒
3𝑥2 −5
by substitution
b. Find
𝑑𝑦
𝑑𝑥
if y = log 𝑠𝑖𝑛2 4𝑥.
𝑑𝑦
Again, y = log 𝑠𝑖𝑛2 4𝑥 takes the form of y = log 𝑢 with u =𝑠𝑖𝑛2 4𝑥. To find 𝑑𝑥 , we use the formula
𝑑
(𝑙𝑜𝑔𝑢)
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
1
𝑢
= log𝑒 .
1
𝑑𝑢
𝑑𝑥
by substitution
𝑑
= 𝑠𝑖𝑛2 4𝑥 log𝑒 . 𝑑𝑥 (𝑠𝑖𝑛2 4𝑥)
log𝑒
𝑑
= 𝑠𝑖𝑛2 4𝑥 . 𝑑𝑥 [(sin 4𝑥)2 ]
log𝑒
by GPR
𝑑
= 𝑠𝑖𝑛2 4𝑥 . 2 (sin 4𝑥)2−1 𝑑𝑥 (sin 4𝑥)
𝑑𝑦
log𝑒
𝑑
= 𝑠𝑖𝑛2 4𝑥 . 2 sin 4x [cos 4𝑥 . 𝑑𝑥 (4𝑥) ]
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
log𝑒
= 𝑠𝑖𝑛2 4𝑥 . 2 sin 4x [cos 4𝑥 . (4)]
=
=
𝑑𝑥
8log𝑒 sin 4𝑥 cos 4𝑥
𝑠𝑖𝑛2 4𝑥
8log𝑒𝑐𝑜𝑠 4𝑥
= 8log𝑒 cot 4𝑥
sin 4𝑥
𝑑𝑦
c. Find 𝑑𝑥 if y = ln( 4x + 3).
𝑑𝑦
y = ln( 4x + 3) takes the form of y = ln u with u = 4x + 3. To find 𝑑𝑥 , we use the formula
𝑑
𝑑𝑥
𝑑𝑦
1
𝑢
(ln 𝑢) = .
1
= 4x + 3.
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑢
𝑑𝑥
𝑑
𝑑𝑥
(4𝑥 + 3)
1
= 4x + 3. (4)
4
= 4x + 3
𝑑𝑥
d. Find f ‘(x) if f(x) = 𝑙𝑛4 (𝑥 + 3)
f(x) = 𝑙𝑛4 (𝑥 + 3) can be expressed in power form as f(x) = [ln(𝑥 + 3)]4
Note: 𝒍𝒏𝟒 (𝒙 + 𝟑) ≠ ln(𝒙 + 𝟑)𝟒
f(x) = [ln(𝑥 + 3)]4 takes the form of f(x) = 𝑢𝑛 where u = ln(𝑥 + 3) and x = 4. To find f ‘(x), we
use the GPR.
𝑑𝑢
f’(x) = n𝑢𝑛−1 . 𝑑𝑥
u = ln(𝑥 + 3)
𝑑𝑢
𝑑𝑥
1
𝑑
= 𝑥+3 𝑑𝑥 (𝑥 + 3)
n=4
𝑑𝑢
1
= 𝑥+3 (1)
𝑑𝑥
𝑑𝑢
𝑑𝑥
1
= 𝑥+3
1
f ‘(x) = 4 [ln(𝑥 + 3)]3 𝑥+3
f ‘(x) =
4𝑙𝑛3 (x+3)
𝑥+3
𝑥2
𝑑𝑦
e. Find 𝑑𝑥 given that y = 𝑙𝑛𝑥
𝑥2
𝑑𝑦
y = 𝑙𝑛𝑥 takes the form of a quotient of two functions, so to find for 𝑑𝑥 , we use the quotient rule.
𝑑𝑦
=
𝑑𝑥
𝑑(𝑥)𝑛′ (𝑥)−𝑛(𝑥)𝑑′(𝑥)
[𝑑(𝑥)]2
𝑥2
y = 𝑙𝑛𝑥
n(x) = 𝑥 2
d(x) = lnx
n’(x) = 2x
d’(x) = 𝑥 𝑑𝑥 (𝑥)
1 𝑑
1
d’(x) = 𝑥
𝑑𝑦
=
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
=
1
𝑥
𝑙𝑛𝑥(2𝑥)−𝑥 2 ( )
(𝑙𝑛𝑥)2
2𝑥𝑙𝑛𝑥−𝑥
𝑙𝑛2 𝑥
or
𝑥(2𝑙𝑛𝑥−1)
𝑙𝑛2 𝑥
Application
Activity 1
𝑑𝑦
Find the 𝑑𝑥 of each function below.
3
1. y = log √√12𝑥
2. y = 𝑙𝑛2 𝑥 − ln(𝑙𝑛𝑥 2 )
3. y = ln(sec 𝑥 + tan 𝑥)
4. y = ln 𝑥 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.
𝒅
𝒅𝒙
𝒅
𝒅𝒙
(𝒂𝒖 ) = 𝒂𝒖 ln a .
(𝒆𝒖 ) = 𝒆𝒖
𝒅𝒖
𝒅𝒙
𝒅𝒖
𝒅𝒙
Where: a = constant and a > 0 and a ≠1
e = 2.718281828… ( e is an irrational number)
EXAMPLE 2
𝑑𝑦
a. Find the 𝑑𝑥 if y = 34𝑥
𝑑𝑦
y = 34𝑥 is in the form of y = 𝑎𝑢 . To find 𝑑𝑥 , we use the formula
𝑑𝑦
𝑑𝑥
𝒅
𝒅𝒖
= 𝒅𝒙 (𝒂𝒖 ) = 𝒂𝒖 ln a . 𝒅𝒙
Where a = 3,
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
u = 4x
𝒅
= 34𝑥 ln 3 . 𝒅𝒙 (4x)
= 34𝑥 ln 3 . 4
= 4(34𝑥 )ln3
𝑑𝑦
b. Given y = 𝑒 2𝑥+3 , find 𝑑𝑥 .
𝑑𝑦
Again, y = 𝑒 2𝑥+3 is in the form of y = 𝑒 𝑢 . To find 𝑑𝑥 , we use the formula
𝑑𝑦
𝑑𝑥
𝒅
= 𝒅𝒙 (𝒆𝒖 ) = 𝒆𝒖
𝒅𝒖
𝒅𝒙
u = 2x + 3
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑
= 𝑒 2𝑥+3 𝑑𝑥 (2x + 3)
= 𝑒 2𝑥+3 (2)
= 2 𝑒 2𝑥+3
c. Find f ‘(x) if f(x) = 𝑒 𝑥 ( 𝑒 2𝑥 - 𝑒 𝑥 )
f(x) = 𝑒 𝑥 ( 𝑒 2𝑥 - 𝑒 𝑥 ) is a product of two functions. Now, instead of using the product rule,
simplify the function using DPMA. That is,
f(x) = 𝑒 𝑥 ( 𝑒 2𝑥 - 𝑒 𝑥 )
using the DPMA
f(x) = 𝑒 𝑥 𝑒 2𝑥 - 𝑒 𝑥 𝑒 𝑥
using the product law of exponents
f(x) = 𝑒 𝑥+ 2𝑥 - 𝑒 𝑥+𝑥
simplifying the exponents
f(x) = 𝑒 3𝑥 - 𝑒 2𝑥
Find f ‘(x) using the sum rule
𝑑
𝑑
f ’(x) = 𝑑𝑥 (𝑒 3𝑥 ) - 𝑑𝑥 (𝑒 2𝑥 )
𝑑
𝑑
f ‘(x) = 𝑒 3𝑥 . 𝑑𝑥 (3x) - 𝑒 2𝑥 . 𝑑𝑥 (2x)
f ‘(x) = 𝑒 3𝑥 (3) - 𝑒 2𝑥 (2)
f ‘(x) = 3𝑒 3𝑥 - 2𝑒 2𝑥
𝑑𝑦
d. Given y = √1 + 𝑒 𝑥 , find 𝑑𝑥 .
1
y = √1 + 𝑒 𝑥 in power form is 𝑦 = (1 + 𝑒 𝑥 )2 . Since the function is in power form y = 𝑢𝑛 ,
𝑑𝑦
we use the GPR to find the 𝑑𝑥 .
𝑑𝑦
𝑑𝑥
𝑑𝑢
= n𝑢𝑛−1 . 𝑑𝑥
1
𝑦 = (1 + 𝑒 𝑥 )2
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
u = 1 + 𝑒𝑥
1
n=2
𝑑𝑢
= n𝑢𝑛−1 . 𝑑𝑥
=
=
=
=
1
−1
1
−1
1
−1
𝑑
(1 + 𝑒 𝑥 ) 2 . (1 + 𝑒 𝑥 )
2
𝑑𝑥
𝑑
(1 + 𝑒 𝑥 ) 2 . (0 + 𝑒 𝑥 (𝑥))
2
𝑑𝑥
(1 + 𝑒 𝑥 ) 2 . 𝑒 𝑥
2
𝑒𝑥
2 √1+ 𝑒 𝑥
𝑒 3𝑥 +1
e. Find f ‘(x) if f(x) = 𝑒 3𝑥 −1
𝑒 3𝑥 +1
f(x) = 𝑒 3𝑥 −1 is in the form of a quotient, so we use the quotient rule.
f ‘(x) =
𝑑(𝑥)𝑛′ (𝑥)−𝑛(𝑥)𝑑′(𝑥)
[𝑑(𝑥)]2
n(x) = 𝑒 3𝑥 + 1
n’(x) = 𝑒 3𝑥
f ‘(x) =
𝑑
𝑑𝑥
(3𝑥) + 0
d(x) = 𝑒 3𝑥 − 1
d’(x) = 𝑒 3𝑥
𝑑
𝑑𝑥
(3𝑥) – 0
n’(x) = 𝑒 3𝑥 (3)
d’(x) = 𝑒 3𝑥 (3)
n’(x) = 3𝑒 3𝑥
d’(x) = 3𝑒 3𝑥
(𝑒 3𝑥 −1)3𝑒 3𝑥 −(𝑒 3𝑥 +1)3𝑒 3𝑥
[𝑒3𝑥 −1]
2
Using DPMA in the numerator
f ‘(x) =
f ‘(x) =
3𝑒 6𝑥 − 3𝑒 3𝑥 −3𝑒 6𝑥 − 3𝑒 3𝑥
[𝑒3𝑥 −1]
2
combining similar terms in the numerator
− 6𝑒 3𝑥
[𝑒3𝑥 −1]
2
Application
Activity 2
𝑑𝑦
Find the 𝑑𝑥 for each function below.
3
1. y = √(𝑒 −2𝑥 + 2𝑒 𝑥 )2
2. y = 4𝑥 ln 4𝑥
3. y = (1 − 𝑒 −𝑥 )2 ( 1 + 𝑒 𝑥 )
4. y = arctan√𝑒 𝑥 − 1
𝑒 2𝑥 −1
5. y = ln (𝑒 2𝑥 +1)
Before we move to the last differentiation technique, observe these three groups of functions.
y = 𝑥4
y = 4𝑥
y = 𝑥𝑥
y = (2𝑥 + 1)𝑒
y = 𝑒 (2𝑥+1)
𝑦 = (2𝑥 + 1)𝑒
y = 𝑥2
y = 2𝑥
y = 𝑥2
𝑦 = (ln 𝑥)2
𝑦 = 2𝑙𝑛𝑥
y = (ln 𝑥)2𝑥
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
𝑥
𝑥
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 = 𝑢𝑣 where u = g(x) and v = h(x). Then y = 𝑔(𝑥)ℎ(𝑥) 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 = 𝑢𝑣 where u = g(x) and v = h(x), then
𝒅𝒚
𝒅𝒙
𝒅
= 𝒖𝒗 𝒅𝒙 (𝒗𝒍𝒏𝒖)
EXAMPLE 2
𝒅𝒚
a. Find 𝒅𝒙 if y = 𝒙𝒙
y = 𝒙𝒙 is in the form of y = 𝑢𝑣 where u = x and v = x. To find
𝒅𝒚
,
𝒅𝒙
we use the logarithmic differentiation.
That is,
𝒅𝒚
𝒅𝒙
= 𝒖𝒗
𝒅
(𝒗𝒍𝒏𝒖)
𝒅𝒙
𝒅𝒚
𝒅𝒙
= 𝒙𝒙 𝒅𝒙 (𝒙𝒍𝒏𝒙)
dy
dx
= x x [x (x . 1) + ln x (1)]
dy
dx
= x x [1 + ln x ]
𝒅
𝑑
using product rule to evaluate 𝑑𝑥 (𝑥𝑙𝑛𝑥)
1
𝒅𝒚
b. Find 𝒅𝒙 if 𝑦 = (2𝑥 + 1)𝑒
𝑥
𝑥
Again, 𝑦 = (2𝑥 + 1)𝑒 takes the form of y = 𝑢𝑣 where u = 2x + 1 and v = 𝑒 𝑥 .
𝒅𝒚
𝒅𝒙
= 𝒖𝒗 𝒅𝒙 (𝒗𝒍𝒏𝒖)
𝒅
dy
dx
= (2𝑥 + 1)𝑒
dy
dx
= (2𝑥 + 1)𝑒 [𝑒 𝑥 (
dy
dx
= (2𝑥 + 1)𝑒 [
𝑥 𝒅
𝒅𝒙
(𝑒 𝑥 ln(2x + 1))
1
𝑥
2𝑥+1
2𝑒 𝑥
𝑥
2𝑥+1
using product rule to evaluate
𝑑
(𝑒𝑥 ln(2x + 1))
𝑑𝑥
. 2) + ln(2x + 1)(𝑒 𝑥 . 1) ]
+ 𝑒 𝑥 ln(2x + 1) ]
𝑑𝑦
c. Find 𝑑𝑥 if 𝑥 𝑦 + 𝑒 𝑥 = 8
𝑑𝑦
𝑥 𝑦 + 𝑒 𝑥 = 8 is an implicit function, so we use implicit differentiation to find for 𝑑𝑥 .
Differentiating both sides with respect to x, we have
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
(𝑥 𝑦 + 𝑒 𝑥 ) = 𝑑𝑥(8)
𝑑
(𝑥 𝑦 ) + 𝑑𝑥 (𝑒 𝑥 ) = 0
𝑑
𝑥 𝑦 𝑑𝑥(y ln x) + 𝑒 𝑥 (1) = 0
1
𝑑
𝑥 𝑦 [𝑦 (𝑥 . 1) + ln 𝑥 . 𝑑𝑥 (𝑦)] + 𝑒 𝑥 = 0
𝑦
𝑥 𝑦 (𝑥 + 𝑙𝑛 𝑥
𝑦(𝑥 𝑦 )
𝑥
𝑑𝑦
𝑑𝑥
)+ 𝑒 𝑥 = 0
𝑑𝑦
+ 𝑥 𝑦 ln x 𝑑𝑥 + 𝑒 𝑥 = 0
𝑑𝑦
𝑥 𝑦 ln x 𝑑𝑥 = -𝑒 𝑥 𝑑𝑦
=
𝑑𝑥
𝑑𝑦
𝑑𝑥
=
−𝑒 𝑥 −
𝑥
𝑥 𝑦 ln x
−𝑥𝑒𝑥 − 𝑦(𝑥𝑦 )
𝑥
𝑥 𝑦 ln x
−𝑥𝑒 𝑥 − 𝑦(𝑥 𝑦 )
(𝑥)𝑥 𝑦 ln x
𝑑𝑦
=
𝑑𝑥
−𝑥𝑒 𝑥 − 𝑦(𝑥 𝑦 )
(𝑥)𝑥 𝑦 ln x
𝑑𝑦
−𝑥𝑒 𝑥 − 𝑦(𝑥 𝑦 )
=
𝑑𝑥
𝑥
𝑦(𝑥𝑦 )
𝑑𝑦
=
𝑑𝑥
𝑦(𝑥 𝑦 )
𝑥 𝑦+1 ln x
Application
Activity 3
Find
𝑑𝑦
𝑑𝑥
for each function below.
1. y = (ln 𝑥)2𝑥
2. y = (sin 𝑥)𝑥
3. y = 𝑥 2
𝑥
4. y = 𝑥 ln( 1+ 𝑒
5. y = 𝑥 √𝑥
𝑥)