Math 112 – Differential Calculus
Chapter 2
The Derivative, Slope and Rate of Change
2.1 The Derivative ( The following discussion was derived on pages 31 - 32 of Differential and
Integral Calculus by Love and Rainville, sixth edition)
Given a continuous function, y = f(x), ①
let us choose some fixed value of x, the corresponding value of y being given. Now consider another value of x
differing from the first one by an amount (positive or negative) which we will call the increment of x, and will
be denote by the symbol Δx. For this value of x, y will have a new value, differing from the original by an
amount Δy. In other words, when x changes to the value x + Δx, y changes to the value y + Δy, and we have
y + Δy = f(x + Δx)
Δy = f(x + Δx) – f(x) ②
𝜟𝒚
Now let us form the ratio , and investigate the behaviour of this ratio when Δx approaches zero.
𝜟𝒙
Since f(x) is continuous, the Δy of equation ② also approaches zero.We have found that when both
numerator and denominator of a fraction approach zero, the fraction itself may, or may not, approach a limit.
In Fig. 2-1, let the curve AB represent the graph
𝜟𝒚
of the given function. The ratio is the slope of the Y
𝜟𝒙
line joining the points P(x,y) and P’ ( x + Δx, y + Δy). As Δx
approaches zero, P’ approaches P along the curve,
and in all ordinary cases, the line PP’ approaches a
certain straight line ( PT in the figure ) as a limiting P′
position. That is, for a sufficiently well-behaved curve, Δy
𝜟𝒚
the ratio approaches a limit, the slope of the line P
𝜟𝒙
PT. This limit is called the derivative of y with respect
to x. PT y
Fundamental Definition: the derivative of y with res- X
𝜟𝒚
pect to x is the ratio when Δx approaches zero. x Δx
𝜟𝒙
𝒅𝒚
The derivative is designated by the symbol ; Fig. 2.1
𝒅𝒙
𝒅𝒚 𝜟𝒚 𝒇(𝒙+ 𝜟𝒙)−𝒇(𝒙)
= 𝐥𝐢𝐦 = 𝐥𝐢𝐦
𝒅𝒙 𝜟𝒙→𝟎 𝜟𝒙 𝜟𝒙→𝟎 𝜟𝒙
The operation of finding the derivative is called differentiation.
2.2 Determination of the Derivative ( Four-step Rule)
1. Replace x by (x + Δx) and y by (y + Δy). The function y = f(x) becomes y+Δy = f(x + Δx).
2. Solve for Δy such that Δy = f(x + Δx) – y.
3. Divide both sides of the resulting equation by Δx.
𝜟𝒚 𝒇(𝒙+ 𝜟𝒙)−𝒚
=
𝜟𝒙 𝜟𝒙
4. Determine the limit as Δx approaches zero.
Before proceeding to examples, knowledge of expanding the expression (𝑥 + 𝛥𝑥)𝑛 or (𝑥 − 𝛥𝑥)𝑛
where n is the exponent is important. Below is a triangle representing the constant corresponding to the
exponent of the expanded expression.
1 The presentation at the left is called the Pascal’s triangle.
1 1 ① The numbers on the respective lines are the values of the
1 2 1 ② constants when the expression (𝑥 + 𝛥𝑥)𝑛 or (𝑥 − 𝛥𝑥)𝑛
1 3 3 1 ③ is expanded.The numbers ①, ② etc. are the respective
1 4 6 4 1 ④ values of the exponent n. For the expression (𝑥 + 𝛥𝑥)𝑛 ,
1 5 10 10 5 1 ⑤ the constants have + sign while for the expression (𝑥 − 𝛥𝑥)𝑛
the constant have alternate + and – signs, the first being +.
Take note that the intermediate values are the sum of the two numbers above. The expression (𝑥 + 𝛥𝑥)3 is
expanded as [ 𝑥 3 + 3𝑥 2 𝛥𝑥 + 3 𝑥(𝛥𝑥)2 + (𝛥𝑥)3 ]. Take note the sum of the exponents of the term is equal to
the exponent. The expression (𝑥 − 𝛥𝑥)4 is expanded as [ 𝑥 4 − 4𝑥 3 𝛥𝑥 + 6 𝑥 2 (𝛥𝑥)2 – 4𝑥(𝛥𝑥)3 +(𝛥𝑥)4 ]. Take
, Math 112 – Differential Calculus
Example 2.1. Using the four-step rule, determine the derivative of the functions.
a) y = 5𝑥 3 + 3𝑥 2 – 6x + 8
Solution:
y + Δy = 5 [x + Δx]3 + 3 [x + Δx]2 − 6 ( x + Δx) + 8
Δy = 5 [ x 3 + 3x 2 Δx + 3x (Δx)2 + (Δx)3 ] + 3[x 2 + 2xΔx + (Δx)2 ] − 6 ( x + Δx) + 8
– (5x 3 + 3x 2 – 6x + 8)
Δy = 5x 3 + 15x 2 Δx + 15x (Δx)2 + 5(Δx)3 +3x 2 + 6xΔx + 3(Δx)2 − 6x −6Δx +8
–5x 3 − 3x 2 + 6x − 8
Δy = 15x 2 Δx + 15x (Δx)2 + 15(Δx)3 + 6xΔx + 3(Δx)2 −6Δx
𝛥𝑦
= 15x 2 + 15x Δx + 15 (Δx)2 + 6x + 3 Δx − 6
𝛥𝑥
dy Δy
= lim = 15x 2 + 15x(0) + 15 (0)2 + 6x + 3(0) −6 = 15x 2 + 6x −6
dx Δx→0 Δx
𝐝𝐲
= 𝟏𝟓𝐱 𝟐 + 6x −6
𝐝𝐱
b. y = 𝑥 5 − 3𝑥 4 + 1000
Solution:
y + Δy = [x + Δx]5 − 3 [x + Δx]4 + 1000
Δy = [ x 5 + 5x 4 Δx + 10x 3 (Δx)2 + 10x 2 (Δx)3 + 5x(Δx)4 + (Δx)5 ]
− 3[ x 4 + 4x 3 Δx + 6x 2 (Δx)2 + 4x(Δx)3 + (Δx)4 ] + 1000
− (x 5 − 3x 4 + 1000)
Δy = x 5 + 5x 4 Δx + 10x 3 (Δx)2 + 10x 2 (Δx)3 + 5x(Δx)4 + (Δx)5
− 3x 4 − 12x 3 Δx− 18x 2 (Δx)2 − 12x(Δx)3 − 3(Δx)4 + 1000 −x 5 +3x 4 − 1000
Δy = 5x 4 Δx + 10x 3 (Δx)2 + 10x 2 (Δx)3 + 5x(Δx)4 + (Δx)5
− 12x 3 Δx− 18x 2 (Δx)2 − 12x(Δx)3 − 3(Δx)4
Δy
= 5x 4 + 10x 3 Δx + 10x 2 (Δx)2 + 5x(Δx)3 + (Δx)4 − 12x 3 − 18x 2 Δx− 12x(Δx)2 − 3(Δx)3
Δx
dy Δy
= lim = 5x 4 + 10x 3 (0) + 10x 2 (0)2 + 5x(0)3
dx Δx→0 Δx
+ (0)4 − 12x 3 − 18x 2 (0)− 12x(0)2 − 3(0)3
𝐝𝐲
= 𝟓𝐱 𝟒 − 12𝐱 𝟑
𝐝𝐱
5x −3
c. y =
4x+2
Solution:
5(x + Δx)−3
y + Δy =
4(x +Δx)+2
5x + 5Δx−3 5x−3 (5x + 5Δx−3)(4x+2)−(4x+4Δx+2)(5x−3)
Δy = − =
4(x +Δx)+2 4x+2 [4(x +Δx)+2](4x+2)
(20x2 + 10x+20xΔx+10Δx−12x −6)−( 20x2 −12x+20xΔx −12Δx+10x −6)
Δy =
[4(x +Δx)+2](4x+2)
20x + 10x+20xΔx+10Δx−12x −6 −20x2 +12x−20xΔx+12Δx−10x+6
2
Δy =
[4(x +Δx)+2](4x+2)
10Δx+ 12Δx 22Δx
Δy = =
[4(x +Δx)+2](4x+2)] [4(x +Δx)+2](4x+2)
Δy 22
=
Δx [4(x +Δx)+2](4x+2)]
dy Δy 22 22
= lim Δx= =
dx Δx→0 [4(x +0)+2](4x+2) [4x+2](4x+2)
𝐝𝐲 𝟐𝟐
=
𝐝𝐱 (𝟒𝐱+𝟐)𝟐
Chapter 2
The Derivative, Slope and Rate of Change
2.1 The Derivative ( The following discussion was derived on pages 31 - 32 of Differential and
Integral Calculus by Love and Rainville, sixth edition)
Given a continuous function, y = f(x), ①
let us choose some fixed value of x, the corresponding value of y being given. Now consider another value of x
differing from the first one by an amount (positive or negative) which we will call the increment of x, and will
be denote by the symbol Δx. For this value of x, y will have a new value, differing from the original by an
amount Δy. In other words, when x changes to the value x + Δx, y changes to the value y + Δy, and we have
y + Δy = f(x + Δx)
Δy = f(x + Δx) – f(x) ②
𝜟𝒚
Now let us form the ratio , and investigate the behaviour of this ratio when Δx approaches zero.
𝜟𝒙
Since f(x) is continuous, the Δy of equation ② also approaches zero.We have found that when both
numerator and denominator of a fraction approach zero, the fraction itself may, or may not, approach a limit.
In Fig. 2-1, let the curve AB represent the graph
𝜟𝒚
of the given function. The ratio is the slope of the Y
𝜟𝒙
line joining the points P(x,y) and P’ ( x + Δx, y + Δy). As Δx
approaches zero, P’ approaches P along the curve,
and in all ordinary cases, the line PP’ approaches a
certain straight line ( PT in the figure ) as a limiting P′
position. That is, for a sufficiently well-behaved curve, Δy
𝜟𝒚
the ratio approaches a limit, the slope of the line P
𝜟𝒙
PT. This limit is called the derivative of y with respect
to x. PT y
Fundamental Definition: the derivative of y with res- X
𝜟𝒚
pect to x is the ratio when Δx approaches zero. x Δx
𝜟𝒙
𝒅𝒚
The derivative is designated by the symbol ; Fig. 2.1
𝒅𝒙
𝒅𝒚 𝜟𝒚 𝒇(𝒙+ 𝜟𝒙)−𝒇(𝒙)
= 𝐥𝐢𝐦 = 𝐥𝐢𝐦
𝒅𝒙 𝜟𝒙→𝟎 𝜟𝒙 𝜟𝒙→𝟎 𝜟𝒙
The operation of finding the derivative is called differentiation.
2.2 Determination of the Derivative ( Four-step Rule)
1. Replace x by (x + Δx) and y by (y + Δy). The function y = f(x) becomes y+Δy = f(x + Δx).
2. Solve for Δy such that Δy = f(x + Δx) – y.
3. Divide both sides of the resulting equation by Δx.
𝜟𝒚 𝒇(𝒙+ 𝜟𝒙)−𝒚
=
𝜟𝒙 𝜟𝒙
4. Determine the limit as Δx approaches zero.
Before proceeding to examples, knowledge of expanding the expression (𝑥 + 𝛥𝑥)𝑛 or (𝑥 − 𝛥𝑥)𝑛
where n is the exponent is important. Below is a triangle representing the constant corresponding to the
exponent of the expanded expression.
1 The presentation at the left is called the Pascal’s triangle.
1 1 ① The numbers on the respective lines are the values of the
1 2 1 ② constants when the expression (𝑥 + 𝛥𝑥)𝑛 or (𝑥 − 𝛥𝑥)𝑛
1 3 3 1 ③ is expanded.The numbers ①, ② etc. are the respective
1 4 6 4 1 ④ values of the exponent n. For the expression (𝑥 + 𝛥𝑥)𝑛 ,
1 5 10 10 5 1 ⑤ the constants have + sign while for the expression (𝑥 − 𝛥𝑥)𝑛
the constant have alternate + and – signs, the first being +.
Take note that the intermediate values are the sum of the two numbers above. The expression (𝑥 + 𝛥𝑥)3 is
expanded as [ 𝑥 3 + 3𝑥 2 𝛥𝑥 + 3 𝑥(𝛥𝑥)2 + (𝛥𝑥)3 ]. Take note the sum of the exponents of the term is equal to
the exponent. The expression (𝑥 − 𝛥𝑥)4 is expanded as [ 𝑥 4 − 4𝑥 3 𝛥𝑥 + 6 𝑥 2 (𝛥𝑥)2 – 4𝑥(𝛥𝑥)3 +(𝛥𝑥)4 ]. Take
, Math 112 – Differential Calculus
Example 2.1. Using the four-step rule, determine the derivative of the functions.
a) y = 5𝑥 3 + 3𝑥 2 – 6x + 8
Solution:
y + Δy = 5 [x + Δx]3 + 3 [x + Δx]2 − 6 ( x + Δx) + 8
Δy = 5 [ x 3 + 3x 2 Δx + 3x (Δx)2 + (Δx)3 ] + 3[x 2 + 2xΔx + (Δx)2 ] − 6 ( x + Δx) + 8
– (5x 3 + 3x 2 – 6x + 8)
Δy = 5x 3 + 15x 2 Δx + 15x (Δx)2 + 5(Δx)3 +3x 2 + 6xΔx + 3(Δx)2 − 6x −6Δx +8
–5x 3 − 3x 2 + 6x − 8
Δy = 15x 2 Δx + 15x (Δx)2 + 15(Δx)3 + 6xΔx + 3(Δx)2 −6Δx
𝛥𝑦
= 15x 2 + 15x Δx + 15 (Δx)2 + 6x + 3 Δx − 6
𝛥𝑥
dy Δy
= lim = 15x 2 + 15x(0) + 15 (0)2 + 6x + 3(0) −6 = 15x 2 + 6x −6
dx Δx→0 Δx
𝐝𝐲
= 𝟏𝟓𝐱 𝟐 + 6x −6
𝐝𝐱
b. y = 𝑥 5 − 3𝑥 4 + 1000
Solution:
y + Δy = [x + Δx]5 − 3 [x + Δx]4 + 1000
Δy = [ x 5 + 5x 4 Δx + 10x 3 (Δx)2 + 10x 2 (Δx)3 + 5x(Δx)4 + (Δx)5 ]
− 3[ x 4 + 4x 3 Δx + 6x 2 (Δx)2 + 4x(Δx)3 + (Δx)4 ] + 1000
− (x 5 − 3x 4 + 1000)
Δy = x 5 + 5x 4 Δx + 10x 3 (Δx)2 + 10x 2 (Δx)3 + 5x(Δx)4 + (Δx)5
− 3x 4 − 12x 3 Δx− 18x 2 (Δx)2 − 12x(Δx)3 − 3(Δx)4 + 1000 −x 5 +3x 4 − 1000
Δy = 5x 4 Δx + 10x 3 (Δx)2 + 10x 2 (Δx)3 + 5x(Δx)4 + (Δx)5
− 12x 3 Δx− 18x 2 (Δx)2 − 12x(Δx)3 − 3(Δx)4
Δy
= 5x 4 + 10x 3 Δx + 10x 2 (Δx)2 + 5x(Δx)3 + (Δx)4 − 12x 3 − 18x 2 Δx− 12x(Δx)2 − 3(Δx)3
Δx
dy Δy
= lim = 5x 4 + 10x 3 (0) + 10x 2 (0)2 + 5x(0)3
dx Δx→0 Δx
+ (0)4 − 12x 3 − 18x 2 (0)− 12x(0)2 − 3(0)3
𝐝𝐲
= 𝟓𝐱 𝟒 − 12𝐱 𝟑
𝐝𝐱
5x −3
c. y =
4x+2
Solution:
5(x + Δx)−3
y + Δy =
4(x +Δx)+2
5x + 5Δx−3 5x−3 (5x + 5Δx−3)(4x+2)−(4x+4Δx+2)(5x−3)
Δy = − =
4(x +Δx)+2 4x+2 [4(x +Δx)+2](4x+2)
(20x2 + 10x+20xΔx+10Δx−12x −6)−( 20x2 −12x+20xΔx −12Δx+10x −6)
Δy =
[4(x +Δx)+2](4x+2)
20x + 10x+20xΔx+10Δx−12x −6 −20x2 +12x−20xΔx+12Δx−10x+6
2
Δy =
[4(x +Δx)+2](4x+2)
10Δx+ 12Δx 22Δx
Δy = =
[4(x +Δx)+2](4x+2)] [4(x +Δx)+2](4x+2)
Δy 22
=
Δx [4(x +Δx)+2](4x+2)]
dy Δy 22 22
= lim Δx= =
dx Δx→0 [4(x +0)+2](4x+2) [4x+2](4x+2)
𝐝𝐲 𝟐𝟐
=
𝐝𝐱 (𝟒𝐱+𝟐)𝟐
