Calculus ebook

Differentiation 1
LIMITS AND CONTINUITY
Let f(x) be a function defined on an interval that contains x = ax = a, except possibly at x
= ax = a. Then we say that,
(lim
𝑥→a
𝑓( x) = L
Examples 1
Find the limits of the following
Solution
We need to factor both numerator and denominator as shown below.
The simplify to obtain,
Example 2. Calculate the limit
We need to look at the limit from the left of 2 and the limit from the right of 2. As x approaches
2 from the left x - 2 < 0 hence
|x - 2| = - (x - 2)
Substitute to obtain the limit from the left of 2 as follows

Differentiation 2
= - 8
As x approaches 2 from the right x - 2 > 0 hence
|x - 2| = x - 2
Substitute to obtain the limit from the right of 2 as follows
= 8
The limit from the right of 2 and the limit from the left of 2 are not equal therefore the given
limit DOES NOT EXIST.
Example 3. Calculate the limit
Solution to Example 4:
As x approaches -1, cube root x + 1 approaches 0 and ln (x+1) approaches - infinity hence an
indeterminate form 0 . infinity
Let us rewrite the limit so that it is of the infinity/infinity indeterminate form.

Differentiation 3
We now use L'hopital's Rule and find the limit.
Example 5: Find the limit
Solution to Example 5:
As x gets larger x + 1 gets larger and e^(1/(x+1)-1) approaches 0 hence an indeterminate form
infinity.0
Let us rewrite the limit so that it is of the 0/0 indeterminate form.
Apply the l'hopital's theorem to find the limit.

Differentiation 4
= - 1
Solution
As x approaches 9, both numerator and denominator approach 0. Multiply both numerator and
denominator by the conjugate of the numerator.
Expand and simplify.
and now find the limit.
= 1 / 6

Differentiation 5
Solution:
The range of the cosine function is.
-1 <= cos x <= 1
Divide all terms of the above inequality by x, for x positive.
-1 / x <= cos x / x <= 1 / x
Now as x takes larger values without bound (+infinity) both -1 / x and 1 / x approaches 0. Hence
by the squeezing theorem the above limit is given by
Solution
As t approaches 0, both the numerator and denominator approach 0 and we have the 0 / 0
indeterminate form. Hence the l'hopital theorem is used to calculate the above limit as follows

Differentiation 6
Solution
We first factor out 16 x 2
under the square root of the denominator and take out of the square root
and rewrite the limit as
Since x approaches larger positive values (infinity) | x | = x. Simplify and find the limt.
= 3 / 4
Solution
Factor x 2
in the denominator and simplify.

Differentiation 7
As x takes large values (infinity), the terms 2/x and 1/x 2
approaches 0 hence the limit is
= 3 / 4
Solution
Factor x 2
in the numerator and denominator and simplify.
As x takes large values (infinity), the terms 1/x and 1/x 2
and 3/x 2
approaches 0 hence the limit
is
= 0 / 2 = 0

Differentiation 8
Solution
Multiply numerator and denominator by 3t.
Use limit properties and theorems to rewrite the above limit as the product of two limits and a
constant.
We now calculate the first limit by letting T = 3t and noting that when t approaches 0 so does T.
We also use the fact that sin T / T approaches 1 when T approaches 0. Hence
The second limit is easily calculated as follows
The final value of the given limit is

Differentiation 9
Solution
Factor x 2
inside the square root and use the fact that sqrt(x2
) = | x |.
Since x takes large values (infinity) then | x | = x. Hence the indeterminate form
Multiply numerator and denominator by the conjugate and simplify
Factor x out of the numerator and denominator and simplify

Differentiation
10
As x gets larger, the terms 1/x and 1/x2
approach zero and the limit is
= 1 / 2
Solution:
Let z = 1 / x so that as x get large x approaches 0. Substitute and calculate the limit as follows.
Exercises
Calculate the following limits
1)
2)

Differentiation
11
3)
DIFFERENTIATION
Differentiation from the first principle
DEFINITION The derivative of the function ƒ(x)with respect to the variable x is the function ƒ′
whose value
is ƒ′(x) = lim
𝑥→0
𝑓(𝑥 + ℎ) − (𝑓(𝑥))
ℎ
Example 1. Differentiate from the first principle
ƒ(x) =
x
𝑥 − 1
Solution
We use the definition of derivative, which requires us to calculate ƒ(x + h) and then subtract ƒ(x)
to obtain the numerator in the difference quotient.
ƒ(x) =
x
𝑥−1
and 𝑓(𝑥 + ℎ) =
(x+h)
(𝑥+ℎ) −1
so,
ƒ′(x) = lim
𝑥→0
𝑓(𝑥 + ℎ) − (𝑓(𝑥))
ℎ
= lim
𝑥→0
{(
(x+h)
(𝑥+ℎ)−1
) −(
x
𝑥−1
)}
1
ℎ
lim
𝑥→0
((x + h)(x − 1)− x(x + h − 1))
(x + h − 1)(x − 1)h
simplify
lim
𝑥→0
−ℎ
(x + h − 1)(x − 1) h
h cancles out
= lim
𝑥→0
−1
(x + h − 1)(x − 1)
= lim
𝑥→0
−1
(x − 1)(x − 1)
=
−1
(x − 1)2

Differentiation
12
Example 2:
Find the derivative of f(x) = 4x3
from first principles.
ƒ′(x) = lim
𝑥→0
𝑓(𝑥 + ℎ) − (𝑓(𝑥))
ℎ
lim
𝑥→0
4(x + h)3
− 4x3
ℎ
lim
𝑥→0
4(x3
+ 3x2
h + 3xh2
+ h3) − 4x3
ℎ
lim
𝑥→0
4x3
+ 12x2
h + 12xh2
+ 4h3
− 4x3
ℎ
lim
𝑥→0
12x2
h + 12xh2
+ 4h3
ℎ
lim
𝑥→0
h(12x2
+ 12xh + 4h2)
ℎ
lim
𝑥→0
(12x2
+ 12xh + 4h2) = 12𝑥2
Other examples
(a)Find the derivative of ƒ(x) = √ 𝑥 for x > 0.
(b) Find the tangent line to the curve y = √ 𝑥 at x = 4.

Differentiation
13
Differentiation Rules
1. Product Rule
If u and v are differentiable at x, then so is their product uv, and
𝑑
𝑑𝑥
(𝑢𝑣) = 𝑣 (
𝑑𝑢
𝑑𝑥
) + 𝑢 (
𝑑𝑣
𝑑𝑥
)
Similarly, this can be written as:
Product Rule:
𝑑
𝑑𝑥
(𝑢𝑣) = 𝑢′
𝑣 + 𝑣′
𝑢
Examples
Find the derivatives of y = (x2
+ 1)(x3
+ 3).
Solution

Differentiation
14
u = x2
+ 1 and v = x3
+ 3, we find
𝑑
𝑑𝑥
(𝑥2
+ 1)(𝑥3
+ 3)
𝑑𝑢
𝑑𝑥
= 𝑢′
= 2𝑥 𝑎𝑛𝑑
𝑑𝑣
𝑑𝑥
= 𝑣′
= 3𝑥2
From product rule,
𝑑
𝑑𝑥
(𝑢𝑣) = 𝑢′
𝑣 + 𝑣′
𝑢
=2𝑥(𝑥3
+ 3) + 3𝑥2(𝑥2
+ 1) expand
= 3x4
+ 3x2
+ 2x4
+ 6x
= 5x4
+ 3x2
+ 6x.
2. Quotient Rule
𝒅
𝒅𝒙
(𝒖𝒗) =
𝒖′
𝒗 − 𝒗′
𝒖
𝒗 𝟐
Nb: Just differentiate u, then v and substitute the result in the quotient rule formula and
simplify

Differentiation
15
CHAIN RULE
dy
dx
=
dy
du
.
du
dx
Example
Differentiate y = (3x2
+ 1)2
Solution
𝒍𝒆𝒕 𝟑𝒙 𝟐
+ 𝟏 = 𝒖, 𝒕𝒉𝒆𝒏 𝒚 = 𝒖 𝟐
𝒅𝒚
𝒅𝒖
= 𝟐𝒖,
𝒅𝒖
𝒅𝒙
= 𝟔𝒙
dy
dx
= 2𝑢. 6𝑥
But u= 𝟑𝒙 𝟐
+ 𝟏, therefore substituting gives
𝟐(𝟑𝒙 𝟐
+ 𝟏)𝟔𝒙 = 𝟏𝟐𝒙(𝟑𝒙 𝟐
+ 𝟏)
= 𝟑𝟔𝒙 𝟑
+ 𝟏𝟐𝒙
Other Examples applying chain rule

Differentiation
16
Implicit Differentiation
1. Differentiate both sides of the equation with respect to x, treating y as a differentiable function
of x.
2. Collect the terms with dy/dx on one side of the equation and solve for dy/dx.
Examples
1. Find
dy
dx
𝑖𝑓 𝑦2
= 𝑥
Solution
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑤ℎ𝑒𝑟𝑒 𝑦𝑜𝑢 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑎𝑡𝑒 𝑦 𝑝𝑢𝑡 𝑦′
𝑜𝑟
𝑑𝑦
𝑑𝑥
.
𝑦2
= 𝑥 𝑜𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑖𝑛𝑔 𝑏𝑒𝑐𝑜𝑚𝑒𝑠 2𝑦𝑦′
= 1
𝑐𝑜𝑙𝑙𝑒𝑐𝑡 𝑡𝑒𝑟𝑚𝑠 𝑎𝑛𝑑 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟
𝑑𝑦
𝑑𝑥
𝑜𝑟 𝑦′
𝑦′
=
1
2𝑦
2. Find the slope of the circle 𝑥2
+ 𝑦2
= 25 at the point (3, -4).
Solution
𝑥2
+ 𝑦2
= 25 Take derivatives on both sides of the equation
𝑑
𝑑𝑥
(𝑥2
) +
𝑑
𝑑𝑥
(𝑦2
) =
𝑑
𝑑𝑥
(25)
2𝑥 + 2𝑦𝑦′
= 0
2𝑦𝑦′
= −2𝑥
𝑦′
= −
2𝑥
2𝑦
= −
𝑥
𝑦
The slope at (3, −4)is −
𝑥
𝑦
=
−3
−4
=
3
4
Other Examples for Higher Derivatives

Differentiation
18
PARAMETRIC DIFFERENTIATION
Examples

Differentiation
25
Differentiation of Exponential and Logarithmic Functions
Exponential functions and their corresponding inverse functions, called logarithmic
functions, have the following differentiation formulas:

Differentiation
26
Examples
Find f’(x) if
Solution

Differentiation
27
More examples

Differentiation
28
Derivatives of Trigonometric functions
- The derivative of the sine function is the cosine function:
d
dx
(sin x) = cos x.
Example
Differentiate y = x2
sin x
Solution
Here we apply product rule
𝒖 = 𝒙 𝟐
𝒂𝒏𝒅 𝒗 = 𝒔𝒊𝒏𝒙
𝒖′
= 𝟐𝒙, 𝒗′
= 𝒄𝒐𝒔𝒙
𝒅
𝒅𝒙
= 𝒖′
𝒗 + 𝒗′
𝒖 = 𝟐𝒙𝒔𝒊𝒏𝒙 + 𝟐𝒙 𝟐
𝒄𝒐𝒔𝒙
The derivative of the cosine function is the negative of the sine function:
d
dx
(cos 𝑥) = − sin 𝑥.

Differentiation
29
Example 1
Differentiate y = 5x + cos x:
Solution
𝑑𝑦
𝑑𝑥
=
𝑑
𝑑𝑥
(5𝑥) +
𝑑
𝑑𝑥
(𝑐𝑜𝑠𝑥)
= 5 − sin 𝑥
Example 2
Differentiate y = sin x cos x:
Solution
Apply product rule
𝑢 = 𝑠𝑖𝑛𝑥, 𝑣 = 𝑐𝑜𝑠𝑥.
𝑢′
= 𝑐𝑜𝑠𝑥, 𝑣′
= −𝑠𝑖𝑛𝑥
𝑑𝑦
𝑑𝑥
= 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑥 = cos2
𝑥 − sin2
𝑥
Example 3
Differentiate y =
𝑐𝑜𝑠𝑥
1−𝑠𝑖𝑛𝑥
Here we apply quotient rule
𝑢 = cos 𝑥 𝑎𝑛𝑑 𝑣 = sin 𝑥 we get u’ and v’ then substitute in the equation

Differentiation
30
The derivatives of the other trigonometric functions:
𝒅
𝒅𝒙
(𝒕𝒂𝒏 𝒙) = 𝒔𝒆𝒄 𝟐
𝒙
𝒅
𝒅𝒙
(𝒔𝒆𝒄 𝒙 = 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙
𝒅
𝒅𝒙
(𝒄𝒐𝒕𝒙) = −𝒄𝒔𝒄 𝟐
𝒙
𝒅
𝒅𝒙
(𝒄𝒔𝒄 𝒙) = −𝒄𝒔𝒄 𝒙 𝒄𝒐𝒕 𝒙
Examples

Differentiation
31
Example when using Chain Rule in trigonometry
Application of Derivatives
Finding the Absolute Extrema of a Continuous Function ƒ on a Finite Closed Interval
1. Find all critical points of ƒ on the interval.
2. Evaluate ƒ at all critical points and endpoints.
3. Take the largest and smallest of these values

Differentiation
32
Examples

Application of differentiation to problems of minima, maxima, curve sketching
Examples:

Differentiation
35
Related Rates

Differentiation
36
Kinematics
Important Expressions
To get velocity, differentiate distance with respect to t
To get acceleration differentiate velocity with respect to t
Similarly, the 2nd
derivative of distance is the acceleration

Differentiation
37
Examples
1. A particle moves along the x-axis. The function x(t) gives the particle's position at any
time t≥0t,
𝑋(𝑡) = 𝑡4
− 2𝑡2
− 4
Find: The velocity and acceleration from the equation above
𝑑𝑥
𝑑𝑡
= 𝑣(𝑡)
=
𝑑
𝑑𝑡
(𝑡4
− 2𝑡2
− 4)
𝑣 = 4𝑡3
− 4𝑡 To get time at this point, equate the equation to 0
To find acceleration, we differentiate v
𝑑𝑣
𝑑𝑡
= 𝑎(𝑡)
=
𝑑
𝑑𝑡
(4𝑡3
− 4𝑡)
= 12𝑡2
− 4
𝑎 = 12𝑡2
− 4
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