Denition approach of learning new topics

Foundation of Calculus
July 15, 2009
Abstract
Lecture 1 : We learn the foundation of calculus using denition
approach Any concept will be dened and then later if required will
be paralleled with a physical interpretation We rst understand what
is there in coordinate geometry for us dene slope of two points and
extend that denition to understand the slope of a line see how slope
keeps changing as the line orientation keeps changing dene function and
then dene what is domain & range of a function plotting of function
graphs
Lecture 2 : Physical interpretation of derivative in terms of whatever
we learnt till now
Physical interpretation of limits with 4 examples dene limits
limits formulae examples
Lecture 3 : Derivatives denition of derivative derivative of some
examples from using denition formulae & its usage lots of interesting
problems
Lecture 4 : Integration Physical interpreation of integration integration
vs derivative (mathematically speaking) problems (circle area,
circumference, volume of cylinder, sphere ) numerical problems formulae
& Use
Denition approach of learning new topics
Denition approach is what we will use here. We won't really try to physically
locate what the denitions are saying but rather will except them as
starting blocks to start understanding of further development of the topic. Calculus
can be best understood using physics. We can have some examples from
physics but we will solve lot of problems that are mathematical and are going
to use the denition and validate that.
1

2 CONTENTS
Contents
I Lecture I 4
1 Cartesian coordinate system 4
1.1 Plotting points . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Dene slope for two points . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Slope of a line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Slopes of line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2 Functions 8
2.1 Examples of a functions . . . . . . . . . . . . . . . . . . . . . . . 8
2.2 Finding values of function in y = f(x) form . . . . . . . . . . . . 8
2.3 Functions and their graphs . . . . . . . . . . . . . . . . . . . . . 8
2.4 Domain and range of a function . . . . . . . . . . . . . . . . . . . 9
II Lecture 2 11
3 Examples of what is a Limit? 11
3.1 Polygon becomes circle . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Bucket is full or overowing?[5] . . . . . . . . . . . . . . . . . . . 12
3.3 Rotating a marble tied to a cord . . . . . . . . . . . . . . . . . . 13
3.4 A numerical example . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.5 Denition of Limit . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.6 Formulae of limits . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3.7 Worked out problems in limits . . . . . . . . . . . . . . . . . . . 15
3.8 Practise Problems in Limits . . . . . . . . . . . . . . . . . . . . . 18
III Lecture 3 19
4 Derivatives 19
4.1 Derivatives of other functions . . . . . . . . . . . . . . . . . . . . 23
4.2 Understanding the Derivative denition geometrically[1] . . . . . 23
4.3 Rate of change . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4.4 Function & its examples in Physics . . . . . . . . . . . . . . . . . 25
4.5 Instantaneous rate of change[3] . . . . . . . . . . . . . . . . . . . 26
4.6 Chain Rule and friends . . . . . . . . . . . . . . . . . . . . . . . . 27
4.7 Composition of functions . . . . . . . . . . . . . . . . . . . . . . 27
4.8 Derivative of a composite function . . . . . . . . . . . . . . . . . 28
IV Lecture 4 30
5 Integration 30
5.1 Dierentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
5.2 Indenite Integration . . . . . . . . . . . . . . . . . . . . . . . . . 30
5.3 Denite Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 31
5.3.1 Way to evaluate Denite Integrals . . . . . . . . . . . . . 31

CONTENTS 3
5.4 Methods of Solving Integrals . . . . . . . . . . . . . . . . . . . . 35
5.4.1 Method of substitution . . . . . . . . . . . . . . . . . . . . 35

4 1 CARTESIAN COORDINATE SYSTEM
Part I
Lecture I
We start are understanding dening the coordinate space. And further will use
this to understand functions and physical implications of function, derivative
and integration.
1 Cartesian coordinate system
Denition. Cartesian coordinate system:
A system of two perpendicular axes, x & y axis as in the adjoining gure.
A sign convention is laid on the axes. Part of x-axis right of y-axis is positive
x axis and left is negative x-axis. Similarly above x-axis, y axis is positive and
below is negative. This convention is laid to uniquely locate a point in space.
Denition. Point in Coordinate space :
Any point in the space is denoted as (a, b) where a is called the x coordinate
& b is called the y coordinate. Now x-coordinate is the distance of the point
from y axis & y-coordinate is the distance of the point from x axis. See in the
above gure.
1.1 Plotting points
Example. Plot the following points
(1,2), (2,0),(0,3),(-1,2),(-1,-1)
To plot the point (1, 2) we rst move x-coordinate value along x axis (i.e. 1
) and then move y-coordinate value parellel to y-axis (i.e. 2 )

1.2 Dene slope for two points 5
1.2 Dene slope for two points
Denition. Slope (m) of line joining two points
If P ≡ (x 1 , y 1 ) & Q ≡ (x 2 , y 2 ) then slope of line joining these two points is
dened as
m = y 2 − y 1
x 2 − x 1
Example. Find the slope of the line joining the pair of points
• (1, 2) & (2, 3)
• (−1, 2) & (3, −1)
• (4, 0) & 10, 0)
• (2, 3) & (2, 10)
Solution
• Slope of (1, 2) & (2, 3) is 3 − 2
2 − 1 = 1
• Slope of (−1, 2) & (3, −1) is
−1 − 2
3 − (−1) = −3 4
• Slope of (4, 0) & 10, 0) is 0 − 0
4 − 10 = 0
• Slope of (2, 3) & (2, 10) is 10 − 3 = Undefined . Means the line joining
2 − 2
these two points is parellel to y-axis
1.3 Slope of a line
Denition. Slope of a line
Take any two points (x 1 , y 1 ) & (x 2 , y 2 ) on the line the slope of the line is
dened as ∆y
∆x = y 2 − y 1
x 2 − x 1
Informally dened as : rate of change of y with respect to corresponding
change in x

6 1 CARTESIAN COORDINATE SYSTEM
Observe one important point here
Slope of line = y 2 − y 1
x 2 − x 1
= tanθ
Example. Find the slope of the following lines
1. 2x + 3y = 6
2. x = 2
3. y = 4
4. x − y = 3
Solution
1. In 2x + 3y = 6, (3, 0) & (0, 2) satisfy the equation. i.e. these points lie on
the line hence Slope = 2 − 0
0 − 3 = − 2 3
2. In x = 2, (2, 0) & (2, 1) saties the equation i.e. these points lie on the
line hence Slope = 1 − 0
2 − 2 = undefined
3. In y=4, (0, 4) & (1, 4) satises the equation i.e. these points lie on the line
hence Slope = 4 − 4
1 − 0 = 0
4. In x − y = 3, (3, 0) & (0, −3) satises the equation i.e. these points lie on
the line hence Slope = 0 − (−3) = 1
3 − 0

1.4 Slopes of line 7
1.4 Slopes of line
Denition. Inclination of a line
Angle (θ) made by a line with positive x-axis such that 0 ≤ θ < π
Once we have dened the angle of inclination then we will see what is the
relation between increase in angle of inclination and slope.
[1]
Can you see that as the inclination of the line increases from 0 to going close
to right angle there is increase in slope.
For example, angle of inclination = {θ/θ = 0 0 , 30 0 , 45 0 , 60 0 . . .} then corresponding
slopes = {m/m = 0, √3 , 1, √ 3, . . .} since slope= tan θ
1
And the same thing happens corespondingly in the second quandrant θ ∈
, π) as θincreases correspondingly sllope too increases.
( π 2
Fact. Can you see the trend that is seen?
1. Slope of a line parellel to x axis is Zero
2. The slope of line increases as is moves away form positive x axis in anticlockwise
sense towards y axis to reach a huge innite value.
3. At angle π/2 , the slope is undened.
4. Again it increases from positive y axis in counterclockwise direction towards
negative x axis.

8 2 FUNCTIONS
2 Functions
Functions are expressions that exhibit the functioning of a particular thing. Like
a mixer grinder, weighing machine. etc.
Denition. A function y = f(x) is dened as a rule which for all values of x
assignes a unique value of y.
So mathematically a function can be seen as f : X → Y as a mapping that
goes from set X to Y moreover the mapping can be one to one or many to one.
2.1 Examples of a functions
1. Linear function : y = 2x + 3, 2x + 3y = 4, x 2 + y = 1 are equations of
3
lines in dierent forms. They are called linear as they are of degree one
in x & y.
2. Trigonometric function : y = sin x, y = cos x, y = sin(π − x) are all
trigonometric functions.
3. Exponential function : y = a x , y = 2 x , y = e x are exponential functions.
Note they are constant variable kind of functions. Students often confuse
them with x 2 kind of functions which are variable constant form.
2.2 Finding values of function in y = f(x) form
Let us work on some examples of evaluating values of functions at particular
value of x.
1. If f(x) = x + 1
x − 1 then what is f(2), f( 1 2 ), f( 1 x )
f(2) = 2 + 1
2 − 1 = 3
1
f( 1 2 ) = 2 + 1 3
1
2 − 1 = 2
− 1 = −3
2
1
f( 1 x ) = x + 1
1
x − 1 = 1 + x
1 − x = −f(x)
2.3 Functions and their graphs
Denition. Graph
A graph is the collection of all points (x, y) that satisfy the equation of the
function.
Example 1. Let us plot the graph of the equation y = 2x + 3
Points (1, 5) & (0, 3) satify the equation and the equation is linear in x & y

2.4 Domain and range of a function 9
Example 2. Plot the graph of the function given as y = x 2 + 1
Plotting points and getting a rough idea of the graph.
Points (0, 1), (1, 2), (−1, 2), (2, 5), (−2, 5). So we get a rough idea of the graph
and what remains is to get the graph smooth.
2.4 Domain and range of a function
Denition. Domain (D f ):
All values that will be taken (input) by the function y = f(x) i.e. all values
of x that keeps the function well dened.
Denition. Range (R f ):
All values of y (output) that will be taken for all values of x ∈ D f i.e. y ∈ R f
i.e. R f = {y | y = f(x) such that x ∈ D f }
Example 3. Domain and range of the following functions
1. f(x) = 1 x
D f = R − {0}
R f = R − {0}
2. f(x) = √ x
D f : [0, ∞)
R f : [0, ∞) (since √ x is positive root of x)

10 2 FUNCTIONS
3. f(x) = √ 1
x
D f : (0, ∞) (since √ can take only positive values)
R f : (0, ∞) (since the fraction cannot become zero! and √ x is positive
root of x)
4. f(x) = 1
1 + x 2
D f = R
R f = (0, 1] ( use the quadratic method to nd R f )
5. f(x) =
D f : R
R f : (0, 1]
1
√
1 + x
2
6. f(x) = sin x
D f : R since sine is welldened for any angle.
R f : [−1, 1] (since any angle given, sin x lies between −1 & 1)

11
Part II
Lecture 2
3 Examples of what is a Limit?
Instruction for this section : In the examples below, you might not understand
how few limits are used. Go through them without worrying if you are not
following something. Going further you will understand what we did in these
examples.
Let us consider an example to understand what Limits is exactly.
3.1 Polygon becomes circle 1
A circle of radius r is constructed. We can construct regular polygons inscribed
in that circle.
Now we can approximate the area of circle to be equal to area of a polygon
where number of sides is a very large number.
Suppose we have a polygon inscribed that has n sides.
Area of the polygon is
A = n · 1
2 r2 sin 2π n
The end to this process of continuously increasing the number of sides of the
inscribed polygon is same as trying to nd the limit (english word meaning) of
this process. The process limit or end would be that the polygon has become a
circle.
So as n → ∞, Area of polygon →Area of the circle.
This is written in notational form as lim Area of polygon = Area of Circle
n→∞
1
Area of Circle = lim
n→∞ 2 r2 n sin 2π n
r 2 sin 2π n
= lim
n→∞ 2
n
= r 2 sin 2π n
π lim
n→∞ 2π
n
= πr 2 · 1
[using
sin θ
lim = 1]
θ→∞ θ
1 You can view the animation of Polygon tending to a circle at TeachingMathematics

12 3 EXAMPLES OF WHAT IS A LIMIT?
We used above lim
θ→0
sin θ
θ
= 1 [1]
So we are proving this now!
In the neighbourhood of θ = 0. We see that
Area of ∆QRP ≤Area of sector QRP≤Area of ∆QRK
1
2 r2 sin θ ≤ 1 2 r2 θ ≤ 1 2 r2 tan θ
⇒ sin θ ≤ θ ≤ tan θ
Assuming we are seeing for θ > 0 without loss of generality, similar steps
and idea will follow for θ < 0
⇒ 1 ≤
θ
sin θ ≤ cos θ
⇒ cos θ ≤ sin θ
θ
≤ 1
sin θ
As θ → 0 ⇒ lim cos θ ≤ lim
θ→0 θ→0
θ
≤ 1
sin θ
⇒ lim
θ→0
θ
= 1 [Using Sandwich theorem]
3.2 Bucket is full or overowing?[5]
Consider this example of lling a bucket with the following rule, Each hour
the bucket is lled by half the amount left empty. So when the process is
started at the beginning the bucket is empty. First hour half the bucket is
half lled ( 1 2 ) , next hour we have half bucket left and hence we ll ( 1 2 + 1 4 ).
Third hour we have lled half of the left lled, So we have ( 1 2 + 1 4 + 1 8 ). And
keep doing this process we get ( 1 2 + 1 4 + 1 8 + · · · = 1/2
1−1/2 = 1).
So this is another example of limit concept. Here the end state of this process
is that the bucket is always lled by half the amount left and that would never
allow the bucket to get full, but if the process is continued till innity the end
state would be lling the bucket.

3.3 Rotating a marble tied to a cord 13
3.3 Rotating a marble tied to a cord
A marble tied to a cord if rotated about the center.
So during this rotation if the marble rotates about the arc specied in the
adjoining diagram. Now if the cord reaches the point A and the cord is cut
what do u think what would be the path of the marble. Think about this!
The marble would move along the tangent to its path at the point. So that
means it will move along the tangent to the path. To nd that path we need to
nd the tangent at that very point.
So tangent at point A can be approximated using the secant joining any
other point to the right of A and point A. Now we keep moving the point to
the right of A i.e. B 1 towards A along the points B 2 , B 3 , B 4 , B 5 , · · · to reach
the point A.
So in limit notation we can write this as
as B i → A
3.4 A numerical example
i.e. lim
i→∞
Secant AB i = T angent at A
Consider this term f(x) = x2 −1
x−1
. Now this term is well dened at all x ∈ R
except x = 1.
We want to see what numerical value does f(x) takes as x → 1 i.e. x goes
closer and closer to 1.

14 3 EXAMPLES OF WHAT IS A LIMIT?
But x can get closer and closer to 1 from two directions, one from left of 1
and other from right of 1.
• x → 1 +
As x approches to 1 from right of 1. what value does f(x) takes?
x 1.1 1.01 1.001 1.0001 1.00001
f(x) 2.1 2.01 2.001 2.0001 2.00001
Can you make out what is happening as x → 1 + then f(x) → 2
• x → 1 −
As x approaches to 1 from left of 1. What values does f(x) takes?
x 0.9 0.99 0.999 0.9999 0.99999
f(x) 1.9 1.99 1.999 1.9999 1.99999
So you can see that here to the limit for f(x) is 2.
Hence we conclude mathematically that
x 2 − 1
lim
x→2 x − 1 = 2

3.5 Denition of Limit 15
3.5 Denition of Limit
If y = f(x) is given and as x → a , f(x) → l then we say that the limit of f(x)
is l
Mathematically that is written as
lim f(x) = l
x→a
Now let us put some thought on the denition of limit
we say as x → a, f(x) → l, now x tending to a means x comes closer to a.
Since a lies on Real line x will come closer to a from left of a and right of a.
For example, x → 2 means x will take values 1.9, 1.99, 1.999, 1.9999, . . . from
left of 2 and 2.1, 2.01, 2.001, 2.0001, 2.00001, . . .. from right of 2.
So the above denition,
lim f(x) exists
x→a
actually means
Note : here the equality in the
mathematical way of writing the
limit is not the usual equality.
It is just a representation that
f(x) is getting closer to l as x
goes closer to a.
3.6 Formulae of limits
LHL = lim f(x) exists
x→a− RHL = lim f(x) exists
x→a +
&
LHL = RHL = l
sin x
1. lim
x→0 x
= 1, lim
tan x
cos x = 1 & lim = 1
x→0 x→0 x
2. lim
x→0
(1 + x) 1/x = e = lim
x→∞ (1 + 1 x )x
a x − 1
3. lim = log a , a > 0
x→0 x
3.7 Worked out problems in limits
1. lim
x→2
x 2 − 4
x − 2
x 2 − 4
lim
x→2 x − 2
=
(x − 2)(x + 2)
lim
x→2 x − 2
= lim (x + 2)
x→2
(since here x → 2 So x − 2 ≠ 0)
= 4
Above method implicitly means this as x → 2 =⇒ x + 2 → 4 & x − 2 →
(x + 2)(x − 2)
0 =⇒ → 4 (since x − 2 is a factor both numertor −
x − 2
denominator)

16 3 EXAMPLES OF WHAT IS A LIMIT?
x 2 − 4
2. lim
x→0 x − 2
Now observe in this problem the function is not undened at x = 0. So
the limit is same as the function value at that point. But why doesnt that
happen in the rst problem above?
lim
x→0
x 2 − 4
x − 2 = 0 − 4
0 − 2
= 2
3. lim
x→0
sin 1 x
As x → 0 here sin gets applied on it to oscillate it more and more often.
And hence it doesnt reach any point.
Check the gure hence the limits remains undened.
4. lim x sin 1
x→0 x
As x → 0 here though sin 1 x
oscillates x term goes closer and closer to
zero. decreasing the osciallation and leading the total product to zero .
Hence the limit is Zero.
5. Find the limit of the following function at respective points
(a) f(x) = |x| at x = 0
{
x x ≥ 0
Now we know that f(x) = |x| =
. Now we need to nd
−x x < 0
the limit of |x| at x = 0. This will be LHL and RHL and both should
be equal for the limit to exist. Remember the story of two villages
that were separated by a forest and a river passing through the center
of the village dividing the villages.

3.7 Worked out problems in limits 17
RHL = lim
x→0 + |x|
= lim
x→0 + x
= 0
LHL = lim
x→0 − |x|
= lim
x→0 −(−x)
= 0
Hence we see that both the RHL and LHL are equal hence we say
that the limit exists for this function at zero. 2
{
1 x ≥ 0
(b) f(x) =
−1 x < 0 at x = 1
LHL = lim
x→1 + f(x)
= lim 1
x→1 + (to the right of 1 f(x) = 1)
= 1
RHL = lim
x→1 − f(x)
= lim 1
x→1− (since f(x) = 1 for x < 1)
= 1
Since LHL=RHL hence the limits exist and is equal to 1.
But did you observe here we needed to nd the limit at the x = 1.
To the left and right of x = 1 the function had the same value that
is f(x) = 1.
Hence instead of nding LHL and RHL we could have just wrote
limit = lim
x→1
f(x)
= lim
x→1
1
(since f(x) takes same value)
2 Now you might have a question as why we didn't try to solve the previous limit problem
as LHL and RHL but just worked out limit? This was purely done since in previous problem
the function denition was not changing on either side of the point of nding limit. But in
the present problem the function is dierent to the left and right of the point x = 0. In the
previous problem nding the LHL and RHL is same as nding the limits in general. But
actually limits is dened as
lim f(x) =
x→a
lim f(x) =
x→a +
lim f(x)
x→a −

18 3 EXAMPLES OF WHAT IS A LIMIT?
{
1 x ≥ 0
(c) f(x) =
−1 x < 0 at x = 0
Now will we just nd limit here without trying to see what is the
RHL and LHL??
Here we have to nd the LHL and RHL separately and check if both
are equal. If both are equal then the function has limits at x = 0 or
else it doesnt have a limit.
RHL = lim
x→0 + f(x)
= lim 1
x→0 +
(since right of 0 f(x) = 1)
= 1
LHL = lim
x→0 − f(x)
= lim (−1)
x→0 − (since left of 0 f(x) = −1)
= −1
And we see that LHL ≠ RHL
Hence the limits doesnt exist in this case.
3.8 Practise Problems in Limits
1. lim
x→0 + sin x
|x|
2. lim
θ→0
tan θ
θ
(Solution : 1)
(Solution : 1)
3. Find the anwers to the following functions, given D f : [0, 3] and is represented
in the adjoining gure
(a) f(1)

19
(b) lim x→1 +
(c) lim x→1 −
(d) lim f(x)
x→1
(e) lim f(x)
x→2 +
(f) lim x→2 −
(g) f(2)
Solutions- a:2,b:1,c:1,d:1,e:0,f:-1,g:0
Part III
Lecture 3
4 Derivatives
Denition. Derivative of a function y = f(x) at a point x = a is dened as
dy
f(x) − f(a)
dx∣ = lim
x→a
x=a
x − a
= lim
h→0
f(a + h) − f(a)
h
Above we found derivative of the function at some particular point x = a.
What if we want to nd the derivative at any point x? We just replace a by x!
Simple right?
Notation : If given y = f(x),
dy
dx = d
dx g(x) = g′ (x)
So the derivative of a function y = f(x) at any point x is given as
dy
f(x + h) − f(x)
dx∣ = lim
at any x
h→0 h
Example. If f(x) = sin x, g(x) = x + 1
x − 1
then we have

20 4 DERIVATIVES
1. f(t) = sin t, g(t) = t + 1
t − 1
2. f(t − 1) = sin(t − 1), g(t) = t − 1 + 1
t − 1 − 1 = t
t − 2
3. f(e x ) = sin(e x ), g(e x ) = ex + 1
e x − 1
4. f(ax 2 + bx + c) = sin(ax 2 + bx + c), g(ax 2 + bx + c) = ax2 + bx + c + 1
ax 2 + bx + c − 1
5. f(cos x) = sin(cos x), g(x) = cos x + 1
cos x − 1
6. f(h(x)) = sin(h(x)), g(x) = h(x) + 1
h(x) − 1
Denition. Composite function
Given functions y = f(x) & y = g(x) then composition of functions f ◦g(x) =
f(g(x))
Example. Given f(x) = sin x & g(x) = e x then nd f ◦ g(x) & g ◦ f(x)
f ◦ g(x) = f(g(x)) = sin(g(x)) = sin(e x ) &
g ◦ f(x) = g(f(x)) = e f(x) = e sin x
Problem. Find the derivative of the function y = f(x) = x
dy
dx = lim f(x + h) − f(x)
h→0 h
x + h − x
= lim
h→0 h
= lim
h
= lim 1
h→0
= 1
h→0
h
That means the derivative of f(x) = x is d
dx (x) = 1
We know the derivative is slope of the tangent at a point x. Now derivative
1 means slope of the tangent to f(x) = x at any point is 1
Problem. Find the derivative of the function y = f(x) = x 2 .
Here in the problem they have not asked at which point of x we want to nd
the derivative means they want us to nd the derivative at any point x. Let us
use the above denition.

21
dy
dx = lim f(x + h) − f(x)
h→0 h
=
(x + h) 2 − x 2
lim
h→0 h
=
2xh + h 2
lim
h→0 h
=
2x + h
lim
h→0 1
(since h → 0, hence h ≠ 0)
= 2x
Problem. Find the derivative of f(x) = x n where n ∈ N
Again using the denition of derivative
dy
dx = lim f(x + h) − f(x)
h→0 h
(x + h) n − x n
= lim
h→0 h
= lim
h→0
h(nx n−1 + c 1 x n−2 h + c 2 x n−3 h 2 + . . . + h n−1 )
h
= nx n−1
Hence we can write
d
dx (xn ) = nx n−1 . Now we can nd any of these
d
dx (x2 ) = 2x
d
dx (x3 ) = 3x 2
d
dx (x4 ) = 4x 3 . So we have ended up nding general formula for derivative of
functions of the form variable constant
Problem. Find the derivative of f(x) = 2
This is a constant function.
dy
dx = lim f(x + h) − f(x)
h→0 h
2 − 2
= lim
h→0 h
= 0
Can you see that derivative of a constant function is coming out to be zero.
That means derivative of any function f(x) = c is Zero.
Problem. Prove the relation d
d
(k · f(x)) = k · (f(x)) where k is a constant.
dx dx
What we mean here is a constant can come out of the derivative operator.
By denition,

22 4 DERIVATIVES
d
k · f(x + h) − k · f(x)
(k · f(x)) = lim
dx h→0 h
= lim
h→0
k ·
= k · lim
= k ·
f(x + h) − f(x)
h
f(x + h) − f(x)
h→0
d
dx (f(x))
Theorem. Derivative of Addition/Subtraction of functions is equal to addition/subtraction
of their derivatives
• d
df(x)
(f(x) ± g(x)) =
dx dx
+ dg(x) 3
dx
• d (f(x) · g(x)) = f(x)df(x)
dx dx
+ g(x)dg(x) dx
• d ( ) f g df
= dx − f dg
dx
dx g g 2
h
3 To nd the derivative of f(x) with respect to x is also called as dierentiating f(x) with
respect to x

4.1 Derivatives of other functions 23
4.1 Derivatives of other functions
We have seen
1.
2.
3.
d
dx (xn ) = nx n−1
(a)
d
dx (k) = 0
(b)
d
dx (x) = x1−1 = x 0 = 1
(c)
d
dx (x2 ) = 2x 2−1 = 2x
(d)
d
dx (x−1 ) = −x −1−1 = −x −2
d
dx (log e x) = 1 x
(a)
d
dx (log a x) = d
we used here :
d
dx (ax ) = a x · log e a
(a)
dx (log e x · log a e) = log a e ·
d
dx (k · f(x)) = k · d
dx (f(x)) )
d
dx (log e x) = log a e
x
( note
d
dx (ex ) = e x (note this is the function whose derivative is the function
itself)
4. Trigonometric functions
(a)
d
(sin x) = cos x
dx
(b)
d
(cos x) = − sin x
dx
(c)
d
dx (tan x) = sec2 x
(d)
d
dx (cot x) = −cosec2 x
(e)
d
(sec x) = sec x cosec x
dx
(f)
d
(cosec x) = −cosec x cot x
dx
4.2 Understanding the Derivative denition geometrically[1]
Given a function y = f(x). Let the curve of this function be as in the adjoining
gure. Let us see what this term means geometrically
f(x + h) − f(x)
h
=
f(x + h) − f(x)
(x + h) − (x)

24 4 DERIVATIVES
Let us locate this term in the gure.
tanθ = ∆y f(x + h) − f(x)
= is the slope of the secant joining the points
∆x h
A & B, where θ is the angle made by the secant with positive x axis.
And the by the denition of the derivative as h → 0 means point B will tend
to come closer and closer to point A i.e. B → A. That means geometrically the
secant joining A & B will become tangent at A. Thats Beautiful! isn't it!
Yes this the geometrical interpretations we are interested in nding the tangent
to the curve y = f(x) and this is a very strong method.
4
Example. I have a ladder of length L = 5m inclined on a wall. Now if the
ladder has no friction with the wall and ground and if it starts sliding under
gravity then
1. What is the displacement of base with oor when the the base with the
wall at 4m height has displaced with 0.1m?
2. What is the velocity of the base with wall when the velocity of the base
with wall is 3m/sec and at a distance of 4m from the ground?
Solution : From the diagram,
x 2 + y 2 = l 2
Also, (x + ∆x) 2 + (y − ∆y) 2 = l 2
Solving both equations we get (∆x) 2 + (∆y) 2 + 2x∆x − 2y∆y = 0
Now we need the instanteneous velocities of both the bases. To nd that we
need to get lim
∆t→0
∆x
∆t & lim
∆t→0
∆y
∆t .
4 Note one thing, in the geometrical interpretation of derivative above point B → A that
is B goes closer and closer to A. But can never reach the point A since if the point B reaches
point A the secant will dissapear instead of determining the tangent. Hence we utilize limit
since the secant if continuously goes near and near to tangent that means the nal limit of
the nearing process is the nal tangent at point A. Interesting hmmm..

4.3 Rate of change 25
Obviously, if we say, very small time ∆t has elapsed and we are still making
it closer to zero by doing ∆t → 0. That means even the displacement ∆x & ∆y
are going closer to zero too. So (∆x) 2 & (∆y) 2 are becoming more smaller and
can be neglected at ∆t → 0.
Hence we have,
x∆x − y∆y = 0
1. We need to nd ∆x when we know y = 4, x = 3 & ∆y = 0.1
∆x = y∆y = 0.133m towards right ( positive signies along positive x
x
axis)
2. From the above equation, x ∆x
∆t − y ∆y
∆t = 0
as ∆t → 0 ∆x
∆t = v x & ∆y
∆t = v y
So we have xv x = yv y
Given, v y = 3m/sec & y = 4m ⇒ x = 3m & v x = 4m/sec
Now we see the solution of this problem using derivatives
From geometry we get x 2 + y 2 = L 2
Dierentiating with respect to time t we get
d
dt (x2 ) + d dt (y2 ) = d dt (L2 )
2x dx dy
+ 2y
dt dt
= 0 (Since Derivative of constant = 0)
xv x + yv y = 0
4.3 Rate of change
If y = f(x) then rate of change in y with respect to change in corresponding
change in x.
Rate of change is dened as ∆y
∆x
Similarly,
In physics we need to monitor what is change in displacement x when there
is a change in time t. Since displacment is dened as a function of time.
i.e. x = f(t)
Average velocity = ∆x
∆t = x 2 − x 1
t 2 − t 1
= f(t 2) − f(t 1 )
t 2 − t 1
lim Average V elocity =
velocity at time t 1
f(t 2 ) − f(t 1 )
lim
t 2→t 1 t 2 − t 1
4.4 Function & its examples in Physics
Let us see some examples of how functions exist in physics.
= dx
dt = v = Instantaneous
1. A person drives his car in a straight line such that he starts from a point
A and movesx a distance of 2m in every one second. That means in t

26 4 DERIVATIVES
seconds will travel 2t metres. So the function relating the time taken t to
travel x distance is given by 5 x = 2t
or x(t) = 2t
So what is the rate of change of displacement with respect to time, between
t = 1 and t = 2.
We know that rate of change of displacement is ∆x
∆t = x 2 − x 1
= 4 − 2
t 2 − t 1 2 − 1 =
2. In physics, rate of change of diplacement is termed as velocity.
Hence the average velocity is 2.
What if we want to nd the instateneous velocity at t = 1. For that we
need to get the secant at t = 1 which is a tangent. And we know that the
tangent is the derivative. So the velocity at t = 1 is dx
dt = d dt (2t) = 2
2. A particle starts moving from a starting point A such that velocity at
every point is given by v = 2t 3 where t is the time taken. Then how is
acceleration related to velocity and time taken.
v = 2t 3
The rate of change of velocity is termed in physics as acceleration i.e.
a = dv . Dierentiating we get,
dt
a = dv
dt = d dt (2t3 ) = 2 d dt (t3 ) = 2 · 3t 2 = 6t 2
4.5 Instantaneous rate of change[3]
Now we know that average rate of change in the function y = f(x) is represented
as ∆y and to get the instantaneous rate we use the limit of average
∆x
∆y
lim
∆x→0 ∆x
5 Since x can be expressed in terms of variable t. Second notation describes that. We have
see if a function f is expresses in terms of x then we write it as y = f(x).

4.6 Chain Rule and friends 27
Similarly,
Instantaneous Velocity
= lim Average V elocity
∆x
= lim
∆t→0 ∆t
f(t 2 ) − f(t 1 )
= lim
∆t→0 t 2 − t 1
= dx
dt
where Displacement is a function of time i.e. x = f(t)
Instantaneous Acceleration = dv
dt = lim ∆v
∆t→0 ∆t
4.6 Chain Rule and friends
If y is a function of t and t is a function of x then
dy
dx = dy
dt · dt
dx
From chain rule we can nd the derivative of composite functions
In y = f(g(x)) let t = g(x) & y = f(t)
Dierentiating,
t = g(x) ⇒ dt
dx = g′ (x)
y = f(t) ⇒ dy
dt = f ′ (t)
Using Chain rule
dy
dx = dy
dt · dt
dx
= f ′ (t) · g ′ (x)
d
dx f(g(x)) = d ∣
∣∣∣t=g(x)
dt (f(t)) d
·
dx g(x)
Parametric form
If we have y = φ(t) and x = ψ(t) two functions which both depend on t.
Then
dy
dx · dx
dt
= dy
dt
dy
dx = dy/dt
dx/dt
4.7 Composition of functions
(using chain rule)
Denition. If f(x) & g(x) are functions of x then composition of f & g are
dened as
(f ◦ g)(x) = f(g(x))
This can be generalised as (f ◦ g ◦ h ◦ · · · ◦ s)(x) = f(g(h(· · · s(x) · · · )))

28 4 DERIVATIVES
Example. A general non-mathematical example of composite function.
Rotation of fan (r) depends on the current passing into the fan (c), mathematically
can be written r = f(c)
And the ux of air (a) made to move by the fan on rotation (r), mathematically
can be written a = g(r)
This composite function current(c) → rotation(r) → air flux(a)
This same example can be converted to parametric form
Rotation of fan (r) depends on current(c) i.e. r = f(c)
Also air ux(a) coming from the fan depends on the current(c) i.e. a = g(c)
Example. Given f(x) = sin x & g(x) = x 2 then (f ◦ g)(x) & (g ◦ f)(x) ?
Solution :
(f ◦ g)(x) = f(g(x)) = f(x 2 ) = sin(x 2 )
(g ◦ f)(x) = g(f(x)) = g(sin x) = (sin x) 2 = sin 2 x
Example. Given f(x) & g(x) as in the above example nd (f ◦ g ◦ f ◦ g)(x)
(f ◦ g ◦ f ◦ g)(x) = f(g(f(g(x))))
= f(g(f(x 2 )))
= f(g(sin x 2 ))
= f(sin 2 x 2 )
= sin(sin 2 x 2 )
Problem. Practice problems
1. If f(x) = x 2 & g(x) = √ x then nd f ◦ g & g ◦ f
2. If f(x) = √ 1 + x 2 & g(x) = √ x 2 − 1 then nd f ◦ g & g ◦ f
4.8 Derivative of a composite function
If y = (f ◦ g)(x) is a composite of functions of f(x) & g(x) then
y = f(x), x = g(t) ⇒
y = f(g(t)) same as composite.
right!
(f ◦ g) ′ (x) = f ′ (g(x)) · g ′ (x)
d(f ◦ g)
dx
This is same as chain rule,
If y = f(x) & x = g(t) then
= df(g(x)
dx
dy
dt = dy
dx · dx
dt
· dg
dx
Example. If the displacement x(t) is given as a function of time t as x(t) =
2t 2 + 1
Instanteneous Velocity is v(t) = d dt (x(t)) = d dt (2t2 + 1) = 4t
Instanteneous Acceleration is a(t) = d dt (v(t)) = d dt (4t) = 4

4.8 Derivative of a composite function 29
Rate of change of Velocity with respect to displacement dv
dx = dv
dt /dx dt =
4/4t = 1 t

30 5 INTEGRATION
Part IV
Lecture 4
5 Integration
5.1 Dierentials
Let us rst understand what do we mean by dierentials
Dierential of x is ∆x and is denoted as dx
lim
∆x→0
So a dierential dx is a small dynamic change in x (Remember dierential
of x is d(x) = dx)
Example. See the following worked examples of dierentials
1. Dierential of y is dy
2. Dierential of f(x) is d(f(x)) = f ′ (x)dx
3. Dierential of sin x is d(sin x) = cos x dx
4. Dierential of sin x 2 is d(sin x 2 ) = cos x 2 2x dx
5. Dierential of f(g(x)) is d(f(g(x))) = f ′ (g(x)) g ′ (x)dx
This can be seen in a dierent way, dierential of f(x) is d(f(x)) this can be
produced by dierenting f(x) with respect to the independent variable here i.e.
x
d df(x)
f(x) =
dx dx
= f ′ (x) ⇒ d(f(x)) = f ′ (x) dx
5.2 Indenite Integration
Denition. It is dened as reverse operation of Dierentiation.
Anti-derivative.
Also called
Notation: ´
f(x)dx : read as integral of f(x) dx which means we are trying to nd
that function whose derivative with respect to x is f(x)
for example: ´ sin x dx = − cos x : means derivative of − cos x is sin x
1.
2.
3.
4.
d
dx (xn+1 ) = (n + 1) · x n ↔ ´ x n dx = xn+1
n + 1
d
dx (ax ) = a x · log e a ↔ ´ a x dx =
ax
log e a
d
dx (log e x) = 1 x ↔´ 1
x dx = log e x
d
dx (ex ) = e x ↔ ´ e x dx = e x
5. Trigonometric functions
where n ≠ −1

5.3 Denite Integration 31
(a)
d
dx (sin x) = cos x ↔ ´ cos x dx = sin x
(b)
d
dx (cos x) = − sin x ↔ ´ sin x dx = − cos x
(c)
d
dx (tan x) = sec2 x ↔ ´ sec 2 x dx = tan x
5.3 Denite Integration
Denition. It is the area under the curve y = f(x) from x varying from a to
b (see the diagram)
Notation :
x=b ´
f(x) : read as integral of f(x) from x = a to x = b.
x=a
Example. We see here if we have a function y = f(x) and want to nd the
area bounded between x = a (point A) and x = b (point B) along x-axis. And
along y-axis its bounded by y = f(x) and x-axis. This is represented by denite
integral
´b
a
f(x) dx
We can nd this area under the curve (as its called) using approximation.
We cover the area with rectangles and keep on increasing the number of such
rectangles (just as we worked out to approximate the area of circle with area
of n-gon). And then we nd limit of this process which turns out to be denite
integral stated above.
In the gure below, we have shown some stages of increasing the number
of rectangles between x = a & x = b. Here n = 4, 7, 12, 26, 60, the limits case
nally.
5.3.1 Way to evaluate Denite Integrals
This is called the Fundamental Law

32 5 INTEGRATION
d
dx F (x) = f(x) then ´ f(x) dx = F (x)
x=b ´
Hence f(x) dx = F (x)| b a
= F (b) − F (a)
x=a
Example. Evaluate
2´
x 2 dx
We know ´ x 2 dx = x3
3 + c 2
1
ˆ
1
x 2 dx = x3
3
= 23
∣
2
1
3 − 13
3
= 7/3
Example. Find the area of the circle with radius R using integration
Method I
Area of the circle is integration of all small circles concentric with the given
circle (as shown in the gure)
We take a random such small thickness circle at a variable distance r from
the center.
6
Now the area of the thin ring of dx thickness at a distance of r (random)
from center = 2πr · dr
Now we collect all such thin rings which are at a distance of zero(0) to R
6 dx = lim ∆x so dx is an every decreasing and shrinking quantity. Hence the limit of
∆x→0
the approximation gets us the integration to give the right answer.

5.3 Denite Integration 33
from the center. And that is done with integration
A =
ˆ
r=R
r=0
2πrdr
= 2π rdr| R 0
= 2π r2
R
2 ∣
0
( R
2
= 2π
2 − 0 )
2
= πR 2
Method II
Now we make the small element in a dierent sense. See the gure
To nd the area of the circle we are collecting all small sectors of angle dθ.
For that we collect all small sectors lying at an angle θ from positive x-axis
where θ = 0 to 2π.
A =
θ=2π ˆ
θ=0
1
2 R2 θdθ
= 1 θ=2π ˆ
2 R2 θdθ
= 1 2
θ=0
2π
θ2
R2 2 ∣
0
= 1 2 R2 ( 4π2
2 − 0 2 )
= πR 2
Method III
Here in this third type of solution, the strip is parellel to y-axis with dx
thickness and at a distance of x (variable) from y-axis.

34 5 INTEGRATION
Area of the thin strip = 2 √ R 2 − x 2 dx
Area of the circle is by collecting all such strips from x = −R to x = R
Area of circle =
x=R ˆ
x=−R
2 √ R 2 − x 2 dx
= 2 · x √
R2 − x
2
2 + R2
2 sin−1 ( x ∣ ∣∣∣
x=R
R )
x=−R
= 2{(0 + R2
2 sin−1 (1)) − (0 + R2
2 sin−1 (−1))}
= 2 R2
2 (π 2 − (−π 2 ))
= πR 2
Example. Derive Kinematical Equations v = u + at, s = ut + 1 2 at2 , v 2 =
u 2 + 2as
In the kinemetical equations, acceleration is assumed to be constant.
a = dv
dt = constant = a
⇒ dv = a ⇒ dv = adt (Note : this is the dierential form)
dt
Integrating,
v´ ´t
dv = adt
u
0
⇒ v| v u = at| t 0
⇒ v − u = at
i.e. v = u + at
Now v = ds = u + at ⇒ ds = udt + atdt
dt
Integrating,
´s
0
ds =
s| s 0 = ut| t 0 + 1 2 at2 | t 0
s = ut + 1 2 at2
´t
0
´t
udt + atdt
0

5.4 Methods of Solving Integrals 35
Finally, dv
dt = a
dv ds
ds dt = a ⇒ dv v = a ⇒ vdv = ads
ds
v´ ´s
vdv = ads
u
0
v 2
2 − u2
2 = as ⇒ v2 − u 2 = 2as
Example. Find the Volume of a sphere.
Let us take a small shell at a random distance r from center of the sphere.
The thickness of this shell is dr. So volume taken by the shell is 4πrdr.
4
3 πR3
Hence volume of the sphere of radius R=
5.4 Methods of Solving Integrals
5.4.1 Method of substitution
Ŕ
0
4πr 2 dr = 4π
Ŕ
0
r 2 dr = 4π r3
3
This is most widely exployed method of integration while solving problems. Here
we substitute a part of the integrand such that the integral problems becomes
a simple integral problem that we might have solved earlier.
Let us learn this with examples
Example. ´ log x
x
dx
We Observe that some part of the integrand,
transform the problem to a much simpler problem.
We see d
dx (log x) = 1 x ⇒ d(log x) = dx x
Let t = log x ⇒ dt = log x · dx
Problem becomes, ´ tdt = t2 2 + c ⇒ log2 x
2
+ c
log x
x
∣
R
0
=
can be substituted to

Index
Cartesian coordinate system, 4
Domain of a function, 9
Examples of limits, 11
function, 8
Inclination of a line, 7
Polygon becomes circle, 11
range of a function, 9
Slope of a line, 5
36

REFERENCES 37
References
[1] Thomas & Finney - Calculus & Analytical Geometry 6ed.
[2] I A Maron - Problems in One Variable Calculus
[3] Piskunov - Dierential & Integral Calculus
[4] Richard Goldberg - Methods of Real Analysis
[5] Bartle & Sherbert - Introduction to Real Analysis