An Insight into Differentiation and Integration
Differentiation
Differentiation is basically a task to find out how one variable is changing in relation
to another variable, the latter is usually taken as a cause of the change.
For instance, y = x 2
We might consider x taking values: 1, 2, 3, ……. n, n + 1
successively by increment 1, then y is increasing by an increment
(n + 1) 2 − n 2 = 2n + 1 at x = n
The average increment over the interval (n, n + 1) of x is
(n + 1) 2 − n 2
= 2n + 1
1
To be more generally,
the average increment over the interval ( x, x + 1) of x is
By decreasing the amount of increment of x to be
( x + 1) 2 − x 2
= 2x +1
1
1
, the average increment over the
2
1
interval ( x, x + ) of x is
2
1 2
( x + ) − x2
1
2
= 2 x + ( )2
1
2
2
Generalizing, the average increment over the interval ( x, x + δ x) of x is
( x + δ x) 2 − x 2
= 2 x + (δ x) 2
δx
Going on this way, we then come to the idea that the instantaneous average increment
at x is 2 x , when the interval becomes widthless. This is defined as the rate of change
of x 2 with respective to x at any value of x .
To express the process of work, which we define as differentiation, and the result, we
write
d 2
( x + δ x)2 − x 2
= 2x
x = lim
δ x →0
δx
dx
We have three fundamental results of differentiation:
d n
x = nx n −1 for positive integer n
dx
sin δ x
d
sin x = cos x (Based on the fundamental limit result lim
= 1 ), when
δ x →0 δ x
dx
x is in radians
d x
e = ex
dx
1
From these, with the helps of a number of rules such as The Product Rules, The
Quotient Rule, The Chain Rule, Implicit Function Rule… we deduce many standard
results including
d n
x = nx n −1 for any rational number n (Finally, for any real number n )
dx
Of course, we should be able to establish all other results from first principles, which
means a work relying only on the definition
d
f ( x + δ x) − f ( x)
f ( x) = lim
δ
x
→
0
dx
δx
Integration
Differentiation is important not only that it is a tool to track the change of one variable
with respect to another, also, knowing of the derivative of a function leads us to the
identification of the function itself.
d
d
For example, if we find that
f ( x) =
g ( x) = u ( x) , then f ( x) and g ( x) can only
dx
dx
differ by a constant
d
So, given
f ( x) = u ( x) , to find f ( x) , we first look for g ( x) which has the known
dx
derivative u ( x)
The job is ∫ u ( x)dx , which gives a result pending a constant
We write f ( x) = g ( x) + c
If we know that f (a) = 0 , then c = − g (a) , f ( x) = g ( x) − g (a) . This is a definite
result, which gives value of f ( x) at any value, for example, at x = b ,
f (b) = g (b) − g (a) , for which we write
∫
b
a
u ( x) dx
The value is definite and the work is called a definite integration
Cases of using derivative to find the original functions
A well-known case is finding distance traveled with knowledge of velocity v . If s is
the distance traveled in time t , we have
ds
= v , based on the genuine definition of v
dt
For some other cases, the derivative of a variable must be reasoned out, though they
are usually taken intuitively. Take the case when we come to such work like finding
area under a curve.
2
Y
y+δ y
y
δA
A(x)
O
a
X
x
δx
If we define A(x) to be the area under the curve from x = a to x
dA
= y , so A= ∫ ydx , Area A(x) can then be found
Intuitively, we take
dx
To be a bit more rigorous
We first claim that δ A lies between yδ x and (y + δ y )δ x ,
δA
So,
lies between y and (y + δ y )
δx
dA
δA
= lim
=y
δ
x
→
0
δx
dx
Take another case, for finding the volume V(h) of a right circular cone of height h
Denoting V(h) to be the volume of the circular cone, of the same shape, with a height
h , considered as a variable, S to be the surface area of the circular cross-section at
height h,
3
A
B
S
r
δh
V(h)
h
O
We claim that δ V lies between Sδ h and (S + δ S)δ h
δV
lies between S and (S + δ S)
δh
dV
δV
so
= lim
=S .
δ
h
→
0
dh
δh
By similarity, r = kh, for a constant k
h3
+c
3
1
1
1
= π (kh) 2 h + c = π r 2 h + c = π r 2 h if we take V = 0 for h = 0
3
3
3
V= ∫ Sdh = ∫ π (kh) 2 dh = π k 2
Exercise (without solution attached)
1.
2.
3
Differentiate tan x with respect to x from first principles.
sin δ x
(You may use the result lim
= 1)
δ x →0 δ x
d n
Base on the result
x = nx n −1 for positive integer n , and the rules of
dx
d 13 1 − 23
differentiation, prove that
x = x
dx
3
2
1
d 3 2 −3
Prove also that
x = x
dx
3
A particle is moving vertically upwards with velocity v = 50t − 5t 2
Show that it is moving downwards after t = 10
Find the distance of the particle from its position at t = 0
4
(i)
(ii)
4.
5.
6.
When t = 8
When t = 12
12
5
s = ∫ (50t − 5t 2 )dt = [25t 2 − t 3 ] 12
0 = 720
0
3
dy
Given that y = sec x 0 , find
dx
Given a reason why in Calculus, radian measure is preferred rather than
degree measure.
A kite is at a horizontal distance l from the flyer and a vertical distance h
above the same. When it is rising up at a velocity v , the flyer let off more
string to keep the horizontal distance constant.
Assuming the string to be straight, and the angle of elevation of the kite to be
θ radians , find the rate of change of θ with respective to time t
x
Let (e* ) x = (1 + ) n , where n is a large number
n
d * x
Show that
(e ) ≈ (e* ) x
dx
5
Exercise (with solution attached)
1.
Differentiate tan x with respect to x from first principles.
sin δ x
(You may use the result lim
= 1)
δ x →0 δ x
Solution:
tan( x + δ x) − tan x sin( x + δ x) cos x − cos( x + δ x) sin x
=
=
δ x[cos( x + δ x) cos x]
δx
sin( x + δ x − x)
sin δ x
1
=
δ x[cos( x + δ x) cos x]
δ x cos( x + δ x) cos x
tan( x + δ x) − tan x
sin δ x
1
= lim[
] = sec2 x
So, lim
δ x →0
δ x →0
δ x cos( x + δ x) cos x
δx
d n
x = nx n −1 for positive integer n , and the rules of
dx
d 13 1 − 32
differentiation, prove that
x = x
dx
3
2
1
−
d 3 2 3
x = x
Prove also that
dx
3
Solution:
1
dy
1
1
1 − 23
3
2 dy
3
Let y = x , then y = x , 3 y
= 1,
=
= 2 = x
3
dx
dx 3 y 2
3x 3
d 23 2 − 13
For proving of
, let z 3 = x 2
x = x
dx
3
2.
Base on the result
A particle is moving vertically upwards with velocity v = 50t − 5t 2
Show that it is moving downwards after t = 10
Find the distance of the particle from its position at t = 0
When t = 8
(iii)
When t = 12
(iv)
Solution:
v = 50t − 5t 2 = 5t (10 − t ) < 0 when t > 10
So, the particle is moving downwards after t = 10
If s is the distance moved upwards after t
(i)
ds
= v = 50t − 5t 2 , noting that this is true for all t ,
dt
that is whether t < 10 or t ≥ 10
8
5
2
s = ∫ (50t − 5t 2 )dt = [25t 2 − t 3 ] 80 = 746
0
3
3
3.
(ii)
ds
= v = 50t − 5t 2 , even though t ≥ 10
dt
12
5
s = ∫ (50t − 5t 2 )dt = [25t 2 − t 3 ] 12
0 = 720
0
3
6
dy
dx
Given a reason why in Calculus, radian measure is preferred rather than
degree measure.
Solution
Let y = sec x 0 = sec u , where u radians = x degrees . Then 180u = xπ
dy dy du
π
π
= (sec x 0 tan x 0 )
=
= (sec u tan u )
180
180
dx du dx
d
π
0
0
sec x = (sec x tan x 0 )
As seen from the above,
180
dx
d
Whereas
sec u rd = sec u tan u
du
4.
Given that y = sec x 0 , find
5.
A kite is at a horizontal distance l from the flyer and a vertical distance h
above the same. When it is rising up at a velocity v , the flyer let off more
string to keep the horizontal distance constant.
Assuming the string to be straight, and the angle of elevation of the kite to be
θ radians , find the rate of change of θ with respective to time t
Solution
h
We have tan θ = , where h is a variable and l is a constant
l
dθ
is what we are looking for, we need not
Though an expression for
dt
h
dθ
.We can choose to do indirectly
write θ = tan −1 ( ) , and do
l
dt
dθ 1 dh 1
Differentiating both sides of this equation, we get sec 2 θ
=
= v
dt l dt l
dθ
1
= (cos 2 θ ) v , though the result is not totally in terms of the variable h .
So
dt
l
x
Let (e* ) x = (1 + ) n , where n is a large number
n
d * x
Show that
(e ) ≈ (e* ) x
dx
Solution
d * x
d
x
x
1
x
x
1
(e ) =
(1 + ) n = n(1 + ) n −1 ( ) = (1 + ) n −1 = (1 + ) n
n (1 + x )
dx
dx
n
n
n
n
n
x n
x
* x
≈ (1 + ) = (e ) , as (1 + ) ≈ 1
n
n
6.
7