Principle of Derivative Calculus - Differential Calculus - An Introduction by Arun Umrao

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CALCULUS
DERIVATIVE-AN INTRODUCTION
Arun Umrao
www.sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING

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Contents
1 Derivatives 9
1.1 Discrete Sampling . . . . . . . . . . . . . . . . . . . . . . . . 9
1.1.1 First Order Difference . . . . . . . . . . . . . . . . . . 9
1.1.2 Second Order Difference . . . . . . . . . . . . . . . . . 11
1.1.3 Sampling & Derivative . . . . . . . . . . . . . . . . . . 25
1.1.4 Derivative Representation . . . . . . . . . . . . . . . . 29
1.2 Numerical Derivative . . . . . . . . . . . . . . . . . . . . . . . 29
1.2.1 Sum of Differences . . . . . . . . . . . . . . . . . . . . 34
1.3 Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 37
1.3.1 Infinitesimal . . . . . . . . . . . . . . . . . . . . . . . . 41
1.3.2 Interpretation of Derivative . . . . . . . . . . . . . . . 45
dy/dx at a Point . . . . . . . . . . . . . . . . . . . . . 46
1.3.3 First Principle Method . . . . . . . . . . . . . . . . . . 49
1.4 Direct Derivative . . . . . . . . . . . . . . . . . . . . . . . . . 57
1.4.1 Partial Derivative . . . . . . . . . . . . . . . . . . . . 57
1.4.2 Total Derivative . . . . . . . . . . . . . . . . . . . . . 58
1.4.3 Properties of Derivatives . . . . . . . . . . . . . . . . . 58
1.4.4 Odd & Even Function . . . . . . . . . . . . . . . . . . 66
1.4.5 Absolute Functions . . . . . . . . . . . . . . . . . . . . 69
1.5 Rules of Derivatives . . . . . . . . . . . . . . . . . . . . . . . 72
1.5.1 Product Rule . . . . . . . . . . . . . . . . . . . . . . . 72
Geometrical Representation . . . . . . . . . . . . . . . 74
1.5.2 Quotient rule . . . . . . . . . . . . . . . . . . . . . . . 78
1.5.3 Reverse Derivative Simplification . . . . . . . . . . . . 83
1.5.4 Derivative by Partial Fraction . . . . . . . . . . . . . . 85
1.5.5 Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . 89
1.5.6 Logarithmic Differential . . . . . . . . . . . . . . . . . 91
1.5.7 Derivatives of Inverse Function . . . . . . . . . . . . . 98
1.5.8 Derivatives of Hyperbolic Function . . . . . . . . . . . 102
1.5.9 Parametric Functions . . . . . . . . . . . . . . . . . . 104
1.5.10 Derivation of Function wrt a Function . . . . . . . . . 106
1.5.11 Advanced Derivative . . . . . . . . . . . . . . . . . . . 107
Meaning of f
′
(x) > 0 and f
′
(x) < 0 . . . . . . . . . . 107
Meaning of f
′′
(x) > 0 and f
′′
(x) < 0 . . . . . . . . . . 111
1.6 Derivative by Trigonometric Manipulation . . . . . . . . . . . 119
1.7 Mean Value Theorems . . . . . . . . . . . . . . . . . . . . . . 122
1.7.1 Roll’s Theorem . . . . . . . . . . . . . . . . . . . . . . 122
1.7.2 Mean Value Theorem (Lagrange Theorem) . . . . . . 125
1.8 Linearization by Derivative . . . . . . . . . . . . . . . . . . . 127

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2 Applications 129
2.1 Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
2.1.1 Linear Motion . . . . . . . . . . . . . . . . . . . . . . 129
2.1.2 Circular Motion . . . . . . . . . . . . . . . . . . . . . 133
2.1.3 Angular Motion . . . . . . . . . . . . . . . . . . . . . 133
2.1.4 Pendulum Oscillation . . . . . . . . . . . . . . . . . . 134
2.1.5 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
2.2 Electronics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
2.2.1 Electricity . . . . . . . . . . . . . . . . . . . . . . . . . 136
2.3 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
2.3.1 Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
2.3.2 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
2.3.3 Volume . . . . . . . . . . . . . . . . . . . . . . . . . . 143
3 Differential Equations 147
3.1 Solution of DE . . . . . . . . . . . . . . . . . . . . . . . . . . 147
3.1.1 First Order DE . . . . . . . . . . . . . . . . . . . . . . 147
By Separation of Variables . . . . . . . . . . . . . . . 147
By Linear Constant Coefficient . . . . . . . . . . . . . 148
3.1.2 Function with Limits . . . . . . . . . . . . . . . . . . . 149
3.1.3 Function with Integration . . . . . . . . . . . . . . . . 150
4 Partial Derivative 153
4.1 Euler’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 153
4.1.1 First Order Derivative . . . . . . . . . . . . . . . . . . 153
4.1.2 Total Differentiation . . . . . . . . . . . . . . . . . . . 153

1.1. DISCRETE SAMPLING 5
1Derivatives
1.1 Discrete Sampling
The value of a function at a given point is called function value. It is also
known as sampled value of the function at that point. For example, function
y = sin(x) has sample value y = sin(90◦
), i.e. ‘1’ at x = 90◦
. Sampling of
function is basis of calculus and communication engineering. Mathemati-
cally, a sampled value of a function, f(x), at a sampling point xn, is given
by
y[n] = f(xn) (1.1)
Here, xn is the nth
instantaneous value of independent variable, which is
given by xn = x0 + nh. Here x0 is first value of independent variable. y[n]
is nth
sampled value of function..
1.1.1 First Order Difference
The difference of sampled values is called sample difference and it is given
by
∆y[n] = y[n] − y[n − 1] (1.2)
Equation (1.2) is equation of backward difference. While forward difference
is given by
∆y[n] = y[n + 1] − y[n] (1.3)
It is also called first order difference equation. The block diagram of differ-
ence equation is given by
+
p0 Delay
x[n] y[n]
p0 × x[n]
x[n]
ℜ × p0 x[n]

6 Derivatives
From the above figure, the output is
y[n] = x[n] + p0 ℜ x[n]
Here ℜ right-shift operator which delays sample by one step. Its value is
any positive integer. See the following table in which impulse signal is right
sifted multiple times.
I = 1, 0, 0, 0, 0, 0, . . .
ℜI = 0, 1, 0, 0, 0, 0, . . .
ℜ2
I = 0, 0, 1, 0, 0, 0, . . .
ℜ3
I = 0, 0, 0, 1, 0, 0, . . .
ℜ4
I = 0, 0, 0, 0, 1, 0, . . .
ℜ5
I = 0, 0, 0, 0, 0, 1, . . .
.
.
.
.
.
.
From the above table, ℜ1
cause delay by one step, hence ℜ x[n] becomes
x[n − 1]. Similarly, ℜ2
cause delay sample by two steps, hence ℜ2
x[n] be-
comes x[n − 2]. So,
y[n] = x[n] + p0 x[n − 1] (1.4)
The following block is similar to the above diagram but it is called feedback
diagram.
+
p0 Delay
x[n] y[n]
p0 × ℜ y[n]
x[n] + p0 ℜ y[n]
ℜ × y[n]
From above figure, we see that the output of the block diagram is
y[n] = x[n] + p0 ℜ y[n]
Solving it, we get the difference equation as
y[n](1 − p0 ℜ) = x[n] (1.5)

Here p0 is pole of the difference equation. The output of above block circuit,
about the pole is give by
y[n] =

pn
0 if n ≥ 1
0 otherwise
(1.6)
If pole (say p0 here) is −1 ≤ p0 ≤ 1, the output is convergent otherwise
divergent. The response of functional equation of difference system to a
sample is exponent, i.e. pn
u(x).
1.1.2 Second Order Difference
The difference of values obtained from first order sample differences is given
by
∆∆y[n] = ∆y[n + 1] − ∆y[n]
Substituting the values of ∆y[n + 1] and ∆y[n], we have
∆2
y[n] = y[n + 2] − y[n + 1] − (y[n + 1] − y[n])
Or
∆2
y[n] = y[n + 2] − 2y[n + 1] − y[n] (1.7)
It is also called second order difference equation. The block diagram of
difference equation is given by
+
p0 Delay
x[n]
y1[n]
+
p1 Delay
y[n]
Equivalent diagram of above block diagram is

8 Derivatives
+
p0
Delay
p1
Delay
x[n] y[n]
From the above figure, the output of first block is
y1[n] = x[n] + p0 ℜ x[n]
The output of first block is input of the second block. The output of second
block is given by
y[n] = y1[n] + p1 ℜ y1[n]
Substituting the value of y1[n] in to above equation, we have
y[n] = (1 + p1 ℜ)(1 + p0ℜ) x[n] (1.8)
Here ℜ is delay, i.e. number of samples delayed. ℜ1
is delayed by one step,
hence ℜ x[n] becomes x[n − 1]. Similarly, ℜ2
is delayed by two steps, hence
ℜ2
x[n] becomes x[n − 2]. So, above relation becomes
y[n] = x[n] + (p0 + p1) x[n − 1] + (p0p1) x[n − 2] (1.9)
The following block is similar to the above diagram but it is called feedback
diagram.
+
p0 Delay
x[n]
y1[n]
+
p1 Delay
y[n]
Equivalent diagram of above block diagram is

+
p0
Delay
p1
Delay
x[n] y[n]
From the above figure, the output of first block is
y1[n](1 − p0 ℜ) = x[n]
The output of first block is input of the second block. The output of second
block is given by
y[n](1 − p1 ℜ) = y1[n]
Substituting the value of y1[n] in to above equation, we have
y[n](1 − p1 ℜ)(1 − p0ℜ) = x[n]
Or
y[n]
x[n]
=
1
(1 − p0ℜ)(1 − p1 ℜ)
(1.10)
The standard form of above relations is
y[n] = x[n] + (p0 + p1)ℜy[n] − p0p1ℜ2
y[n] (1.11)
Using partial fraction method to factorize the right hand side, the equation
becomes
y[n]
x[n]
=
a
(1 − p0 ℜ)
+
b
(1 − p1ℜ)
(1.12)
This is way to represent second order difference equation in form of first
order difference equation. The equation (1.12) represents to the gain of the
system of which the given difference equation is. It is free of the independent
variable. Hence gain of the system is purely a function of gain and it is called
system functional (transform function).
Solved Problem 1.1 To get the sampled values of function f(x) between the
domain of 0 x 1. Only five samples are being obtained. Find the ∆x.

10 Derivatives
Solution The domain of x is 0 x 1. There may be infinite number
of starting and ending values of upper and lower bound of x for sampling
points. For finite computation, let lower and upper bounding values of x are
0.1 and 0.9 respectively. Now, the width between two consecutive sampling
points, when end points are taken as sampling points, is
∆x =
0.9 − 0.1
5 − 1
= 0.2
Solved Problem 1.2 Six sampling points within domain of 0 ≤ x ≤ 1 are
taken. Find the ∆x. Also find the x5.
Solution The lower and upper bound values of x are 0 and 1. The
sampling width, ∆x is given by
∆x =
1 − 0
6 − 1
= 0.2
The starting sampling point is x0 = 0. x5 is 5th
sampling point and it is
given by xn = x0 + nh. So, x5 = 0 + 5 × 0.2 = 1.
Solved Problem 1.3 Find the four sampled values of the function y = x2
within 0 ≤ x ≤ 1 and their first difference.
Solution The four arbitrary selected points within 0 ≤ x ≤ 1, are 0.0,
0.3, 0.6 and 1.0, which are n = 0th
, 1st
, 2nd
and 3rd
values of x respectively.
The function values at these points are
y[0] = y0 = 02
= 0.00
y[1] = y0.3 = 0.32
= 0.09
y[2] = y0.6 = 0.62
= 0.36
y[3] = y1 = 12
= 1.00
These are four sampled values of the function. The first order difference of
these sampled values is shown in the following table. The first difference is
obtained by using relation
∆y[n] = y[n + 1] − y[n]

n xn y[n] ∆y[n]
0 0.0 0.00
0.09
1 0.3 0.09
0.27
2 0.6 0.36
0.64
3 1.0 1.00
1
x
y[n]
b
b
b
b
1
x
∆y[n]
b
b
b
The sampled value of function and their first difference is shown in above
figure.
Solved Problem 1.4 Find the sin sampled values of the function y = x3
within 0 ≤ x ≤ 1 and the first difference.
Solution The six arbitrary points selected within 0 ≤ x ≤ 1 are 0.0, 0.2,
0.4, 0.6, 0.8 and 1.0 which are n = 0th
, 1st
, . . ., 5th
The function values at these points are
y[0] = y0 = 03
= 0.00
y[1] = y0.2 = 0.23
= 0.008
y[2] = y0.4 = 0.43
= 0.064
y[3] = y0.6 = 0.63
= 0.216
y[4] = y0.8 = 0.83
= 0.512
y[5] = y1 = 13
= 1
These are six sampled values of the function. The first difference of these
sampled value is shown in the following table. The first difference is obtained
by using relation
∆y[n] = y[n + 1] − y[n]

12 Derivatives
n xn y[n] ∆y[n]
0 0.0 0.000
0.008
1 0.2 0.008
0.056
2 0.4 0.064
0.152
3 0.6 0.216
0.296
4 0.8 0.512
0.488
5 1.0 1.000
1
x
y[n]
b b b
b
b
b
1
x
∆y[n]
b b
b
b
b
The sampled value of function and their first difference is shown in above
figure.
Solved Problem 1.5 The difference equation, of feedback systems in series,
is given by (1 − 1.6ℜ + 0.63ℜ2
)y[n] = x[n]. Write this difference equation in
standard form.
Solution The given difference equation is
(1 − 1.6ℜ + 0.63ℜ2
)y[n] = x[n]
+
1.6
ℜ
−0.63
ℜ
x[n] y[n]

The equation in standard form is written by
y[n] = x[n] + 1.6R y[n] − 0.63R2
y[n]
This is standard form of the difference equation of a feedback system.
Solved Problem 1.6 Find the p0 and p1 of the difference equation, of a
feedback systems in series, which is given by (1 − 1.6ℜ + 0.63ℜ2
)y[n] = x[n].
(1 − 1.6ℜ + 0.63ℜ2
)y[n] = x[n]
Representing it in standard form
y[n]
x[n]
=
1
1 − 1.6ℜ + 0.63ℜ2
Factorising to denominator, we have
y[n]
x[n]
=
1
(1 − 0.7ℜ)(1 − 0.9ℜ)
On comparing the denominator with standard for of (1 + p ℜ), we get p0 =
0.7 and p1 = 0.9. It tells that there shall be two difference blocks
connected in series.
+
0.9 Delay
x[n]
y1[n]
+
0.7 Delay
y[n]
In above block the order of p0 and p1 has no meaning and both can be
taken in any order.
Solved Problem 1.7 Find the roots of the difference equation of feedback
systems which are in series, and system is represented by difference equation
(1 − 1.6ℜ + 0.63ℜ2
)y[n] = x[n].
Solution The given difference equation is equation of feedback type sys-
tem. The given equation is
(1 − 1.6ℜ + 0.63ℜ2
)y[n] = x[n]

14 Derivatives
y[n]
x[n]
=
1
1 − 1.6ℜ + 0.63ℜ2
y[n]
x[n]
=
1
(1 − 0.7ℜ)(1 − 0.9ℜ)
To get the pole, put ℜ →
1
z
and solve the relation for z. So,
y[n]
x[n]
=
z2
(z − 0.7)(z − 0.9)
To get the pole, denominator should be zero. So, (z − 0.7)(z − 0.9) = 0.
It gives z = 0.7 and z = 0.9. These are the pole of the given difference
equation. Here, a question rises that why have we put ℜ → 1/z here? The
answer is, we know that, ℜ is right shift operator and it is a purely positive
integer value, i.e. 1, 2, 3, etc. Hence if we write (1 − 0.7ℜ) = 0 to get the
pole, it becomes
ℜ =
1
0.7
= 0.3333
This fraction value shall not be acceptable as a value of ℜ. But when it is
written as
0.7 =
1
ℜ
There shall be a positive integer value of the ℜ for which the result is 0.7
and value of ℜ is acceptable. This is why, ℜ is replaced by 1/z.
Solved Problem 1.8 The feedback system blocks of difference equation which
are in series are represented by difference equation (1−1.6ℜ+0.63ℜ2
)y[n] =
x[n]. Convert this difference equation for the equivalent feedback systems,
which are in parallel.
Solution The given difference equation is equation of feedback type sys-
tem. The given equation is
(1 − 1.6ℜ + 0.63ℜ2
)y[n] = x[n]
y[n]
x[n]
=
1
1 − 1.6ℜ + 0.63ℜ2

y[n]
x[n]
=
1
(1 − 0.7ℜ)(1 − 0.9ℜ)
To get the difference relation of feedback systems in parallel equivalent to
the given relation, right hand side of above relation is partially factorized.
Using partial factor method, we shall have
y[n]
x[n]
=
4.5
1 − 0.9ℜ
+
−3.5
1 − 0.7ℜ
Solved Problem 1.9 Find the roots and the convergence of the difference
equation represented by y[n] = −
1
4
y[n − 1] +
1
4
y[n − 2] + x[n − 1] − x[n − 2].
y[n] = −
1
4
y[n − 1] +
1
4
y[n − 2] + x[n − 1] − x[n − 2]
This is difference equation of feedback system. Using the delay relation ℜ1
for one step delay, ℜ2
for two steps delay, above difference equation can be
written as
y[n] = −
1
4
Ry[n] +
1
4
ℜ2
y[n] + ℜ x[n] − ℜ2
x[n]
Multiplying both side by 4 and simplifying this relation, we have
4y[n] = −ℜy[n] + ℜ2
y[n] + 4ℜ x[n] − 4ℜ2
x[n]
Or
(4 + ℜ − ℜ2
)y[n] = (4ℜ − 4ℜ2
)x[n]
Written this equation in general form, we have
y[n]
x[n]
=
4ℜ − 4ℜ2
4 + ℜ − ℜ2
To get the poles of the difference equation of feedback system, put ℜ →
1
z
.
The equation becomes
y[n]
x[n]
=
4z − 4
4z2 + z − 1
The poles of the difference equation are obtained by putting denominator
equal to the zero. So,
4z2
+ z − 1 = 0

16 Derivatives
Solving this equation for z, we have z = −0.64 and z = 0.39. Both poles
of the given difference equation are within −1 ≤ z ≤ 1, hence the difference
equation is convergent to zero. Here, a question rises that why have we put
ℜ → 1/z here? The answer is, we know that, ℜ is right shift operator and
it is a purely positive integer value, i.e. 1, 2, 3, etc. Hence, for example, if a
part of denominator of function operator is (1 − 0.7ℜ), and to get the pole
(1 − 0.7ℜ) = 0, the result becomes
ℜ =
1
0.7
= 0.3333
This fraction value shall not be acceptable as a value of ℜ. But when it is
written as
0.7 =
1
ℜ
There shall be a positive integer value of the ℜ for which the result is 0.7
and value of ℜ is acceptable. This is why, ℜ is replaced by 1/z.
Solved Problem 1.10 Find the difference equation of the block diagram as
shown below.
Solution
+
p
ℜ
x[n] y[n]
Solving this block diagram for the adder block, we have
ℜ(py[n] + x[n]) = y[n]
Expanding this equation, we have
pℜy[n] + ℜx[n] = y[n]
Using right shift Operator ℜ, we have
py[n − 1] + x[n − 1] = y[n]
Shifting the value of n by one step right, as n → n + 1, we get
py[n] + x[n] = y[n + 1]
This is the difference equation of the given block diagram.

Solved Problem 1.11 Find the response function or system functional or
transform function from the difference equation y[n] = αx[n] + βx[n − 2] −
y[n − 2]. Take, α = 2 and β = 4.
y[n] = αx[n] + βx[n − 2] − y[n − 2]
Converting this difference equation in form of right shift operator. So,
y[n] = αx[n] + βℜx[n] − ℜ2
y[n]
Substituting, α = 2 and β = 4 and simplifying this relation in form of
y[n]/x[n], we have
y[n]
x[n]
=
2 + 4ℜ
1 + ℜ2
This is the response function or system functional or transform function of
the given difference equation.
Solved Problem 1.12 Construct the difference table for 1 ≤ n ≤ 10 for given
difference equation y[n] = αx[n] + βx[n − 1]. The x[n] is defined as
x[n] =

1 if n ≥ 0
0 Otherwise
Solution The given difference equation is y[n] = αx[n] + βx[n − 1]. In
the table given below, we have constructed two columns, for n and x[n].
Applying the difference equation for the range 1 ≤ n ≤ 10 to get the values
of samples, i.e. y[n].
y[1] = αx[1] + βx[0] = α + β
y[2] = αx[2] + βx[1] = α + β
y[3] = αx[3] + βx[2] = α + β
y[4] = αx[4] + βx[3] = α + β
y[5] = αx[5] + βx[4] = α + β
y[6] = αx[6] + βx[5] = α + β
y[7] = αx[7] + βx[6] = α + β
y[8] = αx[8] + βx[7] = α + β
y[9] = αx[9] + βx[8] = α + β

18 Derivatives
y[10] = αx[10] + βx[9] = α + β
n x[n] y[n] ∆y[n]
0 1 α + β
0
1 1 α + β
0
2 1 α + β
0
3 1 α + β
0
4 1 α + β
0
5 1 α + β
0
6 1 α + β
0
7 1 α + β
0
8 1 α + β
0
9 1 α + β
0
10 1 α + β
The difference of sampled values, i.e. ∆y[n] is shown in fourth column
of the above table.
difference equation y[n] = αx[n] + βx[n − 1] − y[n − 1]. The x[n] is defined
as
x[n] =

1 if n ≥ 0
0 Otherwise
y[n] = αx[n] + βx[n − 1] − y[n − 1]
In the table given below, we have constructed two columns, for n and x[n].

of samples, i.e. y[n]. Note that, when n 0, x[n] is zero, therefore, all the
values of y[n] for n 0 are also zero. The sampled values are
y[0] = αx[0] + βx[−1] − y[−1] = α
y[1] = αx[1] + βx[0] − y[0] = β
y[2] = αx[2] + βx[1] − y[1] = α
y[3] = αx[3] + βx[2] − y[2] = β
y[4] = αx[4] + βx[3] − y[3] = α
y[5] = αx[5] + βx[4] − y[4] = β
y[6] = αx[6] + βx[5] − y[5] = α
y[7] = αx[7] + βx[6] − y[6] = β
y[8] = αx[8] + βx[7] − y[7] = α
y[9] = αx[9] + βx[8] − y[8] = β
y[10] = αx[10] + βx[9] − y[9] = α
n x[n] y[n] ∆y[n]
0 1 α
β − α
1 1 β
α − β
2 1 α
β − α
3 1 β
α − β
4 1 α
β − α
5 1 β
α − β
6 1 α
β − α
7 1 β
α − β
8 1 α
β − α
9 1 β
α − β
10 1 α

20 Derivatives
of the above table.
difference equation y[n] = αx[n] + βx[n − 2] − y[n − 2]. The x[n] is defined
as
x[n] =

1 if n = 0
0 Otherwise
y[n] = αx[n] + βx[n − 2] − y[n − 2]
In the table given below, we have constructed two columns, for n and x[n].
of samples, i.e. y[n]. Note that, when n 6= 0, x[n] is zero, therefore, all the
values of y[n] for n 0 are also zero. The sampled values are
y[−2] = αx[−2] + βx[−4] − y[−4] = 0
y[−1] = αx[−1] + βx[−3] − y[−3] = 0
y[0] = αx[0] + βx[−2] − y[−2] = α
y[1] = αx[1] + βx[−1] − y[−1] = 0
y[2] = αx[2] + βx[0] − y[0] = β − α
y[3] = αx[3] + βx[1] − y[1] = 0
y[4] = αx[4] + βx[2] − y[2] = α − β
y[5] = αx[5] + βx[3] − y[3] = 0
y[6] = αx[6] + βx[4] − y[4] = β − α
y[7] = αx[7] + βx[5] − y[5] = 0
y[8] = αx[8] + βx[6] − y[6] = α − β
y[9] = αx[9] + βx[7] − y[7] = 0
y[10] = αx[10] + βx[8] − y[8] = β − α

n x[n] y[n] ∆y[n]
-2 0 0
0
-1 0 0
α
0 1 α
−α
1 0 0
β − α
2 0 β − α
α − β
3 0 0
α − β
4 0 α − β
β − α
5 0 0
β − α
6 0 β − α
α − β
7 0 0
α − β
8 0 α − β
β − α
9 0 0
β − α
10 0 β − α
of the above table.
1.1.3 Sampling Derivative
The ratio of difference of sampled values to the difference of two consecutive
values of independent parameter is called first order derivative of the function
and it is given by
∆y
∆x

xn=x0+nh
=
y[n + 1] − y[n]
h
(1.13)

22 Derivatives
Here, h is the difference between two consecutive values of independent vari-
able x as ∆x = h. In case of second order derivative of the function, above
relation becomes
∆2
y
(∆x)2

xn=x0+nh
=
∆y[n + 1]
∆x
−
∆y[n]
∆x
Substituting the values of ∆y[n + 1], ∆y[n] and ∆x = h, we have
d2
y
dx2

xn=x0+nh
=
y[n+2]−y[n+1]
h
h
−
y[n+1]−y[n]
h
h
It gives
d2
y
dx2

xn=x0+nh
=
y[n + 2] − 2y[n + 1] + y[n]
h2
(1.14)
Solved Problem 1.15 Find the sin sampled values of the function y = x3
within 0 ≤ x ≤ 1. Also find the first and second order derivatives of the
function.
Solution The six arbitrary points selected within 0 ≤ x ≤ 1 are 0.0, 0.2,
0.4, 0.6, 0.8 and 1.0 which are n = 0th
, 1st
, . . ., 5th
The difference between two consecutive values of x is 0.2, i.e. ∆x = h = 0.2.
The function values, i.e. samples, at these points are
y[0] = y0 = 03
= 0.00
y[1] = y0.2 = 0.23
= 0.008
y[2] = y0.4 = 0.43
= 0.064
y[3] = y0.6 = 0.63
= 0.216
y[4] = y0.8 = 0.83
= 0.512
y[5] = y1 = 13
= 1
These are six sampled values of the function. The first order and second
order derivatives of these sampled value is shown in the following table. The
first order derivative is obtained by using relation
dy
dx

xn=x0+nh
= y′
[n] =
y[n + 1] − y[n]
h
and second order derivative is given by
d2
y
dx2

xn=x0+nh
= y′′
[n] =
y[n + 2] − 2y[n + 1] + y[n]
h2

n xn y[n] y′
[n] y′′
[n]
0 0.0 0.000
0.040
1 0.2 0.008 1.20
0.280
2 0.4 0.064 2.40
0.760
3 0.6 0.216 3.60
1.480
4 0.8 0.512 4.80
2.440
5 1.0 1.000
1
2
3
4
1
n
y[n]
b b b
b
b
b
1
2
3
4
1
n
y′
[n]
b
b
b
b
b
1
2
3
4
1
n
y′′
[n]
b
b
b
b
To check the values, we shall use the first and second order derivative
relations
y′
[n] =
y[n + 1] − y[n]
h
Substitute n = 2, we have
y′
[n] =
y[3] − y[2]
h
On solving it, we have
y′
[2] =
0.216 − 0.064
0.2
= 0.76

24 Derivatives
And second order derivative is given by
y′′
[n] =
y[n + 2] − 2y[n + 1] + y[n]
h2
Substitute n = 2, we have
y′′
[2] =
y[4] − 2y[3] + y[2]
h2
On solving it, we have
y′′
[2] =
0.512 − 2 × 0.216 + 0.064
0.22
= 3.6
Thus we see that, the direct method is same as sampling and difference
method is.
Solved Problem 1.16 Assume an algebraic series operator
O = 1 − pℜ + p2
ℜ2
− p3
ℜ3
+ . . .
If, f(x) = xn
then find O[f(x)]. Assume that ℜ is derivative operator of x.
Solution The given operator is
O = 1 − pℜ + p2
ℜ2
− p3
ℜ3
+ . . .
The O[f(x)] will be given as
O[f(x)] = [1 − pℜ + p2
ℜ2
− p3
ℜ3
+ . . .]f(x)
Or
O[f(x)] = [1 − pℜ + p2
ℜ2
− p3
ℜ3
+ . . .]xn
It shall give result
O[f(x)] = xn
− pℜxn
+ p2
ℜ2
xn
− p3
ℜ3
xn
+ . . .
Or
O[f(x)] = xn
− pnxn−1
+ p2
n(n − 1)xn−2
− . . . + n(n − 1) · 1
This is desired result.

1.2. NUMERICAL DERIVATIVE 25
1.1.4 Derivative Representation
The derivative or differentiation of a function say f(x) = ax2
+ bx + c can
be represented by three ways.
df/dx: In this type, numerator, df as a whole represent deviation in
value of f with respect to deviation in value of x. Denominator dx tells that
the independent variable is x about which function being derivated.
f
′
(x): The quote sign tells the derivative of the function. Here, the
reader must known the independent variable about which derivative is being
done. Number of quote signs tells the number of times the function being
derivated. For example, f
′′
(x) means function f is derivated two times.
f(1)
(x): The exponent number inside parentheses tells the number of
times the function is derivated. For example, f(2)
(x) means function f is
derivated two times about its independent variable.
˙
f(x): The dot sign tells the derivative of the function. Here, the reader
must known the independent variable about which derivative is being done.
Number of dots tells the number of times the function being derivated. For
example, ¨
f(x) means function f is derivated two times.
1.2 Numerical Derivative
Numerical derivative of a function is given by
f
′
(x) =
f(x + h) − f(x)
(x + h) − (x)
=
f(x + h) − f(x)
h
(1.15)
Here f(x) and f(x + h) are the function values at two consecutive points, x
and (x+ h) respectively. To understand the derivative of a function, assume
an experimental function f(x) = sin(x). Now, function table within domain
of the sin, i.e. from x = 0 to x = π and having a common difference h = 0.1
is given below.

26 Derivatives
x f(x) f
′
(x) f
′′
(x) f
′′′
(x) x f(x) f
′
(x) f
′′
(x) f
′′′
(x)
0.0 0.000 1.5 0.997
0.998 0.021
0.1 0.100 -0.100 1.6 1.000 -0.999
0.988 -0.988 -0.079 0.079
0.2 0.199 -0.199 1.7 0.992 -0.991
0.969 -0.968 -0.178 0.178
0.3 0.296 -0.295 1.8 0.974 -0.973
0.939 -0.938 -0.275 0.275
0.4 0.389 -0.389 1.9 0.946 -0.946
0.900 -0.899 -0.370 0.370
0.5 0.479 -0.479 2.0 0.909 -0.909
0.852 -0.851 -0.461 0.460
0.6 0.565 -0.564 2.1 0.863 -0.862
0.796 -0.795 -0.547 0.547
0.7 0.644 -0.644 2.2 0.808 -0.808
0.731 -0.731 -0.628 0.627
0.8 0.717 -0.717 2.3 0.746 -0.745
0.660 -0.659 -0.702 0.702
0.9 0.783 -0.783 2.4 0.675 -0.675
0.581 -0.581 -0.770 0.769
1.0 0.841 -0.841 2.5 0.598 -0.598
0.497 -0.497 -0.830 0.829
1.1 0.891 -0.89 2.6 0.516 -0.515
0.408 -0.408 -0.881 0.880
1.2 0.932 -0.931 2.7 0.427 -0.427
0.315 -0.315 -0.924 0.923
1.3 0.964 -0.963 2.8 0.335 -0.335
0.219 -0.957
1.4 0.985 2.9 0.239
Table 1.1: Data table for function f(x) = sin(x) for definite range of x from
x = 0 to x ≈ π.
The function plot of f(x) = sin(x) by using function data table is given
in following figure.

−1
0
1
0 1 2 3
x
f(x)
The first order difference of the function table by using relation (1.15) is
given in the column f
′
(x). For example, using relation (1.15) for obtaining
first order difference at point x = 2.5
f
′
(2.5) =
f(2.5 + 0.1) − f(2.5)
0.1
=
0.516 − 0.598
0.1
= −0.83
From the table 1.1, data of f
′
(x) column is equal to the plot of cos x. Sym-
bolically, numerical derivative of sin x is similar to
f
′
(x) =
d
dx
sin(x) = cos(x)
Now the plot of function f(x) and f
′
(x), by using function data from table
1.1, is given in following figure.
−1
0
1
0 1 2 3
x
f(x)
−1
0
1
0 1 2 3
x
f
′
(x)
Second order difference of the function table is given by
f
′′
(x) =
f
′
(x + h) − f
′
(x)
h
(1.16)
Again putting the values of f
′
(x + h) and f
′
(x), above relation becomes
f
′′
(x) =
f(x + 2h) − 2 × f(x + h) + f(x)
h2
(1.17)
For example, using relation (1.16), for obtaining second order difference at
point x = 2.5
f
′′
(2.5) =
f
′
(2.5 + 0.1) − f
′
(2.5)
0.1
=
−0.881 − (−0.83)
0.1
= −0.515

28 Derivatives
Or from relation (1.17)
f
′′
(2.5) =
f(2.7) − 2 × f(2.6) + f(2.5)
0.01
=
0.427 − 2 × 0.515 + 0.598
0.01
≈ −0.515
Symbolically, it is similar to the
f
′′
(x) =
d
dx
f
′
(x) =
d
dx
cos(x) = − sin(x)
Now the plot of function f(x), f
′
(x) and f
′′
(x) by using data from table 1.1
is given in following figure.
−1
0
1
0 1 2 3
x
f(x)
−1
0
1
0 1 2 3
x
f
′
(x)
−1
0
1
0 1 2 3
x
f
′′
(x)
Numerical derivative is more precise if the h → 0, i.e. if h is very small
then the derivative result is more accurate.
Solved Problem 1.17 Find the first order numerical derivative of the function
f(x) = x2
+ 1 within limit from x = 0 to x = 1 and h = 0.1. Plot the
function and its derivative in xy plane. Also justify your answer by using
direct method of derivatives.
Solution The limit of x is from 0 to 1. The width of instantaneous values
of x is h = 0.1. So there are 11 sampling points of x where sampled values
of the function are to be obtained. The table consisting function values at
sampling points and their first order difference is shown in the following
table.

x f(x) f
′
(x)
0.0 1.00
0.1
0.1 1.01
0.3
0.2 1.04
0.5
0.3 1.09
0.7
0.4 1.16
0.9
0.5 1.25
1.1
0.6 1.36
1.3
0.7 1.49
1.5
0.8 1.64
1.7
0.9 1.81
1.9
1.0 2.00
Table 1.2: Data table for function f(x) = x2
+ 1 for definite range of x from
x = 0 to x = 1.
Plots of the function and its derivative are
0
1
2
0 1 2
x
f(x)
0
1
2
0 1 2
x
f
′
(x)
f(x)
Here derivative of function f(x) = x2
+ 1 is a straight line with slope 2x.
To verify it we differentiate the function using direct method.
d
dx
f(x) = 2x

30 Derivatives
At x = 0, df(x)/dx = 0 and at x = 1, df(x)/dx = 2. The derivative function
passes through the points (0, 0) and (1, 2) as shown in second part of the
above figure. Here answer is not such precise as h is sufficiently large i.e.,
h = 0.1. If h is more precise, i.e. h = 0.01 then plot of f(x) and f
′
(x) will
be coincide at x = 1 and at x = 1, f(x) = 2 and f
′
(x) = 2.
1.2.1 Sum of Differences
The basic of difference lies in the difference of an element from its previous
element. Take the original data and find the first order difference as given
in the following table. Here ∆n = ni − ni−1.
n 1 5 8 3 6 7 9
∆n 5-1 8-5 3-8 6-3 7-6 9-7
∆n 4 3 -5 3 1 2
The sum of the first order difference element (∆n) is
X
∆n = 4 + 3 − 5 + 3 + 1 + 2 = 8 = (9 − 1)
Which is equal to difference of the last and first element, i.e.
X
∆n = nlast − nfirst
It is due to the cancellation of inner elements except the last and first element
(see below).
X
∆n = (5 − 1) + (8 − 5) + (3 − 8) + (6 − 3) + (7 − 6) + (9 − 7) = 9 − 1
This is basic principle of the differentiation. Similarly, second order differ-
ence is given in the following table as ∆2
n.
n 1 5 8 3 6 7 9
∆n 5-1 8-5 3-8 6-3 7-6 9-7
∆n 4 3 -5 3 1 2
∆2
n 3-4 -5-3 3+5 1-3 2-1
∆2
n -1 -8 8 -2 1

Now the sum of second order difference is equal to the difference of last
and first term of the first order difference table, i.e. 2 − 4 = −2. It gives a
broad spectral that the difference depends only on the first and last element
and it is independent of in between discrete elements.
Absolute Sum Here, summation takes the actual values of differences of
the given array, i.e. difference value is considered with its sign. But what
happens, if the elements are taken as absolute values? The absolute form is
represented as
|x| =

−x, if x 0
x, if x ≥ 0
n 1 5 8 3 6 7 9
∆n 5-1 8-5 (3-8) 6-3 7-6 9-7
∆n 4 3 +5 3 1 2
∆2
n (3-4) (-5-3) 3+5 (1-3) 2-1
∆2
n +1 +8 8 +2 1
In above table, values within parentheses represents their absolute values.
In this case, sum of absolute differences represents to the error and it depends
on each element of the array. The sum of the first order difference element
(∆n) is X
∆n = 4 + 3 + 5 + 3 + 1 + 2 = 18
An error is considered as the difference of measured value from the actual
value. If measured value is less than the actual value, then it is called under-
measured value. Similarly, if measured value is greater than the actual value
then it is called over-measured value. In both cases, it is error. This is why,
the difference method is also used in measurement of the errors occurred in
non-linear functions.
Difference in Coordinate System In coordinate systems, the difference of
coordinate points is measured either along the abscissa or along the ordinate
independently. But in most cases, their relative relation is measured. The
sum of difference along the abscissa depens only on the abscissa values of
last and first coordinate points. Similarly, the sum of difference along the or-
dinate depens only on the ordinate values of last and first coordinate points.
There may be indefinite numbers of coordinates between two extremum of a

32 Derivatives
curve in xy-plane. But the most interesting property of the coordinate dif-
ference is found when we find the ratio of ordinate difference to the abscissa
difference of two consecutive coordinate points.
x
y
b
b
b
b
b
A
B
x
y
A
f(x)
x
B
f(x + h)
x + h
f(x + h) − f(x)
(x + h) − x
Let A and B are two consecutive points taken for getting abscissa and
ordinate difference of first order. Now difference along the abscissa is (x +
h) − x while the difference along the ordinate is f(x + h) − f(x). These
two differences actually form the base and height of a right angle triangle
respectively as shown below.
x
y
A
f(x)
x
B
f(x + h)
x + h
p
b
θ
Where hypoteneous tends to follow the function curve. It means, if we get
the ratio of first order difference of the ordinate to the first order difference
of abscissa, then ratio shall be slope (m) of the line that is tangent to the
curve according to the trigonometric rule of tan θ ∼ θ, where it is equal to
the ratio of perpendicular to the base of a right angle triangle.
tan θ ∼ θ =
f(x + h) − f(x)
(x + h) − x
= f
′
(x) =
p
b
Note that, if p and h are very small then hypoteneous perfectly follow the
function curve. This is why, in differentiation, values of dx and dy are taken
as small as possible.

1.3. DIFFERENTIATION 33
1.3 Differentiation
Differentiation as its meaning, gives difference between two succes-
sive function values. The differentiated form is analyze to find the whole
properties of the given function.
1
2
−1
1 2 3 4
−1
−2
−3
−4
−5
x
y
y =
1
√
3
x + 1
A
B
C
x
y
θ
Figure 1.1: Slope of a line.
For example a line is drawn as shown in above figure. To interpret the
differentiation we use the geometrical method of finding equation of line.
The equation of line is given as
y = mx + c (1.18)
Where m is slope of line. To find m we must know the ordinate value of line
where abscissa is known. Let ordinate is y where abscissa is x then slope of
line is y/x. Now from the definition of slope
m =
y
x
and
y = mx
This is the equation of line. The most prominent part of the above expla-
nation is the slope of line. Assume line is not straight and its y component
varies from point to point. In this case we can not apply the general relation
m = y/x to find the slope but an advanced procedure is followed. Now
consider a small element of the line as shown in figure 1.1. The vertical
component of this line-segment is ∆y and horizontal component is ∆x then
slope is
m =
∆y
∆x
(1.19)

34 Derivatives
If line segment is very very small, then ∆y becomes dy and ∆x becomes dx
and slope would be
m =
dy
dx
(1.20)
This is third form of the slope of line. This is to be noted that the word-
ing statement of right hand side of above slope would be pronounced like
the slope is the ratio of change in y component of line segment
with respect to its x component. This is called differentiation1
as one
component of a function is differentiated with respect to other component.
Here components can be replaced with suitable variables. The derivative of
a function can be found by using above descriptions. From figure 1.2
x
y
f
′
(x) =
f(x + h) − f(x)
h
A
f(x)
x
B
f(x + h)
x + h f(x + h) − f(x)
(x + h) − x
Figure 1.2: Geometrical representation of derivatives.
Let f(x) is a function of x whose value at x is f(x). Now x changes to
x+h and function value changes to f(x+h). From the equation (1.20) slope
of the element of the function is
m =
f(x + h) − f(x)
(x + h) − x
If h is very samll then test element is very small and the equation (1.20)
becomes
f
′
(x) = lim
dx→0
dy
dx
= lim
h→0
f(x + h) − f(x)
h
(1.21)
Here f
′
(x) denotes the derivative of function f(x) with respect to x and also
called the derivative of y with respect to x. Here, x is independent value.
It is remember that the derivative of a function is taken with respect to the
variable on which function depends. Derivative of a function with respect
to an independent variable is always zero. Here a question is expected that
why do we take dx → 0? Its answer is that, derivative works with linear
1
difference in y values at two ends of small line segment with respect to difference in x

functions. This is why, we take smallest possible value of dx for which
function is instantaneously linear. See the behaviour of non linear function
when its dx → 0.
x
y
dx=1.5
x
y
dx=1
x
y
dx=0.5
x
y
dx=0.1
From above figures, the curve approaches to line when dx → 0. The
behaviour of curve is more linear when dx → 0.1 than its other values,
i.e. dx → 1.5, dx → 1 or dx → 0.5. It means, a non-linear function is
transformed into linear function by derivating it.
Solved Problem 1.18 The velocity function is given by v(t) = 2t2
+ 3. Find
the velocity function at the points t = h and t = h + k.
Solution The velocity function v is a function of t. The point values of
the function at t = h and t = h + k respectively are given by
v(h) = 2h2
+ 3
and
v(h + k) = 2(h + k)2
+ 3
Solved Problem 1.19 Profit of a company is a linear function to the number
(n) of employees working in the company. The profit function is P(n) = n3
−
n/6. Find the profit function when number of employees in that company
are n = k and n = k + h respectively.

36 Derivatives
Solution The profit of a company is a function of number of employees
as given by
P(n) = n3
− n/6
The profits of company, when employees are n = k and n = k+h respectively,
are
P(k) = k3
− k/6
and
P(k + h) = (k + h)3
−
k + h
6
Solved Problem 1.20 Loss of a company due to the maintenance of nonfunc-
tional machines is a linear function to its numbers (n). The loss function is
L(n) = n2
− n + 1. Find the loss function when number of nonfunctional
machines are n = k and n = k + h respectively.
Solution The loss function of a company due to nonfunctional machines
is given by
L(n) = n2
− n + 1
The maintenance loss of the company, when nonfunctional machines are
n = k and n = k + h respectively, are
L(k) = k2
− k + 1
and
L(k + h) = (k + h)2
− (k + h) + 1
Solved Problem 1.21 A hospital conducts survey among the children born
in the hospital within one year to get the Infant Mortality Rate (IMR) due
to diseases. Based on the survey, a regression function/formula is generated.
The base for IMR is age (y) of the child in year. The IMR function is
N(y) = 56 − 2y2
. Find the IMR among the children aged 2 years and 3
years. What do you think about the answers obtained by you?
Solution The Infant Mortality Rate (IMR) function generated by the
hospital is
N(y) = 56 − 2y2
Infant Mortality Rate (IMR) prevailing among the children aged 2 years is
N(2) = 56 − 2 × 22
= 48

Similarly, Infant Mortality Rate (IMR) prevailing among the children aged
3 years is
N(3) = 56 − 2 × 32
= 38
Here, it is seen that the IMR among 2 years old children is higher than the
children 3 years old. It predicts that, infants are more prone to the fatal
diseases than younger.
1.3.1 Infinitesimal
In derivatives, it is assumed that the derivative of a function f(x) is valid
only when change in function is very very small with respect to the change
in value. For example assume a square of side t. Its initial area is Ai = t2
.
Now if the length of side changed by dt then its final area is Af = (t + dt)2
.
Now the change in area is
dA = Af − Ai
t
t
dt
dt
t
t
dt
dt
t2
t · dt (dt)2
Substituting the values of initial and final area of the square, we get
relation
dA = t2
+ (dt)2
+ 2t dt − t2
= 2t dt + (dt)2
If dt is not infinitesimally small then in real world problems, (dt)2
can not
be neglected. For infinitesimally small dt, (dt)2
is not significant but its
product with large value of t becomes significant. It mean that the change
in area is
dA = 2t dt (1.22)
In limit form, we can say that
dA = lim
dt→0

d
dt
A

(1.23)

38 Derivatives
Initial area of square is t2
and its sides are changed by dt. If dt is very
small then (dt)2
has no significant effect in result as the element tends to
infinitely small value. It can be neglected in compare to t · dt. If dt is
significantly large then it is included in the solution.
Solved Problem 1.22 A square has sides of one meter. Its sides are increased
by 10%. Find the percentage change in its area.
Solution
s
s
ds
ds
s
s
ds
ds
s2
s · ds (ds)2
Figure 1.3: Geometrical representation of derivatives for infinitesimally small
variation.
We know that the area of a square having side s is given by
A = s2
It is initial area of the square. Now, length of the square is changed by 10%
of its original length, i.e. 0.1m. Let it increases the area of square by dA.
This change in length of square is not negligible quantity with respect to
initial length, i.e. 1m. Hence applying binomial expansion relation for the
side of square s + ds we get final area of the square A + dA as
A + dA = s2
+ (ds)2
+ 2s ds
Now, increase in the area of the square is difference between initial and final
areas of the square.
dA = 2s ds + (ds)2
Now substituting the values of s and ds we have
dA = 2 × 1 × 0.1 + (0.1)2

On simplification dA = 0.21m2
. The change in area of square is 0.21m2
or
21%. If we use simple derivative method then change in area of square is
dA = 2s · ds
On substituting the value of s and ds is
dA = 2 × 1 × 0.1
On simplification
dA = 0.20 = 20%
It differs from the actual value of 21% as we assume that change in dimension
of square is very small, i.e. ds is neglected. If, in the same problem, side
of square is changed by 1% then change in area calculated by both methods
are same as they are shown below:
dAi = 2s ds dAni = 2s ds + (ds)2
dAi = 2 × 1 × 0.01 dAni = 2 × 1 × 0.01 + (0.01)2
dAi = 0.02m2
dAni = 0.02 + 0.0001
dAni = 0.0201 ≈ 0.02m2
It gives us result that when variation in variables are very small, derivative
method approaches to mathematical method. If variations in variables are
large, then derivative method is failed.
Solved Problem 1.23 Use simple calculation method and derivative method
to find the change in volume of an object given by V = πr2
h + πrl/3 when
radii are given by r = 0.5 ± 0.1, r = 0.5 ± 0.05 and r = 0.5 ± 0.01. h and l
takes as unit constants.
Solution In this problem there are three parts, each for change in radii
values.
r = 0.5 ± 0.1 In this case radius value is r = 0.5 unit and change in
radius is dr = ±0.1 unit. Now, change is volume of the object is obtained by
difference in initial and final volumes. Assume here, that change in radius
is positive, i.e. dr = +0.1. So, from simple numerical method
dV = Vf − Vi
Or
dV =

π × 0.62
× 1 +
π × 0.6 × 1
3

−

π × 0.52
× 1 +
π × 0.5 × 1
3

40 Derivatives
It gives, dV = 1.392 cubic units. 1] Using derivative method
dV =

2πrh +
πl
3

dr
Substituting the values, we get, dV = 0.419 cubic units. 2] Both answers
differ largely with each-other as 0.1 unit is not negligible in respect of 0.5
unit.
radius is dr = ±0.05 unit. Now, change is volume of the object is obtained
by difference in initial and final volumes. Assume here, that change in radius
dV = Vf − Vi
Or
dV =

π × 0.552
× 1 +
π × 0.55 × 1
3

−

π × 0.52
× 1 +
π × 0.5 × 1
3

dV =

2πrh +
πl
3

dr
differ slightly with each-other as 0.05 unit may be negligible in respect of
0.5 unit.
radius is dr = ±0.01 unit. Now, change is volume of the object is obtained
by difference in initial and final volumes. Assume here, that change in radius
dV = Vf − Vi
Or
dV =

π × 0.512
× 1 +
π × 0.51 × 1
3

−

π × 0.52
× 1 +
π × 0.5 × 1
3

dV =

2πrh +
πl
3

dr
are approximately same as here 0.01 unit may be neglected in respect of 0.5
unit.

Solved Problem 1.24 A regression function for a sampled data is y = 2x2
+
3x − 9. Find the data when x is given by x = 2 ± 0.1. Also, find the data if
x is given by x = 2 ± 0.2.
Solution The regression function is y = 2x2
+3x−9. The function value
at x = 2 is
y = 2 × 22
+ 3 × 2 − 9 = 5
Gradient of y along x (variation in y with respect to x) is given by derivatives
of the regression function. So,
d
dx
y =
d
dx
(2x2
+ 3x − 9)
= 4x + 3
Substituting the value of x = 0.1, we get the variation in y corresponding to
the variation of x.
dy = (4x + 3) · dx = (4 × 2 + 3) × 0.1 = 1.1
Including variation to y, the value of y is y = 5±1.1. 1] Now, if the variation
in x is x = 0.2, then in this case, variation in y is given by
dy = (4 × 2 + 3) × 0.2 = 2.2
Including variation to y, the y is y = 5 ± 2.2. 2] Comparing these two
answers, we see that if dx becomes larger then answer deviates largely from
its mean value.
1.3.2 Interpretation of Derivative
Assume y is a function of x. The first derivatives of the function is dy/dx.
This derivative has following meanings.
As Slope dy/dx represents the slope of tangent at a point on a curve.
The slope is measured with respect to x axis. dx/dy represents the slope of
normal at a point on a curve. The slope is measured with respect to y axis.
If θ is the slope of line with respect to the x axis then m = tan θ.
As Gradient Gradient is defined as the ratio of variation in dependent
variable to the variation in the independent variable. For example, assume
that u is function of v. Now v is changing with time, then u shall also be
change with time correspondingly. So
G =
∆u
∆v

42 Derivatives
If u and v varies very slowly with time, i.e. du → 0 and dv → 0, then
G =
du
dv
Compressibility Assume a fluid is flowing inside a pipe of length l. Now,
if fluid is compressible, then volume of fluid entering from one end of pipe
is not equal to the volume of fluid emerging out from the other end. In this
case, If volume entering at one end of the pipe is V then volume emerging
out from the pipe at other end is V −dV/dl. Now, loss of the volume of fluid
along the length of pipe is dV/dl. If the fluid is incompressible then volumes
at both ends of the pipe shall be V . In this case, dV/dl = 0. It means, if a
function is compressible
dV
dl
6= 0
and if function is incompressible or solenoidal then
dV
dl
= 0
Parallel to Axes If dy/dx = 0, then the tangent line is parallel to the x
axis. If dx/dy = ∞, then the tangent line is parallel to the y axis.
Maxima Minima For maxima or minima of a function at a point,
dy/dx must be equal to zero. It means slope of tangent must be zero with
respect to x axis.
1
−1
1 2 3
−1
−2
−3
−4
b
a
f
′
(a)
b
b
f
′
(b)
b
c
f
′
(c)
Figure 1.4: Tangent at maxima and minima points.
dy/dx at a Point
The first derivative of a function y, dependent on x, is given by
y
′
(x) =
dy
dx

if y is continuous and differentiable within the domain of [a, b] then y
′
(x)
shall also be continous within the domain of [a, b].
y
y
′
x
=
a
x
=
b
The nature and characteristics of function y are explained according to
the instantaneous value of y
′
. This instanteneous value of y
′
is found by
puting the point value x = x0 in place of x, where [a ≤ x0 ≤ b].
y
′
(x0) =
dy
dx

x=x0
y
b
b
b
b
bb
b
b
b
b
b
b
b
b b b b
b
b
b
b
b
b
b
bb
b
b
b
b
b
b b
b
b
y
′
x
=
a
x
=
b
b
x = x0
b
y
′
(x0)
From the locations of points
dy
dx

x=xk
the function y is increases contin-
uously within the domain of x where value y
′
(xk) is positive and decreases
continously within the domain of x where value of y
′
(xk) is negative. At the
points xk where y
′
(xk) changes its sign (either from positive to negative or
negative to positive), function changes its nature from continuous increasing
to continuous decreasing or vice-versa. Where y
′
(xk) is zero, the function
forms either crest or trough.
Solved Problem 1.25 Find the slope of a line y = 4x + 5 drawn in xy plane.

44 Derivatives
Solution The slope of line is given by dy/dx, which will be obtained by
derivating equation of line about x. So,
m =
dy
dx
=
d
dx
(4x + 5) = 4
This is slope of the line.
Solved Problem 1.26 Find the slope of tangent drawn in the curve l = j2
+2j
in lj plane at point j = 1. Here, j is measured along horizontal axis and l
is measured along vertical axis.
Solution The slope of the tangent drawn in the curve, l = j2
+ 2j, at
any arbitrary point is given by dl/dj. It is obtained by derivating equation
of curve about j. So,
dl
dj
=
d
dj
(j2
+ 2j) = 2j + 2
The slope of tangent at fixed point j = 1 is given by
m =
dl
dj j=1
= 2 × 1 + 2 = 4
This is required answer.
Solved Problem 1.27 A fluid is flowing in a pipe of radius R. Due to high
viscosity of the fluid, velocity of fluid (v) is function of radius (r) of virtual
fluid tube imagined coaxial to the axis of the pipe. The velocity function is
v = aR−brn
. Here, n 1 and a, b are constants. Find the velocity gradient
between two consecutive fluid surfaces.
Solution
R
b
r
b
v
v − dv
dr
From the definition of gradient, velocity gradient is given by v = dv/dr.
So,
∆v =
d
dr
v =
d
dr
(aR − brn
)

Solving this relation, we have ∆v = −bnrn−1
.
Solved Problem 1.28 Find ẏ(2) of the function y = 2x2
− 9. Also find the
nature of function at the point x = 2.
Solution The given function is y = 2x2
−9. ẏ is first derivative of y with
respect to x.
ẏ =
d
dx

2x2
− 9

= 4x
Again from the question ẏ(2) is given by
ẏ(2) = 4x|x=2 = 8
ẏ(2) is positive, hence function y is increasing function at the point x = 2.
Again, slope (mx=2) of the tangent drawn on the function at x = 2 is upward
because dy/dx or ẏ also represent to the slope of tangent on the function at
an arbitrary point x.
1.3.3 First Principle Method
The difference of the function at point x is given by differences of function
values at points x + h and at point x respectively. Here, x + h is immediate
next point to the point x. So, ∆f(x) = f(x + h) − f(x). The rate at which
difference is changing between these two consecutive points is given by ratio
of the ∆f(x) to the distance between these two points, i.e. (x + h) − x = h.
x
y
f
′
(x) =
f(x + h) − f(x)
h
A
f(x)
x
B
f(x + h)
x + h
f(x + h) − f(x)
(x + h) − x
So relative difference, i.e. derivative is given by the relation
∆f(x)
∆x
= lim
h→0
f(x + h) − f(x)
h
is called first principle method of derivation. The relative difference is more
linear if h approaches to zero, i.e. h → 0, i.e. two consecutive points are

46 Derivatives
closer to each other. In general practice, ∆f(x)/∆x is also written as f
′
(x)
or
d f(x)
dx
.
Solved Problem 1.29 Find the derivative of c.
Solution Let f(x) is a function of x and it is equals to c, where c is
constant. Hence
f(x) = c f(x + h) = c
Now using the first principle methods the derivatives of the function is
f
′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
c − c
h
= lim
h→0
0
h
= lim
h→0
0
Taking limit of relation
f
′
(c) = 0
Hence the derivative of c i.e. constant is 0.
d
dx
c = 0 (1.24)
The above relation exhibits that derivative of c or derivative of constant with
respect to x (ie other that c) is zero.
Solved Problem 1.30 Find the derivative of x.
Solution Let f(x) is a function of x and it is equals to x. Hence
f(x) = x f(x + h) = x + h
f
′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
(x + h) − (x)
h
= lim
h→0
h
h
= lim
h→0
1
Taking limit
f
′
(x) = 1
Hence the derivative of x is 1.
d
dx
x = 1 (1.25)

Solved Problem 1.31 Find the derivative of 3x.
Solution Let f(x) is a function of x and it is equals to 3x. Hence
f(x) = 3x f(x + h) = 3(x + h)
f
′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
3(x + h) − 3(x)
h
= lim
h→0
3h
h
= lim
h→0
3
Taking limit f
′
(x) = 3. Hence the derivative of 3x is 3. Generally,
d
dx
cx = c
d
dx
x = c (1.26)
Solved Problem 1.32 Find the derivative of x2
.
Solution Let f(x) is a function of x and it is equals to x2
. Hence
f(x) = x2
f(x + h) = (x + h)2
f
′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
(x + h)2
− (x)2
h
= lim
h→0
x2
+ h2
+ 2xh − x2
h
= lim
h→0
h2
+ 2xh
h
Simplified relation is
f
′
(x) = lim
h→0
h + 2x
Taking limit in right hand side
f
′
(x) = 2x
Hence the derivative of x2
is 2x.
d
dx
xn
= n(x)(n−1)
(1.27)

48 Derivatives
Solved Problem 1.33 Find the derivative of sin(x).
Solution Let f(x) is a function of x and it is equals to sin(x). Hence
f(x) = sin(x) f(x + h) = sin(x + h)
f
′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
sin(x + h) − sin(x)
h
Expanding trigonometric functions
f
′
(x) = lim
h→0
h
(x + h) − (x+h)3
3! + (x+h)5
5! − . . .
i
−
h
x − x3
3! + x5
5! − . . .
i
h
= lim
h→0
h
x + h − (x3
+h3
+3xh2
+3hx2
)
3! + . . .
i
−
h
x − x3
3! + . . .
i
h
= lim
h→0
h
h − (h3
+3xh2
+3hx2
)
3! + . . .
i
h
On simplifying it
f
′
(x) = lim
h→0
1 −
(h2
+ 3xh + 3x2
)
3!
+ . . .
Taking limit of the relation
f
′
(x) = lim
h→0
1 −
(3x2
)
3!
+ . . .
= lim
h→0
1 −
(x2
)
2!
+ . . .
= cos(x)
Hence the derivative of sin(x) is cos(x).
d
dx
sin(x) = cos(x) (1.28)

Solved Problem 1.34 Find the derivative of cos(x).
Solution Let f(x) is a function of x and it is equals to cos(x). Hence
f(x) = cos(x) f(x + h) = cos(x + h)
f
′
(x) = lim
h→0
f(x + h) − f(x)
h
= lim
h→0
cos(x + h) − cos(x)
h
Expanding trigonometric functions
f
′
(x) = lim
h→0
h
1 − (x+h)2
2! + (x+h)4
4! − . . .
i
−
h
1 − x2
2! + x4
4! − . . .
i
h
= lim
h→0
h
1 − (x2
+h2
+2xh)
2! + . . .
i
−
h
1 − x2
2! + . . .
i
h
= lim
h→0
h
−(2xh+h2
)
2! + . . .
i
h
On simplifying it
f
′
(x) = lim
h→0
−
(h + 2x)
2!
+ . . .
Taking limit of h in right hand side
f
′
(x) = −x +
x3
3!
− . . .
= −

x −
x3
3!
+ . . .

= − sin(x)
Hence the derivative of cos(x) is − sin(x).
d
dx
cos(x) = − sin(x) (1.29)
It is derivative of cos(x).

50 Derivatives
Solved Problem 1.35 Find the derivative of tan(x).
Solution From the first principle method, we have
d
dx
tan x = lim
h→0
tan(x + h) − tan x
h
= lim
h→0
tan x+tan h
1−tan x tan h − tan x
h
= lim
h→0
tan h 1 + tan2
x

h (1 − tan x tan h)
On simplification, and separation of terms so that limit may be applied.
d
dx
tan x = lim
h→0
tan h
h
lim
h→0
1 + tan2
x
1 − tan x tan h
Applying the limits, we have
d
dx
tan x = 1 + tan2
x = sec2
x
This is derivative of the tan x.
Solved Problem 1.36 Find the derivative of ex
.
Solution The derivative of the function is
d
dx
ex
=
d
dx

1 + x +
x2
2!
+
x3
3!
+ . . . . . .

= 0 + 1 +
2x
2 · 1
+
3x2
3 · 2 · 1
+ . . . . . .
= 1 + x +
x2
2!
+
x3
3!
+ . . . . . .
= ex
Derivative of an exponential function is exponential itself.
d
dx
ex
= ex
(1.30)
The graph of the function and its derivative is given below.

1 2 3
x
f
f = ex
1 2 3
x
f
f
′
= ex
There are two types of logarithm operators. One is the natural base
logarithm operator and other is the decimal base logarithm operator. Nat-
ural base logarithm is written as ‘ln’ while decimal base logarithm operator
is denoted by ‘log’. Logarithm with bases other than natural and decimal
are written as ‘logbase’. In Indian context, natural base logarithmic
operator is written as ‘log’. The base of logarithm operator other than
natural base is denoted like ‘logbase’. To avoid the confusion, the fol-
lowing table should be memorise.
Type I Type II
ln log
log log10
logb logb
The most confusion case is arises when operator ‘log’ is used as in type
I, it has base ‘10’ while in type II, it has base ‘e’. One should be careful
while using this operator. The plot of logarithm function is given below.
1 2 3
x
f
ln x
1 2 3
x
f
log x
Solved Problem 1.37 Find the derivative of ln(x).

52 Derivatives
Solution We know that from natural logarithm ln(x) = y gives
x = ey
On differentiation
dx = ey
dy
Now on simplification the derivative becomes
dy
dx
=
1
ey
Substitute the value of ey
the result is
dy
dx
=
1
x
(1.31)
This is required answer. The graph of function and its derivative is given
below.
1 2 3
x
f
f = ln x
1 2 3
x
f
f
′
= 1/x
Solved Problem 1.38 Find the derivative of ax
.
Solution Derivative of given function can be calculated by using loga-
rithm of the function. Hence
y = ax
On taking logarithm with natural base e
lne y = x ln a
On differentiation with respect to x
1
y
dy
dx
= ln a

1.4. DIRECT DERIVATIVE 53
Now on simplification the derivative becomes
dy
dx
= y ln a
Substitute the value of y the result is
dy
dx
= ax
ln a (1.32)
1.4 Direct Derivative
Let a function f(x) is defined by f(x) = axn
+ bx + c. Now the derivative
of the function is
f
′
(x) =
d
dx
f(x)
=
d
dx
(axn
+ bx + c)
Using law of distribution
f
′
(x) =
d
dx
axn
+
d
dx
bx +
d
dx
c
= a
d
dx
xn
+ b
d
dx
x +
d
dx
c
Applying direct relation of derivatives
f
′
(x) = anxn−1
+ b + 0
= anxn−1
+ b
Note that, derivative of a function is also a function.
1.4.1 Partial Derivative
The concept of partial derivative is based on the principle that, when a
multidimensional function is derivated along one dimension or parameter,
other parameters are taken as constant. For example, a three dimensional
force
~
F = Fxî + Fyĵ + Fzk̂
is derivated about x only, while y and z are considered as constant. So,
∂ ~
F
∂x
=
∂Fx
∂x
î

54 Derivatives
Partial derivative is helpful in those functions which are compressible, i.e.
function itself varies with distance or time; as well as distance or time also
are variable. For example, a compressible fluid when flows in a pipe, is
compressed along the length of pipe. The gradient of the fluid compression
in volume is ∂V/∂l. This compression in volume increases rapidly if length
of pipe increases. Now for a small element of pipe length dl, the exact
compressible gradient volume of the fluid is
∂V
∂l
× dl
1.4.2 Total Derivative
f is function of t, u, v, w, . . . and u, v, w, . . . are also the functions of t
then total derivative of function does not depend directly only with t but
also depends on u, v, w, . . . indirectly for t. Mathematically derivative of
the function
y = f(t, u, v, w, . . .)
is given as
∂y
∂t
=
∂f
∂t
+
∂f
∂u
∂u
∂t
+
∂f
∂v
∂v
∂t
+
∂f
∂w
∂w
∂t
+ . . . (1.33)
1.4.3 Properties of Derivatives
There are following properties that are used in derivatives.
1. The coefficient of a function can be taken out of the derivative operator.
d
ds
[a f(s)] = a
d
ds
[f(s)]
2. If two functions are in addition then they can be segregated from the
addition sign.
d
ds
[f(s) + g(s)] =
d
ds
[f(s)] +
d
ds
[g(s)]
Solved Problem 1.39 Find the derivative of function f(x) = 3x3
+ 9x with
respect to x.
Solution Differentiating the function with respect to x
f
′
(x) =
d
dx
f(x)
=
d
dx
(3x3
+ 9x)

Applying the additive property
f
′
(x) =
d
dx
3x3
+
d
dx
9x
Now take constant coefficients out of the derivative operator
f
′
(x) = 3 ×
d
dx
x3
+ 9 ×
d
dx
x
Or
f
′
(x) = 3 × 3x2
+ 9
Or
f
′
(x) = 9x2
+ 9
Solved Problem 1.40 Find the derivative of function f(x) = ex
+ 8 with
respect to x.
˙
f(x) =
d
dx
f(x)
=
d
dx
(ex
+ 8)
= ex
Solved Problem 1.41 Find the derivative of function f(x) = ln x + x3
.
˙
f(x) =
d
dx
f(x)
=
d
dx
(ln x + x3
)
=
1
x
+ 3x2

56 Derivatives
Solved Problem 1.42 Find the derivative of function f(x) = tan x + cos x
with respect to t. Here x is a function of t.
Solution Differentiating the function with respect to t
˙
f(x) =
d
dt
f(x)
=
d
dt
(tan x + cos x)
As function and derivative base are different, hence changing the base of
derivative to x from t.
˙
f(x) =
d
dx
f(x) ×
dx
dt
= (sec2
x − sin x)
dx
dt
Here x is a function of t.
Solved Problem 1.43 Find the derivative of function f(x) = ax
+ x.
Solution Given function is f(x) = ax
+ x. Derivation of this function
with respect to ‘x’
d
dx
f(x) =
d
dx
(ax
+ x)
Applying direct method of derivatives in right hand side of the above relation
f
′
(x) = ax
ln(a) + 1
Solved Problem 1.44 Find the first order derivative of the cubic function x3
with respect to x. Also find the point where cubic function and its derivative
intersects to each other.
Solution
18
36
1 2 3
x
f
f = x3
18
36
1 2 3
x
f
f
′
= 3x2

The given function is y = x3
. Getting derivative of this function with
respect to x, we have
y′
=
d
dx
y =
d
dx
(x3
)
It gives, y′
= 3x2
. The point where both function shall intersects, i.e. y = y′
is given by
x3
= 3x2
⇒ x3
− 3x2
= 0
On Solving it, we have x = 0, x = 0 and x = 3. These are the point, where
both y and y′
intersects to each other.
Solved Problem 1.45 Find the first order derivative of the function f(x) =
x3
+ 3x2
− 9 with respect to x. Also find the integer point(s) where this
function and its derivative intersect to each other.
Solution The given function is f = x3
+ 3x2
− 9. Getting derivative of
this function with respect to x, we have
f
′
=
d
dx
f =
d
dx
(x3
+ 3x2
− 9)
It gives, f
′
= 3x2
+ 6x. The point where both function shall intersects, i.e.
f = f
′
is given by
x3
+ 3x2
− 9 = 3x2
+ 6x
Or
x3
− 6x − 9 = 0
Solving this equation by substitution and reduction method to get the roots,
i.e. points. We get x = 3, x = −1.5±0.87i. From all these points, the integer
point is only x = 3.
Solved Problem 1.46 Find the first order derivative of the function f =
x3
+ x2
with respect to x. Also find the integer points where this function
and its derivative intersects to each other.
Solution

58 Derivatives
1
2
−1
1
−1
x
f
f
=
x
3
+
x
2
1
2
−1
1
−1
x
f
f
f
′
10
20
30
1 2
x
f
The given function is f(x) = x3
+ x2
. Getting derivative of this function
with respect to x, we have
f
′
(x) =
d
dx
f(x) =
d
dx
(x3
+ x2
)
It gives, f
′
(x) = 3x2
+ 2x. The point where both function shall intersects,
i.e. f(x) = f
′
(x) is given by
x3
+ x2
= 3x2
+ 2x
Or
x3
− 2x2
− 2x = 0
Solving this equation, we get x = 0, x = 1 ±
√
3. From all these points, the
integer point is only x = 0.
Solved Problem 1.47 Find the derivative of parabolic function y = 4x2
with
respect to x.
Solution
18
36
1 2
x
f
f = 4x2
18
36
1 2
x
f
f
′
= 8x

The given function is y = 4x2
. Getting derivative of this function with
respect to x, we have
y′
=
d
dx
y =
d
dx
(4x2
)
It gives, y′
= 8x.
Solved Problem 1.48 Evaluate Dv [ln(1 + v)].
Solution First we will use derivative formula for logarithm function and
then for first degree algebraic equation of v. So,
Dv [ln(1 + v)] =
d
dv
ln(1 + v) =
1
1 + v
×
d
dv
(1 + v)
It gives result
1
1 + v
.
Solved Problem 1.49 Evaluate Dx [(a + x)n
].
Solution In this problem (a+x)n
has base (a+x). This base is dependent
only to x. So, we can use derivative formula D(xn
) to it. So,
Dx [(a + x)n
] =
d
dx
((a + x)n
)
It gives result as n(a + x)n−1
.
Solved Problem 1.50 Evaluate Du [a + ueu
].
Solution In this problem, independent variable is u and the expression
is sum of two terms. So, first write the relation as
Du [a + ueu
] = Dua + Du(ueu
)
Or
Dua + Du(ueu
) =
d
du
a +
d
du
ueu
It gives result as (u + 1)eu
.
Solved Problem 1.51 Evaluate Dx [x tan x].
Solution There are product of two functions, x and tan x. The derivative
of this problem is obtained by using product rule. So,
Dx [x tan x] = x
d
dx
tan x + tan x
d
dx
x

60 Derivatives
Applying direct derivative formulas for x and tan x, we have
Dx [x tan x] = x sec2
x + tan x
Solved Problem 1.52 Evaluate Dt [ln(t)].
Solution This is derivative of logarithm function having logarithmic base
‘e’. So,
Dt [ln(t)] =
1
t
Solved Problem 1.53 Evaluate Dx

u + sin2
x

.
Solution In this problem, there are two terms, u and sin2
x. The deriva-
tive variable is x, hence u is taken as constant in this problem. So,
Dx

u + sin2
x

=
d
dx
u +
d
dx
sin2
x
Applying chain rule of derivation right hand side for (sin x)2
, we have
Dx

u + sin2
x

= 2 sin x × cos x
It gives Dx

u + sin2
x

= sin 2x.
Solved Problem 1.54 We know that dy/dx is explained by two ways. First,
y is changing with respect to x. Second, it is ratio of variation in y to
variation in x. First is known as gradient, commonly known as position
gradient (velocity), force gradient (viscous force) etc, while second is known
as slope of tangent at a point (m or tan θ). Using second explanation, find
the slope of tangent in a function f(x) = x2
− 2x + 3 at any point.
Solution The slope of a tangent drawn on the function at any point is
df/dx. Now
m =
d
dx
x2
− 2x + 3

Or
m = 2x − 2
It is the slope of tangent drawn on the function at any arbitrary point.

Solved Problem 1.55 We know that dy/dx is explained by two ways. First,
y is changing with respect to x. Second, it is ratio of variation in y to
variation in x. First is known as gradient, commonly known as position
gradient (velocity), force gradient (viscous force) etc, while second is known
as slope of tangent at a point (m or tan θ). Using second explanation, find
the slope of tangent of a function y = x2
− 4x + 2 at any point. Also find
the points where tangent at x = 1 intersects to the x-axis and y-axis.
Solution Function y is dependent to the variable x. Now the slope of
tangent drawn on the function is dy/dx. So, on derivating to the function,
we have slope, m, as
m = 2x − 4
Slope of the tangent on the function at point x = 1
m = 2 × 1 − 4 = −2
Equation of tangent is line equation which is given by, y = mx + c. Using
the value of m, we have slope equation
y = −2x + c
At x = 1, the function gives function value y = −1 and from above relation
c = 1. So, the equation of tangent is
y = −2x + 1
1
2
−1
−2
1 2 3
x
f
f = x2
− 4x + 2 1
2
−1
−2
1 2 3
x
f
b
ytang = −2x + 1 1
2
−1
−2
1 2 3
x
f
b
b
b

62 Derivatives
When Tangent Intersects x-axis In this case, y = 0 and x = 1/2. Hence
point of intersection is (1/2, 0).
When Tangent Intersects y-axis In this case, x = 0 and y = 1. Hence
point of intersection are (0, 1).
1.4.4 Odd Even Function
Odd Function A function is said to be odd function if it becomes negative
when its independent variable is substituted by make itself negative. For
example, if f(x) is a function of x and it is said to be an odd function if
f(x) = f(−x) = −f(x) (1.34)
Odd functions may not be plotted symmetrically about any axis. Odd func-
tion may also be written as
f(x) =
f(x) − f(−x)
2
See the following figure, in which odd function f(x) = x3
− 4x is plotted.
x
f
f
Even Function A function is said to be even function if it remains same
when its independent variable is substituted by make itself negative. For
example if f(x) is a function of x and it is said to be an even function if
f(x) = f(−x) = f(x) (1.35)
Even functions may be plotted symmetrical about the axes. Even function
may also be written as
f(x) =
f(x) + f(−x)
2

See the following figure, in which even function f(x) = x2
− 4 is plotted.
x
f
f
Solved Problem 1.56 Show that derivative of an odd function is an even
function.
Solution The odd function is given by
f(x) =
f(x) − f(−x)
2
Derivating it both side with respect to x.
D[f(x)] =
D[f(x)] − D[f(−x)]
2
Here, D[f
′
(−x)] is equal to the f
′
(−x) × D[(−x)] which is f
′
(−x) × −1.
D[f(x)] =
f
′
(x) − f
′
(−x) × −1
2
Or
D[f(x)] =
f
′
(x) + f
′
(−x)
2
The right hand side represents to an even function, yet it is a derivative form
of another odd function f(x). It shows that, derivative of an odd function is
an even function. In the following figure, odd function f(x) = x3
− 4x and
its derivative f
′
(x) (which is even) are plotted.

64 Derivatives
x
f
f
f
′
Solved Problem 1.57 Show that derivative of an even function is an odd
function.
Solution The even function is given by
f(x) =
f(x) + f(−x)
2
Derivating it both side with respect to x.
D[f(x)] =
D[f(x)] + D[f(−x)]
2
Or
D[f(x)] =
f
′
(x) + f
′
(−x) × −1
2
Or
D[f(x)] =
f
′
(x) − f
′
(−x)
2
The right hand side represents to an odd function, yet it is a derivative form
of another even function f(x). It shows that, derivative of an even function
is an odd function. In the following figure, even function f(x) = x2
− 4 and
its derivative f
′
(x) (which is odd) are plotted.
x
f
f
f
′

1.4.5 Absolute Functions
A function is said to be absolute if its value is always positive irrespective of
values of independent variables. For example, if f(x) = x − 2 is a function
of x of degree one, then g(x) = |f(x)| = |x− 2| is an absolute function, as at
every point −∞ ≤ x ≤ ∞, value of |f(x)| is positive. An absolute function
is always written in a form of piecewise function. To represent so, first we
find the point where function is zero. For the given function
g(x) = |x − 2| = 0 ⇒ x = 2
Now, piecewise function is written about this point like
g(x) = |f(x)| =

−(x − 2) when x 2
x − 2 when x ≥ 2
Here, why −(x − 2) for all values of x 2? This is because, when x 2,
value of x − 2 becomes negative. As function is absolute, then all these
negative values should be positive. To make negative values to positive, we
multiplied them by −1 for all x 2.
1
−1
1 2 3
x
f
1
−1
1 2 3
x
g

66 Derivatives
x f(x) = x − 2 |f(x)| = |x − 2|
1.5 -0.5 0.5
1.6 -0.4 0.4
1.7 -0.3 0.3
1.8 -0.2 0.2
1.9 -0.1 0.1
2.0 0.0 0.0
2.1 0.1 1.0
2.2 0.2 1.0
2.3 0.3 1.0
2.4 0.4 1.0
2.5 0.5 1.0
Table 1.3:
In the above table, values of x are taken from 1.5 ≤ x ≤ 2.5 and corre-
sponding function values f(x) and g(x) = |f(x)| are tabulated in second and
third column respectively. Point being noted here that, in absolute form,
the values of second column are become positive as written in third column.

x |f(x)| = |x − 2| g
′
(x)
1.5 0.5
-1.0
1.6 0.4
-1.0
1.7 0.3
-1.0
1.8 0.2
-1.0
1.9 0.1
-1.0
2.0 0.0
1.0
2.1 0.1
1.0
2.2 0.2
1.0
2.3 0.3
1.0
2.4 0.4
1.0
2.5 0.5
Table 1.4:
In the table 1.4, we find the first difference of the values of |f(x)| as
given in third column. Here, one interesting thing is found here that is,
when x 2, g
′
(x) or ∆|f(x)| is negative and g
′
(x) or ∆|f(x)| is positive
when x 0. At point x = 2, g
′
(x) or ∆|f(x)| does not exits.

68 Derivatives
1
−1
1 2 3 4
x
|f|
|f(x)|
f
′
(x)
Figure 1.5:
It can be verified by using direct derivative method to absolute function.
The absolute function is
g(x) = |f(x)| =

−(x − 2) when x 2
x − 2 when x ≥ 2
Its derivative can be given by
g
′
(x) = ∆|f(x)| =

−1 when x 2
1 when x ≥ 2
That is identical to the third column of table 1.4 and plot shown in graph
1.5. Here, left side derivative (−1) is not equal to the right side derivative
(+1), so, this function g(x) is not differentiable at point x = 2.
1.5 Rules of Derivatives
Derivative of two functions either in product or in division, requires some
rule that are to be used during differentiation. Basic differential rules are
explained below.
1.5.1 Differential By Parts
For two functions which are in addition or subtraction, derivative rule
D(f ± g) = D(f) ± D(g)
is applicable for any type of f and g functions. Here D is differential operator
representing to d/dx. But in case of product, i.e. f ∗ g, above rule is failed.
In case of product of two function
D(f ∗ g) 6= D(f) ∗ D(g)

1.5. RULES OF DERIVATIVES 69
Illustrated Example If f = sin x and g = cos x then
f ∗ g = (sin 2x)/2
Now
D(f ∗ g) = cos 2x
Again D(f) = cos x and D(g) = − sin x and
D(f) ∗ D(g) = − sin x cos x
This way
D(f ∗ g) 6= D(f) ∗ D(g)
Product Rule If f(x) and g(x) are the functions of x and they are in
product, then derivatives of the f(x) · g(x) is given by
d
dx
f(x) · g(x) = f(x)
d
dx
g(x) + g(x)
d
dx
f(x) (1.36)
This methods is used when two trigonometric or trigono-algebraic or algebraic-
logarithmic functions are in product form.
Proof The product of functions is f(x) · g(x) and its derivative with
respect to x, from the first principle of derivation is
d
dx
[f(x) · g(x)] = lim
h→0
f(x + h) · g(x + h) − f(x) · g(x)
h
For reduction of formula for f(x) and g(x), adding and subtracting f(x +
h) · g(x) in right hand side
d
dx
[f(x) · g(x)] = lim
h→0

f(x + h) · g(x + h) − f(x + h) · g(x)
h
+
f(x + h) · g(x) − f(x) · g(x)
h

Rearranging the right hand side of above relation
d
dx
[f(x) · g(x)] = lim
h→0
f(x + h) · g(x + h) − f(x + h) · g(x)
h
+ lim
h→0
f(x + h) · g(x) − f(x) · g(x)
h

70 Derivatives
Taking commons in both terms in right hand side
d
dx
[f(x) · g(x)] = lim
h→0
f(x + h)
g(x + h) − g(x)
h
+ lim
h→0
g(x)
f(x + h) − f(x)
h
Expanding the limit
d
dx
[f(x) · g(x)] = lim
h→0
f(x + h) · lim
h→0
g(x + h) − g(x)
h
+ lim
h→0
g(x) · lim
h→0
f(x + h) − f(x)
h
Applying the differential rule and taking limits
d
dx
[f(x) · g(x)] = f(x) · g′
(x) + g(x) · f
′
(x)
It is the proof of product rule.
Geometrical Representation
The derivative of product of two functions can be represented by geometry.
Let x and y are two functions representing two sides of a rectangle. The
initial area of the rectangle is xy. The sides of rectangle are changed by dx
and dy along x axis and y axis respectively.
b
l
db
dl
lb
l · db
b · dl
dl · db
Figure 1.6: Geometrical representation of product rule.
Now change in area of the rectangle is
dA = xy + x · dy + y · dx + dx · dy − xy
= x · dy + y · dx + dx · dy
On dividing above relation by dt, the result is
dA
dt
=
x · dy + y · dx + dx · dy
dt

If dx and dy are very small then dx dy has no significant meaning. Hence
dA
dt
= x
dy
dt
+ y
dx
dt
(1.37)
Solved Problem 1.58 Evaluate f(x) = x · sin(x).
Solution Applying product rule of derivative
d
dx
f(x) =
d
dx
[x · sin(x)]
= x
d
dx
sin(x) + sin(x)
d
dx
x
= x cos(x) + sin(x)
Solved Problem 1.59 Evaluate f(x) = x2
· sin(x).
Solution Applying product rule of derivative
d
dx
f(x) =
d
dx

x2
· sin(x)

= x2 d
dx
sin(x) + sin(x)
d
dx
x2
= x2
cos(x) + 2x sin(x)
Solved Problem 1.60 Find d(xy).
Solution Applying product rule assuming both x and y are not constant.
d
dx
(xy) = x
d
dx
y + y
d
dx
x
= x
dy
dx
+ y
On simplifying both sides
d(xy) = x dy + y dx (1.38)
This method is very important in solving linear differential equation as all
the terms in right hand side can be written as
x dy + y dx = d(xy) (1.39)

72 Derivatives
and integration of right hand side very easy than gives
Z
d(xy) = xy + c (1.40)
It is required answer.
Solved Problem 1.61 log and ln represents to the logarithmic functions hav-
ing base ‘10’ and ‘e’ respectively. Assuming that ln is logarithmic function of
exponential base ‘e’, find the derivative of the function f(a) = a ln(a) about
‘a’.
Solution Here is a is variable, so
d
da
f(a) =
d
da
(a ln(a))
Using product rule and applying derivative rules for logarithm function, we
have
f
′
(a) = a ×
1
a
+ ln(a) = 1 + ln(a)
This is required result.
Solved Problem 1.62 log and sin represents to the logarithmic and sine func-
tions respectively. Base of logarithmic function is ‘10’. Find the derivative
of the function f(x) = sin(x) log(x) about ‘x’.
Solution Here is x is variable and the function is f(x) = sin(x) log(x).
Changing the base of logarithm ‘10’ to ‘e’, we have
f(x) = sin(x) ×
ln(x)
2.303
Differentiating it both side with respect to ‘x’
d
dx
f(x) =
d
dx

sin(x) ×
ln(x)
2.303

Using product rule and applying derivative rules for logarithm function, we
have
f
′
(x) =
sin(x)
2.303x
+
cos(x) ln(x)
2.303

Solved Problem 1.63 Exponent functions are those functions which have
base and power, represented like ab
where ‘a’ is base and ‘b’ is power. Assume
that ‘a’ is constant and ln is logarithmic function having natural base ‘e’.
Find the derivative of the given function f(x) = ax
ln(x) about variable ‘x’.
Also find the derivative of the function about ‘a’, when ‘x’ is constant?
Solution In the first case ‘x’ is variable and ‘a’ is constant. So,
Dxf(x) = Dx[ax
ln(x)] = ax
Dx ln(x) + ln(x) Dx[ax
]
Applying the direct derivative, we have
f
′
(x) = ax
×
1
x
+ ln(x) × ax
× ln(a)
On simplification, we have result
f
′
(x) =
ax
x
+ ln(a)ax
ln(x)
This is first part of answer. If ‘x’ is constant and ‘a’ is variable then derivative
shall be about ‘a’ and function will be of variable ‘a’. The result will be
Daf(a) = Da[ax
ln(x)] = xax−1
ln(x)
This is second part of answer.
Solved Problem 1.64 Find the derivatives of the following function and find
the relation between them. Can you find the nth
derivative of function,
where n is an positive integer.
f(t) = (t − 2)2
(3 − t)
f(t) = (t − 2)3
(3 − t)
f(t) = (t − 2)4
(3 − t)
f(t) = (t − 2)5
(3 − t)
and
f(t) = (t − 2)n
(3 − t)
Solution Derivative of the function f(t) = (t − 2)2
(3 − t) with respect
to t is
d
dt
f(t) =
d
dt

(t − 2)2
(3 − t)

= 2(t − 2)(3 − t) − (t − 2)2

74 Derivatives
Derivative of the function f(t) = (t − 2)3
(3 − t) with respect to t is
d
dt
f(t) =
d
dt

(t − 2)2
(3 − t)

= 3(t − 2)2
(3 − t) − (t − 2)3
d
dt
f(t) =
d
dt

(t − 2)2
(3 − t)

= 4(t − 2)3
(3 − t) − (t − 2)4
d
dt
f(t) =
d
dt

(t − 2)2
(3 − t)

= 5(t − 2)4
(3 − t) − (t − 2)5
Using the symmetry of above these result, the derivative of function f(t) =
(t − 2)n
(3 − t) is
d
dt
f(t) =
d
dt
[(t − 2)n
(3 − t)]
= n(t − 2)n−1
(3 − t) − (t − 2)n
On simplification, the final required result is (t − 2)n−1
[n(3 − t) − (t − 2)].
1.5.2 Derivative of Quotient
If f(x) and g(x) are the functions of x and they are in fraction, then deriva-
tives of the f(x)/g(x) is given by
d
dx

f(x)
g(x)

=
g(x) d
dx f(x) − f(x) d
dx g(x)
[g(x)]2
(1.41)
This methods is used when two trigonometric or trigono-algebraic or algebraic-
logarithmic functions are in fraction form.
Proof The division of functions is f(x)/g(x) and its differentiation with
respect to x, from the first principle of derivation is
d
dx

f(x)
g(x)

= lim
h→0
f(x+h)
g(x+h) − f(x)
g(x)
h

Or
d
dx

f(x)
g(x)

= lim
h→0
f(x + h) · g(x) − f(x) · g(x + h)
g(x) · g(x + h)h
For reduction of formula for f(x) and g(x), adding and subtracting f(x)·g(x)
in right hand side
d
dx

f(x)
g(x)

= lim
h→0
f(x + h) · g(x) − f(x) · g(x) + f(x) · g(x) − f(x) · g(x + h)
g(x) · g(x + h)h
Rearranging the right hand side of above relation
d
dx

f(x)
g(x)

= lim
h→0
f(x+h)·g(x)−f(x)·g(x)
h + f(x)·g(x)−f(x)·g(x+h)
h
g(x) · g(x + h)
Or
d
dx

f(x)
g(x)

= lim
h→0
f(x+h)·g(x)−f(x)·g(x)
h − f(x)·g(x+h)−f(x)·g(x)
h
g(x) · g(x + h)
Or
d
dx

f(x)
g(x)

= lim
h→0
g(x)f(x+h)−f(x)
h − f(x)g(x+h)−g(x)
h
g(x) · g(x + h)
Applying the differential rule and taking limits
d
dx

f(x)
g(x)

=
g(x) · f
′
(x) − f(x) · g′
(x)
[g(x)]2
It is the proof of quotient rule.
Solved Problem 1.65 Evaluate f(x) =
x
sin x
.
Solution Now taking differentiation of function with respect to x by
applying product rule of derivative
d
dx
f(x) =
d
dx
x
sin x

Or
d
dx
f(x) =
sin x d
dx x − x d
dx sin x
sin2
x
Or
d
dx
f(x) =
sin x − x cos x
sin2
x
It is required answer.

76 Derivatives
Solved Problem 1.66 Find the derivative of elliptical function y2
= 1 − 4x2
with respect to x.
Solution The given function is y2
= 1 − 4x2
. Getting derivative of this
function with respect to x, we have
y′
=
d
dx
y2
=
d
dx
(1 − 4x2
)
2y
dy
dx
= −8x ⇒
dy
dx
=
−4x
y
It gives,
dy
dx
=
−4x
√
1 − 4x2
Solved Problem 1.67 Get the derivative of exponential function f(x) =
ex
x
about the point x.
Solution The given function is fraction. Hence, in derivative, fraction
rules are applied. Now, The given function is f(x) = ex
/x. Getting deriva-
tive of this function with respect to x, we have
f
′
(x) =
d
dx
f(x) =
d
dx

ex
x

Using division rule of derivative, we get
f
′
(x) =
x × d
dx ex
− ex
× d
dx x
x2
Or
f
′
(x) =
ex
(x − 1)
x2
Solved Problem 1.68 Get the derivative of logarithmic function f(x) =
ln(1 + x)
x
about the point x.
Solution Applying division rule of derivative in this function we have
f
′
(x) =
d
dx
f(x) =
d
dx

ln(1 + x)
x

Using division rule of derivative, we get
f
′
(x) =
x × d
dx ln(1 + x) − ln(1 + x) × d
dx x
x2
Or
f
′
(x) =
1
x(1 + x)
−
ln(1 + x)
x2
Solved Problem 1.69 Find the derivative of function f(u) =
u2
+ 2u + 1
u2 − 1
.
Solution The given function is
f(u) =
u2
+ 2u + 1
u2 − 1
This function can be reduced to simplified form.
f(u) =
u2
+ 2u + 1
u2 − 1
=
(u + 1)2
(u − 1)(u + 1)
=
u + 1
u − 1
Derivating this function with respect to ‘u’, we have
Duf(u) = Du

u + 1
u − 1

Applying division rule of derivation, we get
f
′
(u) =
−2
(u − 1)2
Solved Problem 1.70 Find the derivative of function f(x) =
x2
+
√
x
ln(x)
.
Solution From the derivative relation,
Dxf(x) = Dx

x2
+
√
x
ln(x)

Applying, division rule of derivative, we have
f
′
(x) =
ln(x) Dx(x2
+
√
x) − (x2
+
√
x)Dx ln(x)
[ln(x)]2

78 Derivatives
Applying direct derivative method, we have
f
′
(x) =
ln(x) ×

2x + 1
2
√
x

−

x + 1
√
x

ln2
(x)
x2
x2 + a2
.
Solution For the given function, derivating it with respect to ‘x’, we have
Dxf(x) = Dx

x2
x2 + a2

f
′
(x) =
(x2
+ a2
)Dxx2
− x2
Dx(x2
+ a2
)
(x2 + a2)
2
f
′
(x) =
2xa2
(x2 + a2)
2
a2
x2 + a2
.
Solution From the derivative relation,
Dxf(x) = Dx

a2
x2 + a2

f
′
(x) =
(x2
+ a2
)Dxa2
− a2
Dx(x2
+ a2
)
(x2 + a2)
2
f
′
(x) =
−2xa2
(x2 + a2)
2

Solved Problem 1.73 Reduce the given relation y − x
dy
dx
in single term.
Solution Let the given relation is y/x. Derivative of this function with
respect to x
d
dx
y
x

=
xdy
dx − y
x2
Rearranging it
−x2 d
dx
y
x

= y − x
dy
dx
Or
y − x
dy
dx
= −x2

d
dx
y
x

It is the reduction given relation. This relation may be again simplified as
y − x
dy
dx
= −x2 d
dx
y
x

=
d
dx
y
x

d
dx
1
x

1.5.3 Reverse Derivative Simplification
Here, is a question that, do a differntial equation can be simplified before
solving it? Yes, it can. To understand this, take an example of total deriva-
tive
d
dx
z =
d
dx
(xy)
On derivating it, we have
dz
dx
= x
dy
dx
+ y
Or
dz = x dy + y dx
Right side of above relation can be written as
dz = d(xy)
i.e. multi-term of right hand side is simplified before solving it. Reverse
method is sometime very useful in solving differential equation. Now, take
another relation
dz
dx
= y cos(x) + sin(x)
dy
dx

80 Derivatives
In right hand side, cos x is first derivative of sin x. Here, we check the
possibilities of differential relation of function with dy/dx to the
functions present in the term without dy/dx. So, we can write right
hand side as
dz
dx
= y
d
dx
sin(x) + sin(x)
d
dx
y
On simplifying, right hand side of above relation in reverse direction is
dz
dx
=
d
dx
[y sin(x)]
Now, we can solve it easily.
Solved Problem 1.74 Simplify the relation z = −y2
sin x + 2y cos x
dy
dx
.
Solution The given relation is
z = −y2
sin x + 2y cos x
dy
dx
Here, − sin x is first derivative of cos x. Therefore, above relation can be
written as
z = y2 d
dx
cos x + cos x
d
dx
y2
=
d
dx

y2
cos x

This is simplified form of given differential equation.
Solved Problem 1.75 Simplify the relation z = x f
′
(x) + f(x).
z = x f
′
(x) + f(x)
Here, f
′
(x) is first derivative of f(x). Therefore, above relation can be
written as
z = x
d
dx
f(x) + f(x)
d
dx
x =
d
dx
[x f(x)]
Solved Problem 1.76 Simplify the relation z = f
′

x
y

1
y
−
x
y2
dy
dx

.
z = f
′

x
y

1
y
−
x
y2
dy
dx

Simplifying the terms within square brackets.
z = f
′

x
y

y − xdy
dx
y2
#
Or
z = f
′

x
y

y d
dx x − x d
dx y
y2
#
Or
z = f
′

x
y

×
d
dx

x
y

Or
z =
d
dx

f

x
y

1.5.4 Derivative by Partial Fraction
A fraction always represents to an equivalent result of one or more observa-
tions. For example, assume that the cost, (C), of items is 20 rupees. This
cost includes the cost of item 1 (c1), item 2 c2) and item 3 c3). If costs of
each item are equal, then C can be written as
C = c1 + c2 + c3
Similarly, a fraction is also a resultant of one or more fractions written in
sum form. For example
ex
x
=
∞
X
i=0
xi
x · i!
The method of partial fraction is used to convert a complex frac-
tion into sum of simple fraction. Thus this method is used to
convert a series expression into parallel expression. Let f(x)/g(x)
is a fraction, in which numerator, f(x), has degree lesser than denominator,
g(x). It is necessary condition for the conversion of a function into sum
of simple function by method of partial fraction. If this condition is not
met, i.e. numerator has equal or greater degree than denominator, then nu-
merator is divided by denominator, till the degree of quotient not becomes
less than the degree of denominator. Now g(x) is reducible to the partial
fractions. Assume a function
y =
x − 1
(x − 2)(x + 1)

82 Derivatives
The partial fraction of this relation is
y =
1
3
×
1
x − 2
+
2
3
×
1
x + 1
Taking derivative in both side
d
dx
y =
1
3
×
d
dx

1
x − 2

+
2
3
×
d
dx

1
x + 1

It gives
dy
dx
= −
1
3(x − 2)2
−
2
3(x + 1)2
Solved Problem 1.77 Convert the fraction into standard form so that method
of partial fraction can be applied. The fraction is
f(x) =
x5
− 2x3
− 2
x3 − 2x
Use partial fraction to convert the complex fraction into simple fraction.
Find the derivative of the function.
f(x) =
x5
− 2x3
− 2
x3 − 2x
A rational function is said to be in standard form, if highest degree of its nu-
merator is less than the highest degree of its denominator. So, the standard
form of the function is
f(x) = x2
−
2
x3 − 2x
Again to apply partial fraction method, denominator of a fraction should be
in product of its prime factors. So, the standard form of the function is
f(x) = x2
−
2
x(x2 − 2)
Applying partial fraction method in second term of right hand side to sim-
plify it, we have
f(x) = x2
+
1
x
−
x
x2 − 2
Derivating it about x, we have
f
′
(x) = 2x −
1
x2
+
x2
+ 2
(x2 − 2)2
This is required derivative.

Solved Problem 1.78 Convert the fraction into standard form so that method
of partial fraction can be applied. The fraction is
f(x) =
x2
− 2x4
− 2x
x4 − 2x2
Use partial fraction to convert the complex fraction into simple fraction.
Find the derivative of the function.
f(x) =
x2
− 2x4
− 2x
x4 − 2x2
A rational function is said to be standard function, if highest degree of its
numerator is less than the highest degree of its denominator. Therefore,
here we have not to rationalize the fraction function. Again to apply partial
fraction method, denominator of a fraction should be in product of prime
factors form. So, the standard function is
f(x) =
x2
− 2x4
− 2x
x2(x2 − 2)
=
1
x2 − 2
−
2x2
x2 − 2
−
2
x(x2 − 2)
The second term of right hand side is not in standard form. So, converting
it into standard form and simplifying the function, we have
f(x) = −2 −
3
x2 − 2
−
2
x(x2 − 2)
Applying partial fraction method in third term of right hand side, we have
f(x) = −2 −
3
x2 − 2
+
1
x
−
x
x2 − 2
f
′
(x) =
6x
(x2 − 2)2
−
1
x2
+
x2
+ 2
(x2 − 2)2
Or
f
′
(x) = −
1
x2
+
x2
+ 6x + 2
(x2 − 2)2

84 Derivatives
Solved Problem 1.79 Find the derivative of fraction f(x) =
x2
+ 1
x2 − 1
by using
partial fraction method.
f(x) =
x2
+ 1
x2 − 1
A rational function is said to be standard function, if highest degree of
its numerator is less than the highest degree of its denominator. So, on
rationalising the given function, it becomes
f(x) = 1 +
2
x2 − 1
Again to apply partial fraction method, denominator of a fraction should be
in product of its prime factors. So, the standard form of the above function
is
f(x) = 1 +
2
(x − 1)(x + 1)
Applying partial fraction method in second term of right hand side, we have
f(x) = 1 +
1
x − 1
−
1
x + 1
f
′
(x) =
1
(x + 1)2
−
1
(x − 1)2
Solved Problem 1.80 Find the derivative of fraction f(x) =
3x + 1
(x + 1)2(1 − x)
using partial fraction method. Given |x| 1.
f(x) =
3x + 1
(x + 1)2(1 − x)
We have given that |x| 1, it means, x does not exists in the domain of
[−1, 1]. So, the denominator term, 1 − x 0 for all values x −1 and
1 − x 0 for all values of x 1. Therefore, we can take negative sign out
off from the factor (1 − x) in denominator of the given function. So,
f(x) = −
3x + 1
(x + 1)2(x − 1)

Applying partial fraction method to separate terms according the rule,
3x + 1
(x + 1)2(x − 1)
=
A
x + 1
+
Bx
(x + 1)2
+
C
x − 1
On solving it, we have, A = −1, B = 1 and C = 1. It gives
f(x) =
1
x + 1
−
x
(x + 1)2
−
1
x − 1
Taking derivative about x, we have
f
′
(x) = −
1
(x + 1)2
+
x2
− 1
(x + 1)3
+
1
(x − 1)2
This is desired answer.
1.5.5 Chain Rule
If a function is not depend upon the variable with respect to which it is to be
differentiate then variable is substituted by another variable. For example
consider a function sin(x2
) which is to be differentiated with respect to x.
As sine is dependent of x2
, hence at first it is replaced by u and then it
would be differentiated.
Exp. Assume that
y = sin(x2
)
Substitute x2
by u
y = sin(u)
Taking differentiation both sides with respect to x.
dy
dx
=
d
dx
sin(u)
=
d
du
sin(u) ·
du
dx
= cos(u) ·
du
dx
Substituting the value of u.
dy
dx
= cos(x2
) ·
d
dx
x2
= cos(x2
) · 2x
= 2x cos(x2
)

86 Derivatives
This method is called chain rule.
Solved Problem 1.81 Position of a mechanical wave propagation at any
instant of x is represented by y = x sin(x2
). Find the rate at which position
of the mechanical wave is changing with x.
Solution The rate of change of position of the mechanical wave is given
by
r =
dy
dx
=
d
dx
x sin(x2
)
Applying product rule for the given relation
r = x
d
dx
sin(x2
) + sin(x2
)
d
dx
x
On solving it
r = 2x2
cos(x2
) + sin(x2
)
It is the rate of change of position of mechanical wave. The graph of function
and its rate of change are given below.
1 2 3
x
y
f = x sin(x2
)
1 2 3
x
r
r = 2x2
cos(x2
) + sin(x2
)
Solved Problem 1.82 During shock test on an object by a machine, it is found
that impulse is a function of parameter x and is given by F(x) = sin(2x2
).
Find that how does shock force change with parameter x.
Solution The rate of change of shock force is given by
f
′
=
dF
dx
=
d
dx
sin(2x2
)
Applying chain rule for the given relation
f
′
= cos(2x2
)
d
dx
2x2
On solving it
f
′
= 4x cos(2x2
)

It is the rate of change of shock. The graph of shock force and its rate of
chage are given below.
1 2 3
x
F
F = sin(2x2
)
1 2 3
x
F
′
F
′
= 4x cos(2x2
)
1.5.6 Logarithmic Differential
Logarithm method used in those relations in which power of independent
variable is itself. For example xx
can be easily differentiated by using loga-
rithm
y = xx
Taking logarithm both side.
ln(y) = ln xx
In logarithm, ln ab
= b ln a and it will give right hand side of above relation
ln(y) = x ln x
Taking differentiation both sides with respect to x.
d
dx
ln y =
d
dx
(x ln x)
Using product rule in right hand side.
d
dx
ln y = x ×
1
x
+ ln(x)
Changing the derivative base in left hand side
d
dy
ln y ·
dy
dx
= 1 + ln(x)
Or
1
y
·
dy
dx
= 1 + ln(x)
On rationalized simplifications
dy
dx
= y(1 + ln x)

88 Derivatives
Or
dy
dx
= xx
(1 + ln x)
It is the required result. This method can also be used to find the derivative
of multiple terms in product.
Grouping in Exponents
In exponential, the grouping of power is an important factor during
the process of derivative. To clarify it we take a same exponential in
different order of grouping. The function f(x) = xab
can be grouped as
f(x) = x(ab
) and g(x) = (xa
)
b
. Though both functions have same bases
and powers yet they give entirely different results. Assume x = 2, a = 3
and b = 2 then
f(2) = 2(32
) = 29
= 512
and
g(2) = 23
2
= 82
= 64
Here f(x) has higher result with respect to g(x). Ultimately, in derivatives,
grouping order must be maintained.
1
2
3
1 2
x
y
f(x)
g(x)
To verify the concept graphically, plots of functions f(x) and g(x) are
drawn taking same bases and powers in different grouping orders a shown
in above figure. The grouping order of f(x) shows large change in the
slope of function for values of x rather than the grouping order of g(x).
Solved Problem 1.83 A student gathers data about the purchase of item X
during a time being t. He interpolated data into a trigono-algebraic relation
for future observations. The relation is P = t sin(ln t) for his data. Find
that derivative of this relation.

Solution The derivative of relation is
P′
=
dP
dt
=
d
dt
t sin(ln t)
Applying product rule of derivatives for the given relation
P′
= cos(ln t) + sin(ln t)
It is the required answer. The graph of are given below.
1 2 3
t
P
P
=
t
s
i
n
(
l
n
t
)
1 2 3
t
P′
P′
= cos(ln t) + sin(ln t)
Solved Problem 1.84 Find the derivative of function ln[sin(ax2
)].
Solution In this problem, function is dependent of parameter x and a is
constant parameter. So, it shall be derivative for x. Now,
Dxf(x) = Dx ln[sin(ax2
)]
First applying logarithmic rule, then trigonometric properties for derivative
and finally xn
rule. So, we have
f
′
(x) =
1
sin(ax2)
× cos(ax2
) × a × 2x2−1
On simplification, we have
f
′
(x) = 2ax cot(ax2
)
This is the result.
Solved Problem 1.85 Find the derivative of function e4x3
.
Solution Here,
Dxf(x) = Dxe4x3
It gives
f
′
(x) = e4x3
× 12x2
= 12x2
e4x3
This is final answer.

90 Derivatives
Solved Problem 1.86 Find the derivative of function eln(x2
)
.
Solution First we simplify the function before taking derivatives of it.
So, assume
y = eln(x2
)
Taking natural logarithm in both side, we have
ln y = ln
h
eln(x2
)
i
= ln(x2
) × ln(e) = ln(x2
)
t
y
y
=
e
x
p
l
n
(
x
2
)
t
y
y = x2
y
′ =
2x
Taking anti-logarithm both side, we have y = x2
. Now, the derivative of
this function about x is given by y′
= 2x by using xn
rule of derivation.
Solved Problem 1.87 Find the derivative of function e(ln x)2
.
Solution Taking Derivative in both sides, we have
Dxy = Dx
h
e(ln x)2
i
Fist using exponential rule, then logarithm rule and finally xn
rule. We get
the derivative of the function:
y′
= e(ln x)2
× 2 × ln x ×
1
x
On simplification, it will give
y′
=
2e(ln x)2
ln x
x
This is result.

Principle of Derivative Calculus - Differential Calculus - An Introduction by Arun Umrao

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