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Page 1
Differentiability
Let a real valued function f(x) be defined on an I and let x
0
be a point in I. Then, if
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
exists and is equal to ?, then f(x) is said to be differentiable at x
0
and ? is called the
derivative of f(x) at x = x
0
.
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be
differentiable in (a, b).
Properties of differentiation
? Let the functions f and g be differentiable at a point x
0
. Then
i) (cf ?) x
0
= cf ?(x
0
), c any constant.
ii) (f ? g) ?x
0
= f ?(x
0
) ? g ?(x
0
)
iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
)
iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
? If f is differentiable at x
0
and g is differentiable at f(x
0
) then the composite function
h = g(f(x)) is differentiable at x
0
and
h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
)
? If the function y = f(x) is represented in the parametric form as
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then
f ?(x) =
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
Note : If a function is differentiable at x = x
0
, then it is continuous at x = x
0
.
However the converse need not be true.
Page 2
Differentiability
Let a real valued function f(x) be defined on an I and let x
0
be a point in I. Then, if
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
exists and is equal to ?, then f(x) is said to be differentiable at x
0
and ? is called the
derivative of f(x) at x = x
0
.
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be
differentiable in (a, b).
Properties of differentiation
? Let the functions f and g be differentiable at a point x
0
. Then
i) (cf ?) x
0
= cf ?(x
0
), c any constant.
ii) (f ? g) ?x
0
= f ?(x
0
) ? g ?(x
0
)
iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
)
iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
? If f is differentiable at x
0
and g is differentiable at f(x
0
) then the composite function
h = g(f(x)) is differentiable at x
0
and
h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
)
? If the function y = f(x) is represented in the parametric form as
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then
f ?(x) =
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
Note : If a function is differentiable at x = x
0
, then it is continuous at x = x
0
.
However the converse need not be true.
Standard formulae
Function Derivative Function Derivative
k(constant) 0
1
x
sin
a
?
22
1
ax ?
x
n
nx
n ?1 1
x
cos
a
?
22
1
ax
?
?
logx 1/x
1
x
tan
a
?
22
a
ax ?
e
x
e
x 1
x
cot
a
?
22
a
ax
?
?
a
x
a
x
loga
1
x
sec
a
?
22
a
xx a ?
sin x cos x
1
x
cosec
a
?
22
a
xx a
?
?
cos x ? sin x sin h x cos h x
tan x sec
2
x cos h x sin h x
cot x ? cosec
2
x sin h
?1
x
2
1
x1 ?
sec x sec x tan x cos h
?1
x
2
1
x1 ?
cosec x ? cosec x cot x tan h
?1
x
2
1
x1 ?
Solved Example 8 :
Show that the function
f(x) =
2
xcos(1/x), x 0
0x0
? ?
?
?
?
is differentiable at x = 0 but f ?(x) is not
continuous at x = 0.
Solution :
We have
x0
lim f(x) 0 f(0)
?
? ?
? f(x) is continuous at x = 0
Now f ?(0) =
?
?
x0
f(x) f(0)
lim
x
?
?? ??
? ?
?? ??
?? ??
x0
1
lim xcos 0
x
Hence f(x) is differentiable at
x = 0 and f ?(0) = 0
for x ? 0 we have
Page 3
Differentiability
Let a real valued function f(x) be defined on an I and let x
0
be a point in I. Then, if
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
exists and is equal to ?, then f(x) is said to be differentiable at x
0
and ? is called the
derivative of f(x) at x = x
0
.
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be
differentiable in (a, b).
Properties of differentiation
? Let the functions f and g be differentiable at a point x
0
. Then
i) (cf ?) x
0
= cf ?(x
0
), c any constant.
ii) (f ? g) ?x
0
= f ?(x
0
) ? g ?(x
0
)
iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
)
iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
? If f is differentiable at x
0
and g is differentiable at f(x
0
) then the composite function
h = g(f(x)) is differentiable at x
0
and
h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
)
? If the function y = f(x) is represented in the parametric form as
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then
f ?(x) =
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
Note : If a function is differentiable at x = x
0
, then it is continuous at x = x
0
.
However the converse need not be true.
Standard formulae
Function Derivative Function Derivative
k(constant) 0
1
x
sin
a
?
22
1
ax ?
x
n
nx
n ?1 1
x
cos
a
?
22
1
ax
?
?
logx 1/x
1
x
tan
a
?
22
a
ax ?
e
x
e
x 1
x
cot
a
?
22
a
ax
?
?
a
x
a
x
loga
1
x
sec
a
?
22
a
xx a ?
sin x cos x
1
x
cosec
a
?
22
a
xx a
?
?
cos x ? sin x sin h x cos h x
tan x sec
2
x cos h x sin h x
cot x ? cosec
2
x sin h
?1
x
2
1
x1 ?
sec x sec x tan x cos h
?1
x
2
1
x1 ?
cosec x ? cosec x cot x tan h
?1
x
2
1
x1 ?
Solved Example 8 :
Show that the function
f(x) =
2
xcos(1/x), x 0
0x0
? ?
?
?
?
is differentiable at x = 0 but f ?(x) is not
continuous at x = 0.
Solution :
We have
x0
lim f(x) 0 f(0)
?
? ?
? f(x) is continuous at x = 0
Now f ?(0) =
?
?
x0
f(x) f(0)
lim
x
?
?? ??
? ?
?? ??
?? ??
x0
1
lim xcos 0
x
Hence f(x) is differentiable at
x = 0 and f ?(0) = 0
for x ? 0 we have
f ?(x) =
?? ?? ?? ? ?
?? ?
?? ?? ?? ? ?
?? ?? ? ? ??
2
2
111
2xcos x sin
xx x
?? ? ?
??
?? ? ?
?? ? ?
11
2xcos sin
xx
Now ??
x0
lim f x
?
? does not exist as
x0
1
lim sin
x
?
??
??
??
does not exist. Therefore, f ?(x) is not
continuous at x = 0.
Solved Example 9 :
3
yx ? is defined and continuous for all x
investigate whether this function has a
derivative at x = 0.
Solution :
?y =
33
xx x ?? ? at x = 0
?y =
3
x ?
?
??
3
2
3
yx 1
xx
x
??
??
??
?
?
??
2 x0 x0
3
y1
lim lim
x
x
?? ??
?
????
?
?
? There is no finite derivative.
Solved Example 10 :
Investigate the function f(x) = | x | for
differentiability at the point x = 0
Solution :
y = f(x) = | x |
?y = | x + ?x | ? | ?x |
At x = 0, we have ?y = | ?x |
?
x
y
xx
?
?
?
? ?
x0
y
lim 1
x
?
??
?
? ?
?
x0
y
lim 1
x
?
??
?
?
?
Since the right and the left side derivative
are not equal, the function f(x) = |x| is not
differentiable at the point zero.
Solved Example 11 :
Show that
??
nn1
d
xnx
dx
?
?
Solution :
Let y = x
n
Let x receive a small increment ?x and let
the corresponding increment in y be ?y
then
y + ?y = (x + ?x)
n
By subtraction, ?y = (x + ?x)
n
? x
n
?
?? ??
nn
nn
xx x xx x
y
xx xxx
?? ? ?? ?
?
??
? ????
?
??
n
n
n1
x0 x x x
xx x
y
lim lim nx
xxxx
?
?? ???
?? ?
?
??
? ?? ?
?
??
n1 n n1
dy d
nx or x nx
dx dx
? ?
??
Page 4
Differentiability
Let a real valued function f(x) be defined on an I and let x
0
be a point in I. Then, if
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
exists and is equal to ?, then f(x) is said to be differentiable at x
0
and ? is called the
derivative of f(x) at x = x
0
.
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be
differentiable in (a, b).
Properties of differentiation
? Let the functions f and g be differentiable at a point x
0
. Then
i) (cf ?) x
0
= cf ?(x
0
), c any constant.
ii) (f ? g) ?x
0
= f ?(x
0
) ? g ?(x
0
)
iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
)
iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
? If f is differentiable at x
0
and g is differentiable at f(x
0
) then the composite function
h = g(f(x)) is differentiable at x
0
and
h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
)
? If the function y = f(x) is represented in the parametric form as
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then
f ?(x) =
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
Note : If a function is differentiable at x = x
0
, then it is continuous at x = x
0
.
However the converse need not be true.
Standard formulae
Function Derivative Function Derivative
k(constant) 0
1
x
sin
a
?
22
1
ax ?
x
n
nx
n ?1 1
x
cos
a
?
22
1
ax
?
?
logx 1/x
1
x
tan
a
?
22
a
ax ?
e
x
e
x 1
x
cot
a
?
22
a
ax
?
?
a
x
a
x
loga
1
x
sec
a
?
22
a
xx a ?
sin x cos x
1
x
cosec
a
?
22
a
xx a
?
?
cos x ? sin x sin h x cos h x
tan x sec
2
x cos h x sin h x
cot x ? cosec
2
x sin h
?1
x
2
1
x1 ?
sec x sec x tan x cos h
?1
x
2
1
x1 ?
cosec x ? cosec x cot x tan h
?1
x
2
1
x1 ?
Solved Example 8 :
Show that the function
f(x) =
2
xcos(1/x), x 0
0x0
? ?
?
?
?
is differentiable at x = 0 but f ?(x) is not
continuous at x = 0.
Solution :
We have
x0
lim f(x) 0 f(0)
?
? ?
? f(x) is continuous at x = 0
Now f ?(0) =
?
?
x0
f(x) f(0)
lim
x
?
?? ??
? ?
?? ??
?? ??
x0
1
lim xcos 0
x
Hence f(x) is differentiable at
x = 0 and f ?(0) = 0
for x ? 0 we have
f ?(x) =
?? ?? ?? ? ?
?? ?
?? ?? ?? ? ?
?? ?? ? ? ??
2
2
111
2xcos x sin
xx x
?? ? ?
??
?? ? ?
?? ? ?
11
2xcos sin
xx
Now ??
x0
lim f x
?
? does not exist as
x0
1
lim sin
x
?
??
??
??
does not exist. Therefore, f ?(x) is not
continuous at x = 0.
Solved Example 9 :
3
yx ? is defined and continuous for all x
investigate whether this function has a
derivative at x = 0.
Solution :
?y =
33
xx x ?? ? at x = 0
?y =
3
x ?
?
??
3
2
3
yx 1
xx
x
??
??
??
?
?
??
2 x0 x0
3
y1
lim lim
x
x
?? ??
?
????
?
?
? There is no finite derivative.
Solved Example 10 :
Investigate the function f(x) = | x | for
differentiability at the point x = 0
Solution :
y = f(x) = | x |
?y = | x + ?x | ? | ?x |
At x = 0, we have ?y = | ?x |
?
x
y
xx
?
?
?
? ?
x0
y
lim 1
x
?
??
?
? ?
?
x0
y
lim 1
x
?
??
?
?
?
Since the right and the left side derivative
are not equal, the function f(x) = |x| is not
differentiable at the point zero.
Solved Example 11 :
Show that
??
nn1
d
xnx
dx
?
?
Solution :
Let y = x
n
Let x receive a small increment ?x and let
the corresponding increment in y be ?y
then
y + ?y = (x + ?x)
n
By subtraction, ?y = (x + ?x)
n
? x
n
?
?? ??
nn
nn
xx x xx x
y
xx xxx
?? ? ?? ?
?
??
? ????
?
??
n
n
n1
x0 x x x
xx x
y
lim lim nx
xxxx
?
?? ???
?? ?
?
??
? ?? ?
?
??
n1 n n1
dy d
nx or x nx
dx dx
? ?
??
Mean Value Theorems
Rolle's Theorem
The theorem states that
? if f(x) is continuous in the closed interval [a, b] and
? if f ?(x) exists in open interval (a, b) and
? if f(a) = f(b)
Then there exists at least one value c in (a, b) such that f ?(c) = 0
Geometric Interpretation
There exists at last one point at which slope of the tangent is 0 or the tangent is parallel
to x-axis
Lagrange's Mean Value Theorem
The theorem states that
? if f(x) is continuous in the closed interval [a, b] and
? if f ?(x) exists in open interval (a, b)
Then there exists one value c such that
f ?(c) =
??
??
f(b) f(a)
ba
?
?
a c b x
y
f(a) = f(b)
y = f(x)
f ?(c) = 0
Page 5
Differentiability
Let a real valued function f(x) be defined on an I and let x
0
be a point in I. Then, if
?? ? ? ? ? ? ?
0
00 0
xx x 0
0
fx fx fx x fx
lim or lim
xx x
???
????
?? ?? ?? ???
??
exists and is equal to ?, then f(x) is said to be differentiable at x
0
and ? is called the
derivative of f(x) at x = x
0
.
If f(x) is differentiable at every point in the interval (a, b) then f(x) is said to be
differentiable in (a, b).
Properties of differentiation
? Let the functions f and g be differentiable at a point x
0
. Then
i) (cf ?) x
0
= cf ?(x
0
), c any constant.
ii) (f ? g) ?x
0
= f ?(x
0
) ? g ?(x
0
)
iii) (fg) ?(x
0
) = f ?(x
0
) g(x
0
) + f(x
0
) g ?(x
0
)
iv) ??
? ? ?? ?? ??
??
??
00 0 0
00
2
0
gx f x f x g x f
x,gx0
g gx
?
?? ?
??
? ?
??
??
? If f is differentiable at x
0
and g is differentiable at f(x
0
) then the composite function
h = g(f(x)) is differentiable at x
0
and
h ?(x
0
) = g ?(f(x
0
)) f ?(x
0
)
? If the function y = f(x) is represented in the parametric form as
x = ?(t) and y = ?(t), and if ??(t), ??(t) exist, then
f ?(x) =
? ?
??
??
t
dy / dt
t0
dx / dt t
? ?
? ? ??? ??? ? ? ?
? ?
Note : If a function is differentiable at x = x
0
, then it is continuous at x = x
0
.
However the converse need not be true.
Standard formulae
Function Derivative Function Derivative
k(constant) 0
1
x
sin
a
?
22
1
ax ?
x
n
nx
n ?1 1
x
cos
a
?
22
1
ax
?
?
logx 1/x
1
x
tan
a
?
22
a
ax ?
e
x
e
x 1
x
cot
a
?
22
a
ax
?
?
a
x
a
x
loga
1
x
sec
a
?
22
a
xx a ?
sin x cos x
1
x
cosec
a
?
22
a
xx a
?
?
cos x ? sin x sin h x cos h x
tan x sec
2
x cos h x sin h x
cot x ? cosec
2
x sin h
?1
x
2
1
x1 ?
sec x sec x tan x cos h
?1
x
2
1
x1 ?
cosec x ? cosec x cot x tan h
?1
x
2
1
x1 ?
Solved Example 8 :
Show that the function
f(x) =
2
xcos(1/x), x 0
0x0
? ?
?
?
?
is differentiable at x = 0 but f ?(x) is not
continuous at x = 0.
Solution :
We have
x0
lim f(x) 0 f(0)
?
? ?
? f(x) is continuous at x = 0
Now f ?(0) =
?
?
x0
f(x) f(0)
lim
x
?
?? ??
? ?
?? ??
?? ??
x0
1
lim xcos 0
x
Hence f(x) is differentiable at
x = 0 and f ?(0) = 0
for x ? 0 we have
f ?(x) =
?? ?? ?? ? ?
?? ?
?? ?? ?? ? ?
?? ?? ? ? ??
2
2
111
2xcos x sin
xx x
?? ? ?
??
?? ? ?
?? ? ?
11
2xcos sin
xx
Now ??
x0
lim f x
?
? does not exist as
x0
1
lim sin
x
?
??
??
??
does not exist. Therefore, f ?(x) is not
continuous at x = 0.
Solved Example 9 :
3
yx ? is defined and continuous for all x
investigate whether this function has a
derivative at x = 0.
Solution :
?y =
33
xx x ?? ? at x = 0
?y =
3
x ?
?
??
3
2
3
yx 1
xx
x
??
??
??
?
?
??
2 x0 x0
3
y1
lim lim
x
x
?? ??
?
????
?
?
? There is no finite derivative.
Solved Example 10 :
Investigate the function f(x) = | x | for
differentiability at the point x = 0
Solution :
y = f(x) = | x |
?y = | x + ?x | ? | ?x |
At x = 0, we have ?y = | ?x |
?
x
y
xx
?
?
?
? ?
x0
y
lim 1
x
?
??
?
? ?
?
x0
y
lim 1
x
?
??
?
?
?
Since the right and the left side derivative
are not equal, the function f(x) = |x| is not
differentiable at the point zero.
Solved Example 11 :
Show that
??
nn1
d
xnx
dx
?
?
Solution :
Let y = x
n
Let x receive a small increment ?x and let
the corresponding increment in y be ?y
then
y + ?y = (x + ?x)
n
By subtraction, ?y = (x + ?x)
n
? x
n
?
?? ??
nn
nn
xx x xx x
y
xx xxx
?? ? ?? ?
?
??
? ????
?
??
n
n
n1
x0 x x x
xx x
y
lim lim nx
xxxx
?
?? ???
?? ?
?
??
? ?? ?
?
??
n1 n n1
dy d
nx or x nx
dx dx
? ?
??
Mean Value Theorems
Rolle's Theorem
The theorem states that
? if f(x) is continuous in the closed interval [a, b] and
? if f ?(x) exists in open interval (a, b) and
? if f(a) = f(b)
Then there exists at least one value c in (a, b) such that f ?(c) = 0
Geometric Interpretation
There exists at last one point at which slope of the tangent is 0 or the tangent is parallel
to x-axis
Lagrange's Mean Value Theorem
The theorem states that
? if f(x) is continuous in the closed interval [a, b] and
? if f ?(x) exists in open interval (a, b)
Then there exists one value c such that
f ?(c) =
??
??
f(b) f(a)
ba
?
?
a c b x
y
f(a) = f(b)
y = f(x)
f ?(c) = 0
Geometric Interpretation
There exists at last one point at which the tangent is parallel to the secant through the
end points.
Cauchy's Mean Value Theorem
The theorem states that
? if f(x) and g(x) are both continuous in closed interval [a, b] and
? if f ?(x) and g(x) both exist in open interval (a, b)
Then there exists at least one value c such that
??
?? ??
??
f(c) f(b) f(a)
g(c) g(b) g(a)
? ?
?
? ?
Maxima and Minima
A function f(x) is said to have a maximum value at x = a if f(a) is larger than any other
values of f(x) in the immediate neighbourhood of ‘a’. It has a minimum value if f(a) is less
than any other value of f(x) sufficiently near ‘a’.
Finding Maximum and Minimum values of y = f(x)
? Get
dy
dx
and
2
2
dy
dx
. Solve
dy
dx
= 0 and consider its roots. These are the values of x
which make
dy
dx
= 0.
? For each of these values of x, calculate the corresponding value of y and examine
the sign of
2
2
dy
dx
.
? If the sign is ?ve the corresponding value of y is a maximum.
If the sign is +ve, the corresponding value of y is minimum.
a c
b
f(a)
f(b)
f ?(c) =
f(b) f(a)
ba
?? ?
?
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FAQs on Short notes on Differentiation - Engineering Mathematics - Civil Engineering (CE)
| 1. What is the main purpose of differentiation in mathematics? |  |
Ans. The main purpose of differentiation in mathematics is to find the rate at which a function is changing at a particular point.
| 2. How is differentiation used in real-life applications? | |
Ans. Differentiation is used in various real-life applications such as calculating the speed of a moving object, determining the slope of a curve, and optimizing functions in economics and engineering.
| 3. What are the basic rules of differentiation that one should know? | |
Ans. Some basic rules of differentiation include the power rule, product rule, quotient rule, and chain rule, which are essential for finding the derivative of functions.
| 4. How does differentiation relate to integration in calculus? | |
Ans. Differentiation and integration are inverse processes in calculus. Differentiation finds the rate of change, while integration finds the accumulation of change. The Fundamental Theorem of Calculus establishes a connection between the two.
| 5. Can differentiation be used to find maximum and minimum points of a function? | |
Ans. Yes, differentiation can be used to find maximum and minimum points of a function by setting the derivative equal to zero and solving for critical points. The first derivative test or second derivative test can then be applied to determine the nature of these points.
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