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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