Chap 04 Real Analysis: Differentiation
Chap 04 Real Analysis: Differentiation
Chap 04 Real Analysis: Differentiation
v Derivative of a function:
Let f be defined and real valued on [ a, b] . For any point c Î [ a, b ] , form the
quotient
f ( x ) - f (c )
x-c
and define
f ( x ) - f (c )
f ¢(c ) = lim
x ®c x-c
provided this limit exits.
We thus associate a function f ¢ with the function f , where domain of f ¢ is the
set of points at which the above limit exists.
The function f ¢ is so defined is called the derivative of f .
(i) If f ¢ is defined at point x, we say that f is differentiable at x.
(ii) f ¢(c) exists if and only if for a real number e > 0 , $ a real number d > 0
such that
f ( x ) - f (c )
- f ¢(c) < e whenever x - c < d
x-c
(iii) If x - c = h then we have
f (c + h ) - f ( c )
f ¢(c) = lim
h®0 h
(iv) f is differentiable at c if and only if c is a removable discontinuity of the
f ( x ) - f (c )
function j ( x) = .
x-c
v Example
(i) A function f : ¡ ® ¡ defined by
ì x 2 sin 1x ; x¹0
f ( x) = í
î 0 ; x=0
This function is differentiable at x = 0 because
f ( x) - f (0) x 2 sin 1x - 0
lim = lim
x ®0 x-0 x ®0 x-0
2
x sin 1x
= lim = lim x sin 1x = 0
x ®0 x x ®0
v Theorem
Let f be defined on [ a, b] , if f is differentiable at a point x Î [ a, b] , then f is
continuous at x. (Differentiability implies continuity)
Proof
We know that
f (t ) - f ( x)
lim = f ¢( x) where t ¹ x and a < t < b
t®x t-x
Now
æ f (t ) - f ( x) ö
lim ( f (t ) - f ( x) ) = lim ç ÷ lim (t - x )
t®x t®x è t-x ø t®x
= f ¢( x ) × 0
=0
Þ lim f (t ) = f ( x ) .
t ®x
Which show that f is continuous at x.
Note
(i) The converse of the above theorem does not hold.
ìx if x ³ 0
Consider f ( x) = x = í
î- x if x < 0
f ¢(0) does not exists but f ( x) is continuous at x = 0
and
f ( x) - f (1)
D- f (1) = lim
x ®1- h
h®0
x -1
f (1 - h) - f (1) (1 - h)3 - 1
= lim = lim
h ®0 1 - h -1 h ®0 -h
1 - 3h + 3h - h - 1
( )
2 3
= lim = lim 3 - 3h + h 2 = 3
h ®0 -h h ®0
Chap 4 – Differentiation 3
Since D+ f (1) ¹ D- f (1) Þ f ¢(1) does not exist even though f is continuous at
x = 1 . f ¢( x ) exist for all other values of x.
v Theorem
Suppose f and g are defined on [a, b] and are differentiable at a point
f
x Î [a, b] , then f + g , fg and are differentiable at x and
g
(i) ( f + g )¢( x ) = f ¢( x ) + g ¢( x )
(ii) ( fg )¢( x ) = f ¢( x ) g ( x) + f ( x ) g ¢( x)
æ f ö¢ g ( x) f ¢( x) - f ( x ) g ¢( x)
(iii) ç ÷ ( x) =
ègø g 2 ( x)
The proof of this theorem can be get from any F.Sc or B.Sc text book.
Note
The derivative of any constant is zero.
And if f is defined by f ( x ) = x then f ¢( x ) = 1
And for f ( x) = x n then f ¢( x ) = n x n-1 where n is positive integer, if n < 0 we
have to restrict ourselves to x = 0 .
Thus every polynomial Pn ( x) = a0 + a1x + a2 x 2 + .......... + an x n is differentiable
every where and so every rational function except at the point where denominator is
zero.
v Theorem (Chain Rule)
Suppose f is continuous on [a, b] , f ¢( x ) exists at some point x Î [a, b] . A
function g is defined on an interval I which contains the range of f , and g is
differentiable at the point f ( x) .
If h(t ) = g ( f (t ) ) ; a £ t £ b
Then h is differentiable at x and h¢( x ) = g ¢ ( f ( x) ) × f ¢( x) .
Proof
Let y = f ( x)
By the definition of the derivative we have
f (t ) - f ( x ) = (t - x) [ f ¢( x) + u (t )] ………... (i)
and g ( s) - g ( y ) = ( s - y ) [ g ¢( y ) + v( s )] ……….. (ii)
where t Î [ a, b] , s Î I and u (t ) ® 0 as t ® x and v( s ) ® 0 as s ® y .
Let us suppose s = f (t ) then
h(t ) - h( x) = g ( f (t ) ) - g ( f ( x) )
= [ f (t ) - f ( x )][ g ¢( y ) + v( s) ] by (ii)
= (t - x) [ f ¢( x ) + u (t )][ g ¢( y ) + v (s )] by (i)
or if t ¹ x
h(t ) - h( x)
= [ f ¢( x) + u (t )][ g ¢( y ) + v( s )]
t-x
taking the limit as t ® x we have
h¢( x) = [ f ¢( x) + 0][ g ¢( y ) + 0]
= g ¢ ( f ( x ) ) × f ¢( x) Q y = f ( x)
which is the required result.
It is known as chain rule.
………………………………………………..
4 Chap 4 – Differentiation
v Example
Let f be defined by
ì 1
ï x sin ; x¹0
f ( x) = í x
ïî 0 ; x=0
1 1 1
Þ f ¢( x) = sin - cos where x ¹ 0 .
x x x
1
Q at x = 0 , is not defined.
x
\ Applying the definition of the derivative we have
1
t sin
f (t ) - f (0) t = lim sin 1
f ¢(0) = lim = lim
t ®0 t -0 t ®0 t t ®0 t
which does not exit.
The derivative of the function f ( x) does not exist at x = 0 but it is continuous at
x = 0 (i.e. it is not differentiable although it is continuous at x = 0 )
Same the case with absolute value function.
v Example
Let f be defined by
ì 2 1
ï x sin ; x¹0
f ( x) = í x
ïî 0 ; x=0
1 1
We have f ¢( x ) = 2 x sin - cos where x ¹ 0 .
x x
1
Q at x = 0 , is not defined.
x
\ Applying the definition of the derivative we have
f (t ) - f (0) 1
= t sin £ t , (t ¹ 0)
t -0 t
Taking limit as t ® 0 we see that f ¢(0) = 0
Thus f is differentiable at points x but f ¢ is not a continuous function, since
1
cos does not tend to a limit as x ® 0 .
x
v Local Maximum
Let f be a real valued function defined on a metric space X , we say that f
has a local maximum at a point p Î X if there exist d > 0 such that f (q ) £ f ( p )
" q Î X with d ( p, q ) < d .
Local minimum is defined likewise.
…………………………………………..
Chap 4 – Differentiation 5
v Theorem
Let f be defined on [ a, b] , if f has a local maximum at a point x Î [ a, b] and
if f ¢( x ) exist then f ¢( x ) = 0 .
(The analogous for local minimum is of course also true)
Proof
Choose d such that
a < x -d < x < x + d < b
Now if x - d < t < x then f(x)
f (t ) - f ( x)
³0
t-x
Taking limit as t ® x we get
f ¢( x ) ³ 0 …………. (i) a x – d t x t x + d b
If x < t < x + d
Then
f (t ) - f ( x)
£0
t-x
Again taking limit when t ® x we get
f ¢( x ) £ 0 ……………. (ii)
Combining (i) and (ii) we have
f ¢( x ) = 0
v Generalized Mean Value Theorem
If f and g are continuous real valued functions on closed interval [ a, b] , then
there is a point x Î ( a, b ) at which
[ f (b) - f (a)] g ¢( x) = [ g (b) - g (a)] f ¢( x)
The differentiability is not required at the end point.
Proof
Let
h(t ) = [ f (b) - f (a)] g (t ) - [ g (b) - g (a) ] f (t ) ( a £ t £ b)
Q h involves f and g therefore h is
i) Continuous on close interval [ a, b] .
ii) Differentiable on open interval (a, b) .
iii) and h (a) = h(b) .
To prove the theorem we have to show that h¢( x ) = 0 for some x Î (a, b )
There are two cases to be discussed
(i) h is constant function.
Þ h¢( x) = 0 " x Î (a, b)
(ii) If h is not constant.
then h (t ) > h(a ) for some t Î (a, b)
Let x be the point in the interval (a, b) at which h attain its maximum,
then h¢( x ) = 0
Similarly,
if h (t ) < h(a ) for some t Î (a, b) then $ a point x Î (a, b ) at which the
function h attain its minimum and since the derivative at a local minimum is
zero therefore we get h¢( x ) = 0
Hence
h¢( x ) = [ f (b) - f (a)] g ¢( x ) - [ g (b) - g (a)] f ¢( x) = 0
This gives the desire result.
6 Chap 4 – Differentiation
f ( x) - f ( y )
- ( x - y) £ £ ( x - y) when x > y
x- y
and
f ( x) - f ( y )
- ( x - y) ³ ³ ( x - y ) when x < y
x- y
Taking limit as x ® y , we get
0 £ f ¢( y ) £ 0 ù
Þ f ¢( y ) = 0
0 ³ f ¢( y ) ³ 0 úû
which shows that function is constant.
v Question
If f ¢( x ) > 0 in (a, b) then prove that f is strictly increasing in (a, b) and let g
be its inverse function, prove that the function g is differentiable and that
1
g ¢ ( f ( x) ) = ; a< x<b
f ( x)
Solution
Let y Î ( f (a), f (b) )
Þ y = f ( x) for some x Î (a, b )
g ( z) - g ( y)
Þ g ¢( y ) = lim
z®y z-y
g ( f ( xz ) ) - g ( f ( x) )
= lim g ( f ( xz ) ) =
xz ® x f ( xz ) - f ( x )
f -1 ( f ( xz ) ) - f -1 ( f ( x) )
= lim
xz ® x f ( xz ) - f ( x )
xz - x 1 1
= lim = =
xz ® x f ( x ) - f ( x ) f ( xz ) - f ( x ) f ¢( x)
z lim
xz ® x xz - x
v Question
Suppose f is defined and differentiable for every x > 0 and f ¢( x ) ® 0 as
x ® +¥ put g ( x) = f ( x + 1) - f ( x) . Prove that g ( x ) ® 0 as x ® +¥ .
Solution
Since f is defined and differentiable for x > 0 therefore we can apply the
Lagrange’s M.V. T. to have
f ( x + 1) - f ( x) = ( x + 1 - x) f ¢( x1 ) where x < x1 .
Q f ¢( x) ® 0 as x ® ¥
\ f ¢( x1 ) ® 0 as x ® ¥
Þ f ( x + 1) - f ( x ) ® 0 as x ® 0
Þ g ( x) ® 0 as x ® 0
……………………………
Chap 4 – Differentiation 9
f (t ) - f ( x) t-x
= lim ×
t®x t-x g (t ) - ( x)
f (t ) - f ( x) 1
= lim × lim
t®x t-x t ® x g (t ) - ( x )
t-x
f (t ) - f ( x) 1 1 f ¢( x)
= lim × = f ¢( x) × =
t®x t-x g (t ) - ( x) g ¢( x) g ¢( x)
lim
t®x t-x
Q.E.D.
v Question
Suppose f is defined in the neighborhood of a point x and f ¢¢( x) exists.
f ( x + h) + f ( x - h ) - 2 f ( x)
Show that lim = f ¢¢( x)
h®0 h2
Solution
By use of Lagrange’s Mean Value Theorem
f ( x + h) + f ( x) = hf ¢( x1 ) where x < x1 < x + h …………… (i)
and
- [ f ( x - h) - f ( x)] = hf ¢( x2 ) where x - h < x2 < x …………… (ii)
Subtract (ii) from (i) to get
f ( x + h) + f ( x - h) - 2 f ( x) = h [ f ¢( x1 ) - f ¢( x2 )]
f ( x + h) + f ( x - h) - 2 f ( x) f ¢( x1 ) - f ¢( x2 )
Þ =
h2 h
Q x2 - x1 ® 0 as h ® 0
therefore
f ( x + h) + f ( x - h) - 2 f ( x ) f ¢( x1 ) - f ¢( x2 )
\ lim = lim
h®0 h 2 x1 ® x2 x1 - x2
= f ¢¢( x2 )
v Question
c c c c
If c0 + 1 + 2 + ......... + n -1 + n = 0
2 3 n n +1
Where c0 , c1 , c2 ,......., cn are real constants.
Prove that c0 + c1 x + c2 x 2 + ......... + cn x n = 0 has at least one real root between 0
and 1.
Solution
c c
Suppose f ( x ) = c0 x + 1 x 2 + .......... + n x n+1
2 n +1
c c c
Then f (0) = 0 and f (1) = c0 + 1 + 2 + .......... + n = 0
2 3 n +1
Þ f (0) = f (1) = 0
Q f ( x) is a polynomial therefore we have
10 Chap 4 – Differentiation
i) It is continuous on [0,1]
ii) It is differentiable on (0,1)
iii) And f (a ) = 0 = f (b)
Þ the function f has local maximum or a local minimum at some point x Î ( 0,1)
Þ f ¢( x) = c0 + c1 x + c2 x 2 + ......... + cn x n = 0 for some x Î ( 0,1)
Þ the given equation has real root between 0 and 1.
v Riemann Differentiation of Vector valued function
If f (t ) = f1 (t ) + i f 2 (t )
f ¢(t ) = f1¢(t ) + i f 2¢ (t )
where f1 (t ) and f 2 (t ) are the real and imaginary part of f (t ) .
The Rule of differentiation of real valued functions are valid in case of vector valued
function but the situation changes in the case of Mean Value Theorem.
v Example
Take f ( x ) = eix = cos x + i sin x in (0,2p ) .
Then f (2p ) = cos 2p + i sin 2p = 1
f (0) = cos(0) + i sin(0) = 1
Þ f (2p ) - f (0) = 0 but f ¢( x ) = i eix
f (2p ) - f (0)
Þ ¹ i eix (there is no such x )
2p - 0
Þ the M.V.T. fails.
In case of vector valued functions, the M.V.T. is not of the form as in the case of
real valued function.
v Theorem
Let f be a continuous mapping of the interval [a, b] into a space ¡ k and f be
differentiable in (a, b) then $ x Î (a, b ) such that f (b) - f (a) £ (b - a ) f ¢( x ) .
Proof
Put z = f (b) - f (a)
And suppose j (t ) = z × f (t ) (a £ t £ b)
j (t ) so defined is a real valued function and it possess the properties of f (t ) .
Þ M.V.T. is applicable to j (t ) .
We have j (b) - j (a ) = (b - a )j ¢( x)
i.e. j (b) - j (a) = (b - a) z × f ¢( x ) for some x Î (a, b ) ……….. (i)
Also j (b) = z × f (b) and j (a) = z × f (a)
( )
Þ j (b) - j (a ) = z × f (b) - f (a ) …………. (ii)
from (i) and (ii)
z × z = (b - a ) z × f ¢( x )
£ (b - a ) z f ¢( x )
Þ z £ (b - a ) z f ¢( x)
2
Þ z £ (b - a ) f ¢( x )
i.e. f (b) - f (a) £ (b - a ) f ¢( x ) Q z = f (b) - f (a)
which is the required result.
Chap 4 – Differentiation 11
v Question
If f ( x ) = x 3 , then compute f ¢( x ), f ¢¢( x ) and f ¢¢¢( x) , and show that f ¢¢¢(0) does
not exist.
Solution
ì x 3 if x ³ 0
f ( x) = x = í 3
3
î- x if x < 0
f ( x ) - f (0) x3 - 0
Now D+ f (0) = lim = lim = lim x 2 = 0
x ®0 + 0 x-0 x ® 0 + 0 x-0 x ®0+ 0
f ( x) - f (0) - x3 - 0
& D- f (0) = lim = lim = lim (- x 2 ) = 0
x®0- 0 x-0 x®0- 0 x - 0 x ® 0 -0
Q D+ f ( x) = D- f ( x)
\ f ¢( x ) exists at x = 0 & f ¢(0) = 0 .
Now if x ¹ 0 and x > 0 then
f ( x) = x 3 Þ f ¢( x) = 3x 2
and if x ¹ 0 and x < 0 then
f ( x ) = - x3 Þ f ¢( x) = -3x 2
ì 3x2 if x > 0
ï
i.e. f ¢( x) = í 0 if x = 0
ïî -3 x 2 if x < 0
f ¢( x) - f ¢(0) 3x 2 - 0
Now D+ f ¢(0) = lim = lim
x ®0 + 0 x-0 x ®0+ 0 x - 0
= lim 3 x = 0
x ®0+ 0
f ¢( x ) - f ¢(0) -3 x 2 - 0
And Now D- f ¢(0) = lim = lim
x ®0- 0 x-0 x ® 0 -0 x-0
= lim (-3x ) = 0
x ®0 + 0
Q D+ f ¢( x) = D- f ¢( x)
\ f ¢¢( x) exists at x = 0 & f ¢¢(0) = 0 .
Now if x ¹ 0 and x > 0 then
f ¢( x ) = 3x 2 Þ f ¢¢( x) = 6 x
and if x ¹ 0 and x < 0 then
f ¢( x) = -3x 2 Þ f ¢¢( x) = - 6 x
ì 6x if x > 0
ï
i.e. f ¢¢( x ) = í 0 if x = 0
ïî- 6 x if x < 0
f ¢¢( x) - f ¢¢(0) 6x - 0
Now D+ f ¢¢(0) = lim = lim =6
x ®0 + 0 x-0 x ® 0+ 0 x - 0
f ¢¢( x) - f ¢¢(0) - 6x - 0
And D- f ¢¢(0) = lim = lim = -6
x®0- 0 x-0 x ® 0- 0 x - 0
Q D+ f ¢¢(0) ¹ D- f ¢¢(0)
\ f ¢¢¢(0) doest not exist.
But f ¢¢¢(0) exist if x ¹ 0 , and equal to 6 if x > 0 and equal to - 6 if x < 0 .
……………………………….