Capter 5 Differentiation Course Title: Real Analsis 1 Course Code: MTH31 Course instrutor: Dr Atiq ur Reman Class: MS-II Course URL: wwwmatitorg/atiq/fa15-mt31 Derivative of a funtion: Let f be defined and real valued on, quotient f ( ) f ( ) We fi point and stud te beaviour of tis quotient as Page 1 of 7 ab For an point a,, form te Definition Let f be defined on an open interval ab,, and assume tat ( a, b) Ten f is said to be differentiable at wenever te it f ( ) f ( ) eists Tis it is denoted b f () and is alled te derivative of f at point If, ten we ave Eample (i) A funtion f : defined b f ( ) f ( ) f( ) 0 sin 1 ; 0 ( ) f 0 ; 0 Tis funtion is differentiable at 0 beause f ( ) f (0) sin 1 0 0 0 0 0 sin 1 sin 1 0 0 0 n (ii) Let f ( ) ; n 0 (n is integer), Ten n n f ( ) f ( ) n1 n n n1 ( )( ) 1 1 ( n n n n ) n 1 n implies tat f is differentiable ever were and n 1 f ( ) n Capter 5: Differentiation
Teorem Let f be defined on, a, b ontinuous at (Differentiabilit implies ontinuit) We know tat f ( t) f ( ) f( ) t t were t and a t b Now f ( t) f ( ) f ( t) f ( ) t t t t t f( ) 0 0 ( ) f ( ) t Tis sow tat f is ontinuous at ab, if f is differentiable at a point Remarks (i) Te onverse of te above teorem does not old if 0 Consider f ( ) if 0 f (0) does not eists but f( ) is ontinuous at 0 (ii) If f is disontinuous at D ten f () does not eist f eg 1 if 0 f( ) 0 if 0 is disontinuous at 0 terefore it is not differentiable at 0, ten f is Teorem Suppose f and g are defined on [ ab, ] and are differentiable at a point [ a, b], ten f g, fg and f g are differentiable at and (i) ( f g) ( ) f ( ) g( ) (ii) ( fg) ( ) f ( ) g( ) f ( ) g( ) ( ) ( ) ( ) ( ) (iii) f ( ) g f f g g g ( ) Te proof of tis teorem an be get from an FS or BS tet book Page of 7 Capter 5: Differentiation
Teorem (Cain Rule) Suppose f is ontinuous on [ ab, ], f () eists at some point [ a, b] A funtion g is defined on an interval I wi ontains te range of f, and g is differentiable at te point f( ) If ( t) g f ( t) ; a t b Ten is differentiable at and ( ) g f ( ) f ( ) Let f ( ) B te definition of te derivative we ave f ( t) f ( ) ( t ) f ( ) u( t) (i) g( s) g( ) ( s ) g( ) v( s) (ii) and were t a,, s I and ( ) 0 ut as t and vs ( ) 0 as s Let us suppose s f () t ten ( t) ( ) g f ( t) g f ( ) f ( t) f ( ) g( ) v( s) b (ii) ( t ) f ( ) u( t) g( ) v( s) b (i) or if t ( t) ( ) f ( ) u( t) g( ) v( s) t taking te it as t we ave ( ) f ( ) 0 g( ) 0 g f ( ) f ( ) f ( ) wi is te required result It is known as ain rule Eample Let us find te derivative of sin( ), One wa to do tat is troug some trigonometri identities Indeed, we ave sin( ) sin( )os( ) So we will use te produt formula to get ( sin( ) ) ( sin ( )os( ) sin( )os ( ) ) wi implies sin( ) os ( ) sin ( ) ( ) ( ) Using te trigonometri formula os( ) os ( ) sin ( ), we ave ( sin( ) ) os( ) One tis is done, ou ma ask about te derivative of sin(5 )? Te answer an be found using similar trigonometri identities, but te alulations are not as eas as before We will see ow te Cain Rule formula will answer tis question in an elegant wa Page 3 of 7 Capter 5: Differentiation
Let us find te derivative of sin(5 ) We ave ( ) f g( ), were g 5 implies tat '( ) eists and and f ( ) sin( ) Ten te Cain rule [ ] ( ) 5 os(5 ) 5os(5 ) Loal Maimum Let f be a real valued funtion defined on a metri spae X, we sa tat f as a loal maimum at a point p X if tere eist 0 su tat f ( ) f ( p) X wit p Loal minimum is defined likewise Teorem Let f be defined on, f () eist ten f( ) 0 (Te analogous for loal minimum is of ourse also true) Coose su tat a b Now if t ten f ( t) f ( ) 0 t Taking it as t we get f( ) 0 (i) a If t Ten f ( t) f ( ) 0 t Again taking it wen t we get f( ) 0 (ii) Combining (i) and (ii) we ave f( ) 0 ab, if f as a loal maimum at a point a, f() and if t t + b Lagrange s Mean Value Teorem Let f be i) ontinuous on [ ab, ] ii) differentiable on ( ab, ) Ten tere eists a point ( a, b) su tat f ( b) f ( a) f() a Page 4 of 7 Capter 5: Differentiation
We are skipping te proof as it is inluded in BS alulus book Generalized Mean Value Teorem If f and g are ontinuous real valued funtions on losed interval ab, and f is differentiable on ab,, ten tere is a point a, at wi f ( b) f ( a) g( ) g( b) g( a) f ( ) Te differentiabilit is not required at te end point We are skipping te proof as it is inluded in BS alulus book Teorem (Intermediate Value Teorem or Darbou,s Teorem) Suppose f is a real differentiable funtion on some interval ab, and suppose f ( a) f ( b) ten tere eist a point ( a, b) su tat f() A similar result olds if f ( a) f ( b) Put g( t) f ( t) t Ten g( t) f ( t) If t a we ave g( a) f ( a) f( a) 0 g( a) 0 implies tat g is monotoniall dereasing at a a a point t 1 ( a, b) su tat g( a) g( t1) t 1 t b Similarl, g( b) f ( b) f( b) 0 g( b) 0 implies tat g is monotoniall inreasing at b a point t ( a, b) su tat g( t) g( b) te funtion attain its minimum on ( ab, ) at a point (sa) su tat g( ) 0 f( ) 0 f( ) ---------------------------------------------------- Question Let f be defined for all real and suppose tat f ( ) f ( ) ( ) Sine Terefore real & Prove tat f is onstant f ( ) f ( ) ( ) ( ) f ( ) f ( ) ( ) Dividing trougout b, we get Page 5 of 7 Capter 5: Differentiation
( ) ( ) ( ) f ( ) wen and ( ) ( ) ( ) f ( ) wen Taking it as, we get 0 f( ) 0 0 f( ) 0 f( ) 0 wi sows tat funtion is onstant Question Suppose f is defined and differentiable for ever 0 and f( ) 0 as put g( ) f ( 1) f ( ) Prove tat g ( ) 0 as Sine f is defined and differentiable for 0 terefore we an appl te Lagrange s MV T to ave f ( 1) f ( ) ( 1 ) f ( 1) were 1 f( ) 0 as f( 1) 0 as f ( 1) f ( ) 0 as 0 g ( ) 0 as 0 Question (L Hospital Rule) Suppose f ( ), g ( ) eist, g( ) 0 and f ( ) g( ) 0 f ( t) f ( ) Prove tat t g ( t ) g ( ) f ( t) f ( t) 0 f ( t) f ( ) t g( t) t g( t) 0 t g ( t ) ( ) f ( ) g( ) 0 f ( t) f ( ) t t t g( t) ( ) f ( t) f ( ) 1 t t t g ( t ) ( ) t f ( t) f ( ) 1 1 f( ) f( ) t t g( t) ( ) g( ) g( ) t t Question Suppose f is defined in te neigborood of a point and f ( ) eists f ( ) f ( ) f ( ) Sow tat f( ) 0 Page 6 of 7 Capter 5: Differentiation
B use of Lagrange s Mean Value Teorem f ( ) f ( ) f ( 1) were 1 (i) and [ f ( ) f ( )] f ( ) were (ii) Subtrat (ii) from (i) to get f ( ) f ( ) f ( ) [ f ( 1) f ( )] f ( ) f ( ) f ( ) f ( 1) f ( ) Sine 1 0 as 0, terefore f ( ) f ( ) f ( ) f ( 1) f ( ) 0 1 1 f( ) Question If 1 n 1 n 0 0 3 n n1 Were 0, 1,,, n are real onstants n Prove tat 0 1 n 0 as at least one real root between 0 and 1 1 1 Suppose ( ) n n f 0 n 1 1 Ten f (0) 0 and f(1) 0 n 0 3 n 1 f(0) f(1) 0 f( ) is a polnomial terefore we ave i) It is ontinuous on [0,1] ii) It is differentiable on (0,1) iii) And f ( a) 0 f ( b) te funtion f as loal maimum or a loal minimum at some point 0,1 n f( ) for some 0,1 0 1 n 0 te given equation as real root between 0 and 1 Page 7 of 7 Capter 5: Differentiation