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Derivatives

Advanced Level Pure Mathematics









Advanced Level Pure Mathematics







Calculus I 4

Chapter 4 Derivatives



4.1 Introduction 2



4.2 Differentiability 4



Continuity and Differentiability 6



4.3 Rules of Differentiation 10



4.5 Higher Derivatives 17



4.6 Mean Value Theorem 22









Prepared by K. F. Ngai(2003)

Page 1

Derivatives

Advanced Level Pure Mathematics





4.1 INTRODUCTION









Let A( x0 , y 0 ) be a fixed point and P( x, y ) be a variable point on the curve y  f (x) as shown on about

y  y0 f ( x)  f ( x 0 )

figure. Then the slope of the line AP is given by or . When the variable point P

x  x0 x  x0

moves closer and closer to A along the curve y  f (x) , i.e. x  x0 . the line AP becomes the tangent line of

f ( x)  f ( x 0 )

the curve at the point A. Hence, the slope of the tangent line at the point A is equal to lim .

x  x0 x  x0

This term is defined to be the derivative of f (x) at x  x0 and is usually denoted by f ' ( x0 ) . The

definition of derivative at any point x may be defined as follows.



Definition Let y  f (x) be a function defined on the interval a, b and x0  a, b  .

f (x) is said to be differentiable at x 0 ( or have a derivative at x 0 ) if the limit



f ( x)  f ( x 0 ) dy

lim exists. This lime value is denoted by f ' ( x 0 ) or and is called the

x  x0 x  x0 dx x x0



derivative of f (x) at x 0 .

If f (x) has a derivative at every point x in a, b  , then f (x) is said to be differentiable on

a, b .



Remark As x  x0 , the difference between x and x 0 is very small, i.e. x  x0 tends to zero. Usually,

this difference is denoted by h or x . Then the derivative at x 0 may be rewritten as

f ( x 0  h)  f ( x 0 )

lim . ( First Principle )

h 0 h



Example Let f ( x)  x 2  1 . Find f ' (2) .









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Derivatives

Advanced Level Pure Mathematics

 2 1

 x cos , x  0

Example Let f ( x)   x . Find f ' (0) .

 0,

 x0









Example If f ( x)  ln x , find f ' (x) .









Example Let f be a real-valued function defined on R such that for all real numbers x and y,

f ( x  y)  f ( x)  f ( y) . Suppose f is differentiable at x 0 , where x0  R .

f ( h)

(a) Find the value of lim .

h 0 h

(b) Show that f is differentiable on R and express f ' (x) in terms of x 0 .









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Derivatives

Advanced Level Pure Mathematics





4.2 DIFFERENTIABILITY



 1

 x sin , x  0

Example Let f ( x)   x . Show that x  0 , f is continuous but not differentiable.

 0,

 x0

Solution







f (0  h)  f (0) f (h)  f( 0 )

By definition, f ' (0) = lim = lim

h 0 h h 0 h

1

h sin

= lim h

h 0 h

1

= lim sin .

h 0 h

1

Since lim sin does not exist, f is not differentiable at x  0 .

h 0 h



1



Example If f ( x )  x , show that x  0 , f is continuous but not differentiable.

3









 x2 , if x  1

Example Let f ( x)   .

2 x  1 , if x  1



Find f '  ( 1 ) and f '  ( 1 ) by definition. Does f ' (1) exist? Why?









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Derivatives

Advanced Level Pure Mathematics



Example Show that f ( x)  x is not differentiable at x  0 . Find also the derivative of f (x) when



x  0.









x 2  x x0

Example Let f ( x)   , Find f ' (0) .

 x x0









Example Let f ( x)  x 3 . Find f ' (0) .









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 x2 ,x  c

Example A function f is defined as f ( x)   .

ax  b , x  c



Find a, b ( in term of c ) if f ' (c) exists.









CONTINUITY AND DIFFERENTIABILITY









Theorem Let y  f (x) . If f ' ( x) exist, then x  0 implies that y  0 .

y y

Proof Since y  ( )x , we have lim y ( )x

x x 0 x



  lim

y 



 lim x

 x 0 x  x 0

 ( Since both limit exist )





 f ' ( x) lim x .

x 0



As f ' ( x) exists, f ' ( x) is bounded. Futhermore, lim x  0 and so lim y  0 .

x0 x 0





Theorem If y  f (x) is differentiable at x0 , then f (x) is continuous at x 0 .



Proof By Theorem, lim y  0 , i.e. lim [ f ( x0  x)  f ( x0 )]  0 .

x 0 x0









Hence, f (x) is continuous at x 0 .



Prepared by K. F. Ngai(2003)

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Derivatives

Advanced Level Pure Mathematics

Remark We should have a clear concept about the difference between

(a) f (x) is well-defined at x 0 .

(b) the limit of f (x) at x 0 exists.

(c) f (x) is continuous at x 0 .

(d) f (x) is differentiable at x 0 .





D  C  L



 x2  a2

 ,0  x  a

 a



Example Let f ( x)   0 ,x  a .



 a( x  a )

2 2

,x  a



 x2

Show f (x) is continuous at x  a and discuss the continuity of f ' ( x) at x  a .









Example Prove that if f satisfy f ( x  y)  f ( x) f ( y) x, y  R and f ( x)  1  xg ( x) where



lim g ( x)  1 , then f ' ( x) exist x  R and f ' ( x)  f ( x) . Find f (x).

x 0









Prepared by K. F. Ngai(2003)

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Derivatives

Advanced Level Pure Mathematics

f '(1 )

Example Prove that if f satisfies f ( xy)  f ( x)  f ( y) , then f '(x)  , where x  0 and

x

find f (x) .









 1

 x if x  c, c  0



Example Find a, b in terms of c for f ' (c) exists where f ( x)  

a  bx 2 if x  c













Example Let f be a real-valued function such that



f ( x)  f (a)  ( x  a) 2 (x, a  R)



Show that f ' ( x)  0 for all x  R .









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Advanced Level Pure Mathematics





Exercise 4A Name: ________________________

1. Determine which of following functions are differentiable at x  0 .

 1

 x sin

2

( x  0)

(a) f ( x)  x 3

(b) f ( x)   x

 0

 ( x  0)









2. Find f ' ( x) of the following functions. Also, locate the points at which the function is not differentiable.

 4x ( x  3)

(a) f ( x)   2 (b) f ( x)  x 3  8

2 x  6 ( x  3)









 x3 ,x 1

3. Let f ( x)   .

ax  b , x  1



Find the values of a and b so that f is differentiable at x  1 .









Prepared by K. F. Ngai(2003)

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Derivatives

Advanced Level Pure Mathematics





4.3 RULES OF DIFFERENTIATION



Composite functions



d du d du dv d dv du

ku  k (u  v)   uv  u  v

dx dx dx dx dx dx dx dx



du dv

uv

d u dy dy du dy 1

   dx 2 dx   

dx  v  v dx du dx dx dx

dy





Algebraic functions





d k

x  kxk 1 where k must be independent of x (usually a constant)

dx





Inverse functions (esp.: inverse of trigo func)





dy 1

If y  f 1 ( x) the n 

dx df ( y )

dy





Trigonometric functions



d d d

sin x  cos x cos x   sin x tan x  sec2 x

dx dx dx



d d d

sec x  sec x tan x csc x   csc x cot x cot x   csc2 x

dx dx dx



Logarithmic functions



d x d 1

e  ex ln x 

dx dx x



d x d 1

a  a x ln a log a x 

dx dx x ln a



Parametric functions (commonly use in Rate of change)

dy

dy dt



dx dx

dt





Prepared by K. F. Ngai(2003)

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Derivatives

Advanced Level Pure Mathematics

Theorem Chain Rule

If h  g  f , i.e. h( x)  g ( f ( x)) and f, g are differentiable, then h' ( x)  g ' ( f ( x)) f ' ( x) .





Example h( x)  tan e sin x , h' ( x ) =





=



= sec2 e sin x (e sin x cos x)





Example Find the derivatives of the following functions:

x

(a) y  x ln(1  x) (b) y

1 x2









1

sin

(c) y  (3x  100 ) 60 (d) ye x

, where x  0









dy

Example Let sin( xy)  xy, find .

dx









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Derivatives

Advanced Level Pure Mathematics

Example ( Derivatives of inverse function )

d 1

Prove (sin 1 x) 

dx 1 x2

Solution Let y  sin 1 x . x = sin y

dy

1 = cos y

dx

dy 1

=

dx cos y

d 1

 (sin 1 x) =

dx 1 x2



d -1

Example Prove cos1 x =

dx 1 x2

Solution









d 1 d -1

Remark (sin 1 x) = , cos1 x =

dx 1 x2 dx 1 x2

d 1 d -1

tan 1 x = , cot1 x =

dx 1 x2 dx 1 x2



d 1 d -1

sec1 x = , csc1 x =

dx x x2 1 dx x x2 1



d d x

Example* (a) Find ln x (b) Find e

dx dx









Prepared by K. F. Ngai(2003)

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Derivatives

Advanced Level Pure Mathematics

dy

Example Find if

dx

(a) y  a x 1 , where a is a constant.

(b) y  x x  sin x .









Example Find the derivative of following functions

(a) x 2 sin 3 (1  3x) (b) (ln (cos 2 x)) 3









dy

Example If x  cost and y  sin t , find

dx









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Advanced Level Pure Mathematics





Exercise 4B Name ________________________



1. Find the derivatives of the following functions.



(a) y  e sin x (b) y  tan e x (c) 

y  ln( x 2  1) 

3









sin x

(d) y  ln(tan e x ) (e) y

sin 1 x









 2x 

(f) y  tan 1  2 

(g) y  (cos x) x

1 x 









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Derivatives

Advanced Level Pure Mathematics

1

tan

(h) y  3x e x (i) y3 x









1 x



(j) yx x

(k) y4 ln x









d 1

2. By considering the inverse of cos x , show that (cos 1 x) 

dx 1  x2









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Derivatives

Advanced Level Pure Mathematics

dy

3. Find if

dx

1

(a) tan 1 y  xy  x 2  (b) x 2  y 2  x cos y  1

y









dy

4. Find if

dx



 x  t  cost  x  cos3 t

(a)  (b) 

 y  1  sin t  y  sin t

3









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Derivatives

Advanced Level Pure Mathematics





4.5 HIGHER DERIVATIVES



Definition If y  f (x) is a function of x , then the nth derivatives of y w.r.t. x is defined as



d n y d  d n1 y  d n 1 y

  n 1  if is differentiable.

dx n dx  dx 

  dx n 1



dny

Symbolically, the nth derivatives of y w.r.t. x is denoted by y ( n ) , f (n)

( x) or .

dx n

dy 1 d2y 1

Remark 1.  but 2



dx dx dx d 2x

dy dy 2





d 2 y d  dy 

  

dx 2 dx  dx 





dy dy dz dy dz

2. If y is function of z , z  f (x) , then   (.  f ' ( x)  ).

dx dz dx dx dx

d 2 y d  dy  dy d dy

=   = f ' ' ( x)  f ' ( x) ( )

dx 2 dx  dx  dz dx dz

dy dz d  dy 

= f ''(x)  f '(x)   

dz dx dz  dz 

2

dy 2 d y

= f ''(x)  ( f '(x))

dz dz 2





d2y

Example Let y  sin 1 x . Find .

dx 2









Example Prove that y  e  kx (a cos x  b sin x) satisfies the equation y ' '2ky'(k 2  1) y  0 .









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Derivatives

Advanced Level Pure Mathematics

( n 1)

Example If y  xu , where u is a function of x, Prove y (n)

 xu (n)

 nu where y (n ) and u (n ) are

the nth derivative of y and u respectively.

1 1



Hence, if y  x n 1e x , prove y ( n )  (1) n x ( n 1) e x .









Example Find a general formula for the n th derivative of

(a) y  e ax ( a  R )

(b) y  sin x

(c) y  xm ( m  R)









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Advanced Level Pure Mathematics

Theorem Let f (x) and g (x) be two functions which are both differentiable up to nth order. Then

dn dn

(a) kf ( x)  k n f ( x)

dx n dx

dn dn dn

(b) [ f ( x)  g ( x)]  n f ( x)  n g ( x)

dx n dx dx



Theorem Leibniz's Theorem

Let f and g be two functions with nth derivative. Then

dn n

[ f ( x) g ( x)]   C rn f (r )

( x)g ( n  r ) ( x) where f (0)

( x)  f ( x) .

dx n r 0







Example Let y  xe ax , where a is a real constant. Find y ( 20) .









dn 3

Example Find ( x ln x) , x  0 .

dx n









Example Let y  tan 1 x . Show that for n  1 , (1  x 2 ) y ( n  2)  2(n  1) xy ( n 1)  n(n  1) y ( n )  0 .









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Derivatives

Advanced Level Pure Mathematics

2

d y dy

Example Prove that if y  x 2 cos x, then x 2 2

 4 x  ( x 2  6) y  0.

dx dx

dny d n2 y

Deduce that when x  0 (n  2)( n  3) n  n(n  1) n  2  0

dx dx









(a) Prove if y  e  x , y ( n  2)  2 xy ( n 1)  2(n  1) y ( n )  0 .

2

Example



d n  x2  x2 d2 d

(b) Show if n

(e )  e f ( x) , then 2

f ( x)  2 x f ( x)  2nf ( x)  0 .

dx dx dx









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Advanced Level Pure Mathematics





Exercise 4D Name ____________________



 x  1

r



1. Let r be a real number. Define y    for x  1 .

 x 1



dy  2ry

(a) Show that  .

dx x 2  1

(b) For n  1, 2, 3, , show that

( x 2  1) y ( n 1)  2(nx  r ) y ( n )  (n 2  n) y ( n 1)  0

dky

where y ( 0)  y and y ( k )  for k  1 .

dx k









2. Let f ( x)  x n e ax , where a is real and n is a positive integer.

Evaluate f ( 2 n ) (0) .









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Derivatives

Advanced Level Pure Mathematics





4.6 MEAN VALUE THEOREM



Definition Let y  f (x) be a function defined on an interval I. f is said to have an absolute maximum at

c if f (c)  f ( x), x  I and f (c) is called the absolute maximum value.

Similarly, f is said to have an absolute minimum at d if f (d )  f ( x), x  I and f(d) is

called the absolute minimum value.





Theorem Fermat's Theorem

Let y  f (x) be defined and differentiable on an open interval (a, b). If f (x) attains its

absolute maximum or absolute minimum (both are called absolute extremum) at x  c ,

where c  (a, b) , then f ' (c)  0 .





Proof For any x  (a, b), there exists a real number h such that ( x  h)  (a, b) and ( x  h)  (a, b) .

Now, suppose f (x) attains its absolute maximum at x  c . Then we have (c  h)  (a,b)

and (c  h)  (a,b) , and so f (c  h)  f (c) and f (c  h)  f (c) . Now, the left and right

hand derivatives are given by

f(c  h)  f(c)

f'(c)  lim 0, ( since f(c  h)  f(c)  0 )

h 0 h

f(c  h)  f(c)

and f'(c)  lim  0. ( since f(c  h)  f(c)  0 )

h 0 h

Since f (x) is differentiable at x  c , the left and right hand derivatives must be equal,

i.e. f '  (c)  f '  (c) . This is possible only if f ' (c)  0 .

The proof for f (x) attaining its absolute minimum at x  c is similar and is left as an exercise.





Remark 1. f ' ( p)  0 NOT IMPLIES absolute max. or min. at x  p .

e.g. f ( x)  x 3 at x  0 , not max. and min.









figure

2. Fermat's Theorem can't apply to function in closed interval. absolute max. or min may be

attained at the end-points. As a result, one of the left and right hand derivatives at c may

not exist.

e.g. f ( x)  ( x  2) 2  1defined on [ 0, 5] attains its absolute max. at x  5 but its right

hand derivative does not exist.

3. Fermat' s Theorem can't apply to function which are not differentiable.



e.g. f ( x)  x . Not differentiable at x  0 but min. at x  0 .









figure

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Theorem Rolle's Theorem

If a function f (x) satisfies all the following three conditions:

(1) f (x) is continuous on the closed interval [a, b] ,

(2) f (x) is differentiable in the open interval (a, b) ,

(3) f (a)  f (b) ;

then there exists at least a point   (a, b) such that f ' ( )  0 .





Proof Since f (x) is continuous on [a, b]  f (x) is bounded

(i) m  M , where m (min), M (Max) are constant.

 m  f ( x)  M , x  [a, b]

 f ( x)  M , x  [a, b]

 f ' ( x)  0, x  (a, b)

(ii) m  M , the max. and min. cannot both occur at the end points a, b.

 p  (a, b) such that f ( p)  M

i.e. f ( p)  f ( x) x sufficiently closed to p.

By Fermat's Theorem, f ' ( p) exist and equal to 0.





Example Define f ( x)  ( x  2) 2  1 on [0,4]. Note that f (0)  f (4)  5 .

We have f ' ( x)  2( x  2) and so f ' (2)  0 . Since 2  (0,4) , Rolle's Theorem is verified.

The geometric significance of Rolle's theorem is illustrated in the following diagram.









If the line joining the end points (a, f (a)) and (b, f (b)) is horizontal (i.e. parallel to the x-axis)

then there must be at least a point  (or more than one point) lying between a and b such that the

tangent at this point is horizontal.









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Derivatives

Advanced Level Pure Mathematics

Theorem Mean Value Theorem

If a function f (x) is

(1) continuous on the closed interval [a, b] and

(2) differentiable in the open interval (a, b) ,

then there exists at least a point   (a, b) such that

f (b)  f (a)

 f ' ( ) .

ba





Proof Consider the function g defined by

f (b)  f (a)

g ( x)  ( x  a)  f (a)  g (x) is differentiable and continuous on (a, b) .

ba

Let h( x)  g ( x)  f ( x)

 h(x) is also differentiable and continuous on (a, b) .

We have h(a)  0, h(b)  0.

 By Rolle's Theorem,   (a, b) such that h ' ( )  0

 g ' ( )  f ' ( )  0

f (b)  f (a)

 f ' ( ) 

ba

f (b)  f (a)

Remark: 1. The Mean Value Theorem still holds for a  b . f ' ( p)  .

ba

2. Another form of Mean Value Theorem f (b)  f (a)  f ' ( p)(b  a)

3. The value of p can be expressed as p  a   (b  a) , 0    1.

 f (b)  f (a)  f ' (a   (b  a))(b  a)





Example Use the Mean Value Theorem. show a  b



(a) sin a  sin b  a  b



cos ax  cos bx

(b)  a b , x  0

x

sin px

(c)  p, p  0, x  0 .

x









Prepared by K. F. Ngai(2003)

Page 24

Derivatives

Advanced Level Pure Mathematics

Example By using Mean Value Theorem, show that

e y  e a  e a ( y  a)

for all real values y and a .





Solution Let f ( x)  e x .

Case (i) y  a









Case (ii) y  a





Case (iii) y  a









Example Let a, b  R such that a  b and f (x) be a differentiable function on R such that

f (a)  0 , f (b)  0 and f ' ( x) is strictly decreasing. Show that f ' (b)  0 .









Example Let f (x) be a continuous function defined on [ 3, 6 ]. If f (x) is differentiable on ( 3, 6 ) and



f ' ( x)  9  3 . Show that 18  f (6)  f (3)  36 .









Prepared by K. F. Ngai(2003)

Page 25

Derivatives

Advanced Level Pure Mathematics



Example Let P ( x)  a n x n  a n 1 x n 1    a0 be a polynomial with real coefficients.



an a

If  n1    a0  0, by using Mean Value Theorem, show that the equation P( x)  0 has

n 1 n

at least one real root between 0 and 1.









Example Let f be a real-valued function defined on (0, ) . If f ' (t ) is an increasing function,

show that f (n)  f ' (n)(t  n)  f (t )  f (n)  f ' (n  1)(t  n) (t (n, n  1))









Example Let f be a real-valued function such that



f ( x)  f ( y)  ( x  y) 2 , (x, y  R)



Show that f is a differentiable function.

Hence deduce that f ( x)  k for all x  R , where k is a real constant.









Prepared by K. F. Ngai(2003)

Page 26

Derivatives

Advanced Level Pure Mathematics

Example Let f (x) be a function such that f ' ( x) is strictly increasing for x  0 .

(a) Using Mean Value Theorem, or otherwise, show that

f ' (k )  f (k  1)  f (k )  f ' (k  1) , k 1

(b) Hence, deduce that

f ' (1)  f ' (2)    f ' (n  1)  f (n)  f (1)  f ' (2)  f ' (3)   f ' (n) , n  2









Theorem Generalized Mean Value Theorem

Let f (x) and g (x) such that

(i) f (x) and g (x) are continuous on [ a, b ].

(ii) f (x) and g (x) are differentiable on ( a, b ).

Then there is at least one points p  (a, b) such that

[ f (b)  f (a)] g ' ( p)  [ g (b)  g (a)] f ' ( p) .





Proof Let h( x)  [ f (b)  f (a)] g ( x)  [ g (b)  g (a)] f ( x) , a  x  b .

 (i) h(x) is continuous on [ a, b ].

(ii) h(x) is differentiable on ( a, b ).

h(b)  h(a)

By Mean Value Theorem, p  (a, b) such that h ' ( p)  , hence the result is obtained.

ba

( Why ? )



Remark: Suppose that f (x) and g (x) are differentiable on ( a, b ) and that f ' ( x)  g ' ( x)  0 ,





f (b)  f (a) f ' ( p)

x  (a, b) then  .

g (b)  g (a) g ' ( p)



This is useful to establish an inequality by using generalized mean value theorem.









Prepared by K. F. Ngai(2003)

Page 27

Derivatives

Advanced Level Pure Mathematics

Example (a) Let f and g be real-valued functions continuous on [a, b] and differentiable in (a, b) .



(i) By considering the function



h( x)  f ( x)[g (b)  g (a)]  g ( x)[ f (b)  f (a)] on [a, b] , or otherwise,



show that there is c  (a, b) such that f ' (c)[g (b)  g (a)]  g ' (c)[ f (b)  f (a)]



(ii) Suppose g ' ( x)  0 for all x  (a, b) . Show that g ( x)  g (a)  0 for any x  (a, b) .

f ' ( x) f ( x)  f (a)

If, in addition, is increasing on (a, b) , show that P ( x)  is also

g ' ( x) g ( x)  g (a )

Increasing on (a, b) .

 e x cos x  1  

 if x   0, 

 4

 sin x  cos x  1  

(b) Let Q( x)  

 1 if x0





 

Show that Q is continuous at x  0 and increasing on 0,  .

 4

  x

Hence or otherwise, deduce that for x  0,  ,  Q (t )dt  x .

 4 0









Prepared by K. F. Ngai(2003)

Page 28


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