Derivatives
Advanced Level Pure Mathematics
Advanced Level Pure Mathematics
4
Calculus I
Chapter 4
Derivatives
4.1
Introduction
2
4.2
Differentiability
4
Continuity and Differentiability
6
4.3
Rules of Differentiation
9
4.5
Higher Derivatives
13
4.6
Mean Value Theorem
17
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Page 1
Derivatives
4.1
Advanced Level Pure Mathematics
INTRODUCTION
Let A( x0 , y0 ) be a fixed point and P ( x, y ) be a variable point on the curve y  f (x) as shown on about
f ( x)  f ( x 0 )
y  y0
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
figure. Then the slope of the line AP is given by
f ( x)  f ( x 0 )
.
x  x0
This term is defined to be the derivative of f (x) at x  x0 and is usually denoted by f ' ( x 0 ) . The
the curve at the point A. Hence, the slope of the tangent line at the point A is equal to lim
x  x0
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 x0 ( or have a derivative at x0 ) if the limit
lim
x  x0
f ( x)  f ( x 0 )
dy
exists. This lime value is denoted by f ' ( x 0 ) or
dx
x  x0
and is called the
x  x0
derivative of f (x) at x0 .
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 x0 is very small, i.e. x  x0 tends to zero. Usually,
this difference is denoted by h or x . Then the derivative at x0 may be rewritten as
lim
h0
Example
f ( x0  h)  f ( x0 )
. ( First Principle )
h
Let f ( x)  x 2  1 . Find f ' (2) .
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Derivatives
Advanced Level Pure Mathematics
Example
1
 2
 x cos , x  0
Let f ( x)  
. Find f ' (0) .
x
 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 x0 , where x0  R .
(a)
(b)
f ( h)
.
h 0
h
Show that f is differentiable on R and express f ' (x) in terms of x0 .
Find the value of lim
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Derivatives
4.2
Example
Advanced Level Pure Mathematics
DIFFERENTIABILITY
1

 x sin , x  0
Let f ( x)  
. Show that x  0 , f is continuous but not differentiable.
x
 0,
x0
Solution
By definition, f ' (0)
=
lim
h 0
f (0  h)  f (0)
h
=
lim
h 0
f (h)  f( 0 )
h
h sin
Since lim sin
h 0
=
lim
h 0
h
=
lim sin
1
.
h
h 0
1
h
1
does not exist, f is not differentiable at x  0 .
h
1
Example
If f ( x)  x 3 , show that x  0 , f is continuous but not differentiable.
Example
 x2
, if x  1
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
Example
Advanced Level Pure Mathematics
Show that f ( x)  x
is not differentiable at x  0 . Find also the derivative of f (x) when
x  0.
Example
x 2  x x  0
Let f ( x)  
, Find f ' (0) .
x0
 x
Example
Let f ( x)  x 3 . Find f ' (0) .
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Derivatives
Example
Advanced Level Pure Mathematics
 x2
,x  c
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 .
Proof
Since y  (
y
)x , we have lim y
x 0
x
(
y
) x
x

y 

  lim
 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
Theorem
If y  f (x) is differentiable at x0 , then f (x) is continuous at x0 .
Proof
By Theorem, lim y  0 , i.e. lim [ f ( x0  x)  f ( x0 )]  0 .
x  0
x  0
x 0
Hence, f (x) is continuous at x 0 .
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Derivatives
Remark
Advanced Level Pure Mathematics
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
Example

C

L
 x2  a2
,0  x  a

a


Let f ( x)  
0
,x  a .

2
2
 a( x  a )
,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
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Derivatives
Example
Advanced Level Pure Mathematics
Prove that if f satisfies f ( xy)  f ( x)  f ( y ) , then f '(x) 
f '( 1 )
, where x  0 and
x
find f ' ( x) .
Example
 1
 x
Find a, b in terms of c for f ' (c ) exists where f ( x)  
a  bx 2

Example
Let f be a real-valued function such that
f ( x)  f ( a )  ( x  a ) 2
if x  c, c  0
if x  c
(x, a  R)
Show that f ' ( x)  0 for all x  R .
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Derivatives
4.3
Advanced Level Pure Mathematics
RULES OF DIFFERENTIATION
Composite functions
d
du
ku  k
dx
dx
v
u
 
dx  v 
d
du
u
dx
d
dv
du
uv  u  v
dx
dx
dx
dy dy du


dx du dx
dy
1

dx dx
dy
dv
dx
v
d
du dv
(u  v) 

dx
dx dx
2
Algebraic functions
d k
x  kx k 1
dx
where k must be independent of x (usually a constant)
Inverse functions (esp.: inverse of trigo func)
If
y  f 1 ( x )
then
dy
1

df
( y)
dx
dy
Trigonometric functions
d
sin x  cos x
dx
d
cos x   sin x
dx
d
tan x  sec 2 x
dx
d
sec x  sec x tan x
dx
d
csc x   csc x cot x
dx
d
cot x   csc2 x
dx
Logarithmic functions
d x
e  ex
dx
d
1
ln x 
dx
x
d x
a  a x ln a
dx
d
1
log a x 
dx
x ln a
Parametric functions (commonly use in Rate of change)
dy
dy dt

dx dx
dt
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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 ) =
=
=
Example
Find the derivatives of the following functions:
(a)
y  x ln( 1  x)
(c) y  (3x  100)
Example
sec 2 e sin x (e sin x cos x)
(b)
60
(d)
y
ye
x
1 x2
sin
1
x
, where x  0
( Derivatives of inverse )
Let sin( xy)  xy, find
dy
.
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 .

Example
x
=
sin y
1
=
cos y
dy
=
dx
1
cos y
d
(sin 1 x) =
dx
1
d
cos 1 x
dx
Prove
dy
dx
1 x2
-1
=
1 x2
Solution
Remark
d
dx
d
( s in 1 x )
1
ta n
1
x
dx
d
se c
1
x
dx
Example*
1
=
(a) Find
=
x
1
1
x
d
ln x
dx
x
1
cos

x
dx
d
,
1x
=
d
,
1
cot
1
x
=
x
=
dx
d
,
csc
dx
(b) Find
-1
=
1
x
-1
1x
-1
x
x
1
d x
e
dx
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Derivatives
Example
Advanced Level Pure Mathematics
Find
(a)
(b)
dy
if
dx
y  a x 1 , where a is a constant.
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
Example
If x  cos t and y  sin t , find
dy
dx
Exercise 4B 4(c), (d), (i), (k), (n), (q), (n), (v), (w), (x), (y), (z)
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Derivatives
4.5
Advanced Level Pure Mathematics
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


if
is differentiable.

dx n 1
dx n dx  dx n1 
Symbolically, the nth derivatives of y w.r.t. x is denoted by y ( n ) , f ( n ) ( x) or
Remark
1.
dy

dx
1
dx
dy
but
dny
.
dx n
d2y
1

2
dx
d 2x
dy 2
d 2 y d  dy 
  
dx 2 dx  dx 
2.
If y is function of z , z  f (x) , then
d 2 y d  dy 
=  
dx 2 dx  dx 
dy
dz
dy dy dz


 f ' ( x)  ).
(.
dx dz dx
dx
dx
dy
d dy
 f ' ( x) ( )
dz
dx dz
=
f ' ' ( x)
=
f ''(x)
=
2
dy
2 d y
f ''(x)  ( f '(x))
dz
dz 2
dy
dz d  dy 
 f '(x)   
dz
dx dz  dz 
d2y
.
dx 2
Example
Let y  sin 1 x . Find
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
Example
Advanced Level Pure Mathematics
If y  xu , where u is a function of x, Prove y
( n)
 xu
( n)
 nu
( n 1)
where y (n ) and u ( n ) are
the nth derivative of y and u respectively.
n 1
1
x
Hence, if y  x e , prove y
Example
(n)
 (1) x
n
( n 1)
1
x
e .
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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Derivatives
Theorem
Theorem
Advanced Level Pure Mathematics
Let f (x) and g (x ) be two functions which are both differentiable up to nth order. Then
(a)
dn
dn
kf
(
x
)

k
f ( x)
dx n
dx n
(b)
dn
dn
dn
[
f
(
x
)

g
(
x
)]

f
(
x
)

g ( x)
dx n
dx n
dx n
Leibniz's Theorem
Let f and g be two functions with nth derivative. Then
n
dn
[
f
(
x
)
g
(
x
)]

Crn f ( r ) ( x)g ( nr ) ( x) where f ( 0) ( x)  f ( x) .

n
dx
r 0
Example
Let y  xeax , where a is a real constant. Find y ( 20) .
Example
Find
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 .
dn 3
( x ln x) , x  0 .
dx n
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Derivatives
Example
Advanced Level Pure Mathematics
Prove that if y  x 2 cos x, then x 2
2
d y
dy
 4 x  ( x 2  6) y  0.
2
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
Example
(a) Prove if y  e  x , y ( n2)  2 xy( n1)  2(n  1) y ( n )  0 .
2
d n  x2
d2
d
 x2
(b) Show if
(e )  e f ( x) , then
f ( x)  2 x
f ( x)  2nf ( x)  0 .
n
2
dx
dx
dx
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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)  lim
h 0
f(c  h)  f(c)
 0,
h
( since f(c  h)  f(c)  0 )
f(c  h)  f(c)
 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,
and
f'  (c)  lim
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.
2.
3.
figure
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.
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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Derivatives
Advanced Level Pure Mathematics
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] 
(i)
f (x ) is bounded
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
Theorem
Advanced Level Pure Mathematics
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 )
( x  a )  f (a )  g (x ) is differentiable and continuous on (a, b) .
ba
Let h( x)  g ( x)  f ( x)
g ( x) 

h(x) is also differentiable and continuous on (a, b) .

We have h(a )  0, h(b)  0.
  (a, b) such that h ' ( )  0
By Rolle's Theorem,
 g ' ( )  f ' ( )  0

Remark:
1.
2.
3.
Example
f ' ( ) 
f (b)  f (a )
ba
f (b)  f (a)
.
ba
Another form of Mean Value Theorem f (b)  f (a)  f ' ( p)(b  a)
The Mean Value Theorem still holds for a  b . f ' ( p ) 
The value of p can be expressed as p  a   (b  a) , 0    1.
 f (b)  f (a)  f ' (a   (b  a))(b  a)
Use the Mean Value Theorem. show a  b
(a)
sin a  sin b  a  b
(b)
cos ax  cos bx
 a b , x 0
x
(c)
sin px
 p,
x
p  0, x  0 .
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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)
Case (ii)
Case (iii)
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 .
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Derivatives
Example
Advanced Level Pure Mathematics
Let P( x)  a n x n  a n 1 x n 1    a0 be a polynomial with real coefficients.
an
a
 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.
If
Example
Let f be a real-valued function defined on (0, ) . If f ' (t ) is an increasing function,
f (n)  f ' (n)(t  n)  f (t )  f (n)  f ' (n  1)(t  n)
(t (n, n  1))
show that
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.
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Derivatives
Example
Advanced Level Pure Mathematics
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)
(b) Hence, deduce that
f ' (1)  f ' (2)    f ' (n  1)  f (n)  f (1)  f ' (2)  f ' (3)   f ' (n)
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 ).
By Mean Value Theorem, p  (a, b) such that h ' ( p ) 
h(b)  h(a )
, 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 ,
x  (a, b) then
f (b)  f (a)
f ' ( p)
.

g (b)  g (a)
g ' ( p)
This is useful to establish an inequality by using generalized mean value theorem.
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Derivatives
Example
Advanced Level Pure Mathematics
(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) .
If, in addition,
f ( x)  f (a )
f ' ( x)
is increasing on (a, b) , show that P( x) 
is also
g ( x)  g (a )
g ' ( x)
Increasing on (a, b) .
 e x cos x  1
if

 sin x  cos x  1
(b) Let Q( x)  

1
if

 
x   0, 
 4
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
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