Differentiability - Continuity and Differentiability, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


2.3 Dierentiability of functions
Denition 2.14 Suppose f is a (real valued) function dened on an open interval
I and x
0
2I. Then f is said to be dierentiable at x
0
if
lim
x!x
0
f(x)f(x
0
)
xx
0
exists, and in that case the value of the limit is called the derivative of f at x
0
.
The derivative of f at x
0
, if exists, is denoted by
f
0
(x
0
) or
df
dx
(x
0
):

Exercise 2.13 Show that f : I ! R is dierentiable at x
0
2 I if and only if
lim
h!0
f(x
0
+h)f(x
0
)
h
exists. J
Exercise 2.14 Suppose f is dened on an open interval I and x
0
2I. Show that
f is dierentiable at x
0
2 I if and only if there exists a continuous function (x)
such that
f(x) =f(x
0
) + (x)(xx
0
);
and in that case (x
0
) =f
0
(x
0
). J
Exercise 2.15 Let  be as in Exercise 2.14. Then f is dierentiable at x
0
, if and
only if for every sequence (x
n
) inInfx
0
g which converges tox
0
, the sequence (x
n
)
converges, and in that case f
0
(x
0
) = lim
n!1
(x
n
). J
Exercise 2.16 Suppose f and g dened on I are dierentiable at x
0
2 I and
2R. Show that the functions '(x) :=f(x) +g(x) and (x) :=f(x), x2I are
dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
) +g
0
(x
0
); 
0
(x
0
) ='
0
(x
0
):
J
2.3.1 Some properties of dierentiable functions
Theorem 2.28 (Dierentiability implies continuity) Suppose f dened on I
is dierentiable at x
0
2I. Then f is continuous at x
0
.
Proof. Note that
f(x
0
+h)f(x
0
) =
f(x
0
+h)f(x
0
)
h
h!f
0
(x
0
):0 = 0 as h! 0:
Thus, f is continuous at x
0
.
Page 2


2.3 Dierentiability of functions
Denition 2.14 Suppose f is a (real valued) function dened on an open interval
I and x
0
2I. Then f is said to be dierentiable at x
0
if
lim
x!x
0
f(x)f(x
0
)
xx
0
exists, and in that case the value of the limit is called the derivative of f at x
0
.
The derivative of f at x
0
, if exists, is denoted by
f
0
(x
0
) or
df
dx
(x
0
):

Exercise 2.13 Show that f : I ! R is dierentiable at x
0
2 I if and only if
lim
h!0
f(x
0
+h)f(x
0
)
h
exists. J
Exercise 2.14 Suppose f is dened on an open interval I and x
0
2I. Show that
f is dierentiable at x
0
2 I if and only if there exists a continuous function (x)
such that
f(x) =f(x
0
) + (x)(xx
0
);
and in that case (x
0
) =f
0
(x
0
). J
Exercise 2.15 Let  be as in Exercise 2.14. Then f is dierentiable at x
0
, if and
only if for every sequence (x
n
) inInfx
0
g which converges tox
0
, the sequence (x
n
)
converges, and in that case f
0
(x
0
) = lim
n!1
(x
n
). J
Exercise 2.16 Suppose f and g dened on I are dierentiable at x
0
2 I and
2R. Show that the functions '(x) :=f(x) +g(x) and (x) :=f(x), x2I are
dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
) +g
0
(x
0
); 
0
(x
0
) ='
0
(x
0
):
J
2.3.1 Some properties of dierentiable functions
Theorem 2.28 (Dierentiability implies continuity) Suppose f dened on I
is dierentiable at x
0
2I. Then f is continuous at x
0
.
Proof. Note that
f(x
0
+h)f(x
0
) =
f(x
0
+h)f(x
0
)
h
h!f
0
(x
0
):0 = 0 as h! 0:
Thus, f is continuous at x
0
.
For the following theorem, we may recall that if' is continuous at a pointx
0
and
'(x
0
)6= 0, then there exists an open interval I
0
containing x
0
such that '(x)6= 0
for all x2I
0
.
Theorem 2.29 (Products and quotient rules) Suppose f and g dened on I
are dierentiable at x
0
2I. Then the function '(x) :=f(x)g(x) is dierentiable at
x
0
, and
'
0
(x
0
) =f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
): ()
If g(x
0
)6= 0, then the function (x) :=f(x)=g(x) is dierentiable at x
0
, and
 
0
(x
0
) =
g(x
0
)f
0
(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
: ()
Proof. Note that
'(x
0
+h)'(x
0
) = f(x
0
+h)g(x
0
+h)f(x
0
)g(x
0
)
= [f(x
0
+h)f(x
0
)]g(x
0
+h) +f(x
0
)[g(x
0
+h)g(x
0
)]
so that
'(x
0
+h)'(x
0
)
h
=
f(x
0
+h)f(x
0
)
h
g(x
0
+h) +f(x
0
)
g(x
0
+h)g(x
0
)
h
! f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
) as h! 0:
Hence, ' is dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
):
Also, since
 (x
0
+h) (x
0
) =
f(x
0
+h)g(x
0
)f(x
0
)g(x
0
+h)
g(x
0
+h)g(x
0
)
=
[f(x
0
+h)f(x
0
)]g(x
0
)f(x
0
)[g(x
0
+h)g(x
0
)]
g(x
0
+h)g(x
0
)
;
we have
 (x
0
+h) (x
0
)
h
=
f(x
0
+h)f(x
0
)
h
g(x
0
)f(x
0
)
g(x
0
+h)g(x
0
)
h
g(x
0
+h)g(x
0
)
!
f
0
(x
0
)g(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
as h! 0:
Thus, is dierentiable at x
0
, and 
0
(x
0
) =
g(x
0
)f
0
(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
.
Theorem 2.30 (Composition rule) Suppose f is dened on an open interval I
and x
0
2I, and g is dened in an open interval containing y
0
:=f(x
0
). Let
' =gf:
Then we have the following.
Page 3


2.3 Dierentiability of functions
Denition 2.14 Suppose f is a (real valued) function dened on an open interval
I and x
0
2I. Then f is said to be dierentiable at x
0
if
lim
x!x
0
f(x)f(x
0
)
xx
0
exists, and in that case the value of the limit is called the derivative of f at x
0
.
The derivative of f at x
0
, if exists, is denoted by
f
0
(x
0
) or
df
dx
(x
0
):

Exercise 2.13 Show that f : I ! R is dierentiable at x
0
2 I if and only if
lim
h!0
f(x
0
+h)f(x
0
)
h
exists. J
Exercise 2.14 Suppose f is dened on an open interval I and x
0
2I. Show that
f is dierentiable at x
0
2 I if and only if there exists a continuous function (x)
such that
f(x) =f(x
0
) + (x)(xx
0
);
and in that case (x
0
) =f
0
(x
0
). J
Exercise 2.15 Let  be as in Exercise 2.14. Then f is dierentiable at x
0
, if and
only if for every sequence (x
n
) inInfx
0
g which converges tox
0
, the sequence (x
n
)
converges, and in that case f
0
(x
0
) = lim
n!1
(x
n
). J
Exercise 2.16 Suppose f and g dened on I are dierentiable at x
0
2 I and
2R. Show that the functions '(x) :=f(x) +g(x) and (x) :=f(x), x2I are
dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
) +g
0
(x
0
); 
0
(x
0
) ='
0
(x
0
):
J
2.3.1 Some properties of dierentiable functions
Theorem 2.28 (Dierentiability implies continuity) Suppose f dened on I
is dierentiable at x
0
2I. Then f is continuous at x
0
.
Proof. Note that
f(x
0
+h)f(x
0
) =
f(x
0
+h)f(x
0
)
h
h!f
0
(x
0
):0 = 0 as h! 0:
Thus, f is continuous at x
0
.
For the following theorem, we may recall that if' is continuous at a pointx
0
and
'(x
0
)6= 0, then there exists an open interval I
0
containing x
0
such that '(x)6= 0
for all x2I
0
.
Theorem 2.29 (Products and quotient rules) Suppose f and g dened on I
are dierentiable at x
0
2I. Then the function '(x) :=f(x)g(x) is dierentiable at
x
0
, and
'
0
(x
0
) =f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
): ()
If g(x
0
)6= 0, then the function (x) :=f(x)=g(x) is dierentiable at x
0
, and
 
0
(x
0
) =
g(x
0
)f
0
(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
: ()
Proof. Note that
'(x
0
+h)'(x
0
) = f(x
0
+h)g(x
0
+h)f(x
0
)g(x
0
)
= [f(x
0
+h)f(x
0
)]g(x
0
+h) +f(x
0
)[g(x
0
+h)g(x
0
)]
so that
'(x
0
+h)'(x
0
)
h
=
f(x
0
+h)f(x
0
)
h
g(x
0
+h) +f(x
0
)
g(x
0
+h)g(x
0
)
h
! f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
) as h! 0:
Hence, ' is dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
):
Also, since
 (x
0
+h) (x
0
) =
f(x
0
+h)g(x
0
)f(x
0
)g(x
0
+h)
g(x
0
+h)g(x
0
)
=
[f(x
0
+h)f(x
0
)]g(x
0
)f(x
0
)[g(x
0
+h)g(x
0
)]
g(x
0
+h)g(x
0
)
;
we have
 (x
0
+h) (x
0
)
h
=
f(x
0
+h)f(x
0
)
h
g(x
0
)f(x
0
)
g(x
0
+h)g(x
0
)
h
g(x
0
+h)g(x
0
)
!
f
0
(x
0
)g(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
as h! 0:
Thus, is dierentiable at x
0
, and 
0
(x
0
) =
g(x
0
)f
0
(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
.
Theorem 2.30 (Composition rule) Suppose f is dened on an open interval I
and x
0
2I, and g is dened in an open interval containing y
0
:=f(x
0
). Let
' =gf:
Then we have the following.
(i) Suppose f is dierentiable at x
0
and g is dierentiable at y
0
. Then ' is
dierentiable at x
0
and
'
0
(x
0
) =g
0
(y
0
)f
0
(x
0
):
(ii) Suppose' is dierentiable at x
0
,g is dierentiable at y
0
andg
0
(y
0
)6= 0. Then
f is dierentiable at x
0
and
f
0
(x
0
) =
'
0
(x
0
)
g
0
(y
0
)
:
(iii) Suppose' is dierentiable atx
0
,f is dierentiable atx
0
andf
0
(x
0
)6= 0. Then
g is dierentiable at y
0
and
g
0
(y
0
) =
'
0
(x
0
)
f
0
(x
0
)
:
Proof. For x6=x
0
, let
(x) :=
(gf)(x) (gf)(x
0
)
xx
0
:
Let (x
n
) be a sequence inInfx
0
g which converges tox
0
. Then, takingy
n
:=f(x
n
),
n2N, and y
0
=f(x
0
), we have
(x
n
) =
(gf)(x
n
) (gf)(x
0
)
x
n
x
0
=
g(y
n
)g(y
0
)
x
n
x
0
=
g(y
n
)g(y
0
)
y
n
y
0

f(x
n
)f(x
0
)
x
n
x
0
:
(i) Supposef is dierentiable atx
0
. Now, sincex
n
!x
0
we have, by continuity
of f at x
0
, y
n
!y
0
. Therefore,
f(x
n
)f(x
0
)
x
n
x
0
!f
0
(x
0
);
g(y
n
)g(y
0
)
y
n
y
0
!g
0
(y
0
):
Thus, (x
n
)!g
0
(y
0
)f
0
(x
0
) showing that gf is dierentiable at x
0
and
(gf)
0
(x
0
) =g
0
(y
0
)f
0
(x
0
):
(ii) Suppose g f is is dierentiable at x
0
and g
0
(y
0
) 6= 0. Then we have
(x
n
)! (gf)
0
(x
0
) and
f(x
n
)f(x
0
)
x
n
x
0
=
(x
n
)
g(yn)g(y
0
)
yny
0
!
(gf)
0
(x
0
)
g
0
(y
0
)
:
Hence, f is dierentiable at x
0
and f
0
(x
0
) =
(gf)
0
(x
0
)
g
0
(y
0
)
.
(iii) Proof of this part is analogous to the proof of (ii). Hence, we omit the
details.
Page 4


2.3 Dierentiability of functions
Denition 2.14 Suppose f is a (real valued) function dened on an open interval
I and x
0
2I. Then f is said to be dierentiable at x
0
if
lim
x!x
0
f(x)f(x
0
)
xx
0
exists, and in that case the value of the limit is called the derivative of f at x
0
.
The derivative of f at x
0
, if exists, is denoted by
f
0
(x
0
) or
df
dx
(x
0
):

Exercise 2.13 Show that f : I ! R is dierentiable at x
0
2 I if and only if
lim
h!0
f(x
0
+h)f(x
0
)
h
exists. J
Exercise 2.14 Suppose f is dened on an open interval I and x
0
2I. Show that
f is dierentiable at x
0
2 I if and only if there exists a continuous function (x)
such that
f(x) =f(x
0
) + (x)(xx
0
);
and in that case (x
0
) =f
0
(x
0
). J
Exercise 2.15 Let  be as in Exercise 2.14. Then f is dierentiable at x
0
, if and
only if for every sequence (x
n
) inInfx
0
g which converges tox
0
, the sequence (x
n
)
converges, and in that case f
0
(x
0
) = lim
n!1
(x
n
). J
Exercise 2.16 Suppose f and g dened on I are dierentiable at x
0
2 I and
2R. Show that the functions '(x) :=f(x) +g(x) and (x) :=f(x), x2I are
dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
) +g
0
(x
0
); 
0
(x
0
) ='
0
(x
0
):
J
2.3.1 Some properties of dierentiable functions
Theorem 2.28 (Dierentiability implies continuity) Suppose f dened on I
is dierentiable at x
0
2I. Then f is continuous at x
0
.
Proof. Note that
f(x
0
+h)f(x
0
) =
f(x
0
+h)f(x
0
)
h
h!f
0
(x
0
):0 = 0 as h! 0:
Thus, f is continuous at x
0
.
For the following theorem, we may recall that if' is continuous at a pointx
0
and
'(x
0
)6= 0, then there exists an open interval I
0
containing x
0
such that '(x)6= 0
for all x2I
0
.
Theorem 2.29 (Products and quotient rules) Suppose f and g dened on I
are dierentiable at x
0
2I. Then the function '(x) :=f(x)g(x) is dierentiable at
x
0
, and
'
0
(x
0
) =f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
): ()
If g(x
0
)6= 0, then the function (x) :=f(x)=g(x) is dierentiable at x
0
, and
 
0
(x
0
) =
g(x
0
)f
0
(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
: ()
Proof. Note that
'(x
0
+h)'(x
0
) = f(x
0
+h)g(x
0
+h)f(x
0
)g(x
0
)
= [f(x
0
+h)f(x
0
)]g(x
0
+h) +f(x
0
)[g(x
0
+h)g(x
0
)]
so that
'(x
0
+h)'(x
0
)
h
=
f(x
0
+h)f(x
0
)
h
g(x
0
+h) +f(x
0
)
g(x
0
+h)g(x
0
)
h
! f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
) as h! 0:
Hence, ' is dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
):
Also, since
 (x
0
+h) (x
0
) =
f(x
0
+h)g(x
0
)f(x
0
)g(x
0
+h)
g(x
0
+h)g(x
0
)
=
[f(x
0
+h)f(x
0
)]g(x
0
)f(x
0
)[g(x
0
+h)g(x
0
)]
g(x
0
+h)g(x
0
)
;
we have
 (x
0
+h) (x
0
)
h
=
f(x
0
+h)f(x
0
)
h
g(x
0
)f(x
0
)
g(x
0
+h)g(x
0
)
h
g(x
0
+h)g(x
0
)
!
f
0
(x
0
)g(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
as h! 0:
Thus, is dierentiable at x
0
, and 
0
(x
0
) =
g(x
0
)f
0
(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
.
Theorem 2.30 (Composition rule) Suppose f is dened on an open interval I
and x
0
2I, and g is dened in an open interval containing y
0
:=f(x
0
). Let
' =gf:
Then we have the following.
(i) Suppose f is dierentiable at x
0
and g is dierentiable at y
0
. Then ' is
dierentiable at x
0
and
'
0
(x
0
) =g
0
(y
0
)f
0
(x
0
):
(ii) Suppose' is dierentiable at x
0
,g is dierentiable at y
0
andg
0
(y
0
)6= 0. Then
f is dierentiable at x
0
and
f
0
(x
0
) =
'
0
(x
0
)
g
0
(y
0
)
:
(iii) Suppose' is dierentiable atx
0
,f is dierentiable atx
0
andf
0
(x
0
)6= 0. Then
g is dierentiable at y
0
and
g
0
(y
0
) =
'
0
(x
0
)
f
0
(x
0
)
:
Proof. For x6=x
0
, let
(x) :=
(gf)(x) (gf)(x
0
)
xx
0
:
Let (x
n
) be a sequence inInfx
0
g which converges tox
0
. Then, takingy
n
:=f(x
n
),
n2N, and y
0
=f(x
0
), we have
(x
n
) =
(gf)(x
n
) (gf)(x
0
)
x
n
x
0
=
g(y
n
)g(y
0
)
x
n
x
0
=
g(y
n
)g(y
0
)
y
n
y
0

f(x
n
)f(x
0
)
x
n
x
0
:
(i) Supposef is dierentiable atx
0
. Now, sincex
n
!x
0
we have, by continuity
of f at x
0
, y
n
!y
0
. Therefore,
f(x
n
)f(x
0
)
x
n
x
0
!f
0
(x
0
);
g(y
n
)g(y
0
)
y
n
y
0
!g
0
(y
0
):
Thus, (x
n
)!g
0
(y
0
)f
0
(x
0
) showing that gf is dierentiable at x
0
and
(gf)
0
(x
0
) =g
0
(y
0
)f
0
(x
0
):
(ii) Suppose g f is is dierentiable at x
0
and g
0
(y
0
) 6= 0. Then we have
(x
n
)! (gf)
0
(x
0
) and
f(x
n
)f(x
0
)
x
n
x
0
=
(x
n
)
g(yn)g(y
0
)
yny
0
!
(gf)
0
(x
0
)
g
0
(y
0
)
:
Hence, f is dierentiable at x
0
and f
0
(x
0
) =
(gf)
0
(x
0
)
g
0
(y
0
)
.
(iii) Proof of this part is analogous to the proof of (ii). Hence, we omit the
details.
Remark 2.6 The part (ii) and (iii) of Theorem 2.30 is not available in standard
books on Calculus. I found it useful while discussing derivative of logarithm function
in next section (Section ??). 
Exercise 2.17 Prove part (iii) of Theorem 2.30. J
We shall assume that the students are familiar with the following:
 For c2R, if f(x) =c; x2R, then f
0
(x) = 0 8x2R.
 If f(x) =x; x2R, then f
0
(x) = 1 8x2R.
 If f(x) = sinx; x2R, then f
0
(x) = cosx 8x2R.
 If f(x) = cosx then f
0
(x) = sinx.
From these, using theorems in the last subsection, we obtain the following:
 For n2N, if f(x) =x
n
; x2R, then f
0
(x) =nx
n1
8x2R.
 If f(x) = cosx = 1 2 sin
2
(x=2); x2R, then f
0
(x) = sinx 8x2R.
 Iff(x) = tanx forx2D :=fx2R : cosx6= 0g, thenf
0
(x) =sec
2
x 8x2D.
Example 2.24 The function e
x
is dierentiable for every x2R and
(e
x
)
0
=e
x
8x2R:
We note that for h6= 0,
e
x+h
e
x
h
e
x
=
e
x
h
(e
h
 1h) =
e
x
h
1
X
n=2
h
n
n!
:
Now, ifjhj 1, thenjhj
n
jhj
2
for all n2f2; 3;:::g. Thus,
jhj 1 =)




e
x+h
e
x
h
e
x




e
x
jhj
1
X
n=2
1
n!
=e
x
jhj(e 2):
From this we obtain that e
x
is dierentiable at x and its derivative is e
x
. 
Example 2.25 For a> 0, the function a
x
is dierentiable for every x2R and
(a
x
)
0
=a
x
lna 8x2R:
By the composition rule in Theorem 2.30,
(a
x
)
0
= (e
xlna
)
0
=e
xlna
lna =a
x
lna:
Page 5


2.3 Dierentiability of functions
Denition 2.14 Suppose f is a (real valued) function dened on an open interval
I and x
0
2I. Then f is said to be dierentiable at x
0
if
lim
x!x
0
f(x)f(x
0
)
xx
0
exists, and in that case the value of the limit is called the derivative of f at x
0
.
The derivative of f at x
0
, if exists, is denoted by
f
0
(x
0
) or
df
dx
(x
0
):

Exercise 2.13 Show that f : I ! R is dierentiable at x
0
2 I if and only if
lim
h!0
f(x
0
+h)f(x
0
)
h
exists. J
Exercise 2.14 Suppose f is dened on an open interval I and x
0
2I. Show that
f is dierentiable at x
0
2 I if and only if there exists a continuous function (x)
such that
f(x) =f(x
0
) + (x)(xx
0
);
and in that case (x
0
) =f
0
(x
0
). J
Exercise 2.15 Let  be as in Exercise 2.14. Then f is dierentiable at x
0
, if and
only if for every sequence (x
n
) inInfx
0
g which converges tox
0
, the sequence (x
n
)
converges, and in that case f
0
(x
0
) = lim
n!1
(x
n
). J
Exercise 2.16 Suppose f and g dened on I are dierentiable at x
0
2 I and
2R. Show that the functions '(x) :=f(x) +g(x) and (x) :=f(x), x2I are
dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
) +g
0
(x
0
); 
0
(x
0
) ='
0
(x
0
):
J
2.3.1 Some properties of dierentiable functions
Theorem 2.28 (Dierentiability implies continuity) Suppose f dened on I
is dierentiable at x
0
2I. Then f is continuous at x
0
.
Proof. Note that
f(x
0
+h)f(x
0
) =
f(x
0
+h)f(x
0
)
h
h!f
0
(x
0
):0 = 0 as h! 0:
Thus, f is continuous at x
0
.
For the following theorem, we may recall that if' is continuous at a pointx
0
and
'(x
0
)6= 0, then there exists an open interval I
0
containing x
0
such that '(x)6= 0
for all x2I
0
.
Theorem 2.29 (Products and quotient rules) Suppose f and g dened on I
are dierentiable at x
0
2I. Then the function '(x) :=f(x)g(x) is dierentiable at
x
0
, and
'
0
(x
0
) =f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
): ()
If g(x
0
)6= 0, then the function (x) :=f(x)=g(x) is dierentiable at x
0
, and
 
0
(x
0
) =
g(x
0
)f
0
(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
: ()
Proof. Note that
'(x
0
+h)'(x
0
) = f(x
0
+h)g(x
0
+h)f(x
0
)g(x
0
)
= [f(x
0
+h)f(x
0
)]g(x
0
+h) +f(x
0
)[g(x
0
+h)g(x
0
)]
so that
'(x
0
+h)'(x
0
)
h
=
f(x
0
+h)f(x
0
)
h
g(x
0
+h) +f(x
0
)
g(x
0
+h)g(x
0
)
h
! f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
) as h! 0:
Hence, ' is dierentiable at x
0
, and
'
0
(x
0
) =f
0
(x
0
)g(x
0
) +f(x
0
)g
0
(x
0
):
Also, since
 (x
0
+h) (x
0
) =
f(x
0
+h)g(x
0
)f(x
0
)g(x
0
+h)
g(x
0
+h)g(x
0
)
=
[f(x
0
+h)f(x
0
)]g(x
0
)f(x
0
)[g(x
0
+h)g(x
0
)]
g(x
0
+h)g(x
0
)
;
we have
 (x
0
+h) (x
0
)
h
=
f(x
0
+h)f(x
0
)
h
g(x
0
)f(x
0
)
g(x
0
+h)g(x
0
)
h
g(x
0
+h)g(x
0
)
!
f
0
(x
0
)g(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
as h! 0:
Thus, is dierentiable at x
0
, and 
0
(x
0
) =
g(x
0
)f
0
(x
0
)f(x
0
)g
0
(x
0
)
[g(x
0
)]
2
.
Theorem 2.30 (Composition rule) Suppose f is dened on an open interval I
and x
0
2I, and g is dened in an open interval containing y
0
:=f(x
0
). Let
' =gf:
Then we have the following.
(i) Suppose f is dierentiable at x
0
and g is dierentiable at y
0
. Then ' is
dierentiable at x
0
and
'
0
(x
0
) =g
0
(y
0
)f
0
(x
0
):
(ii) Suppose' is dierentiable at x
0
,g is dierentiable at y
0
andg
0
(y
0
)6= 0. Then
f is dierentiable at x
0
and
f
0
(x
0
) =
'
0
(x
0
)
g
0
(y
0
)
:
(iii) Suppose' is dierentiable atx
0
,f is dierentiable atx
0
andf
0
(x
0
)6= 0. Then
g is dierentiable at y
0
and
g
0
(y
0
) =
'
0
(x
0
)
f
0
(x
0
)
:
Proof. For x6=x
0
, let
(x) :=
(gf)(x) (gf)(x
0
)
xx
0
:
Let (x
n
) be a sequence inInfx
0
g which converges tox
0
. Then, takingy
n
:=f(x
n
),
n2N, and y
0
=f(x
0
), we have
(x
n
) =
(gf)(x
n
) (gf)(x
0
)
x
n
x
0
=
g(y
n
)g(y
0
)
x
n
x
0
=
g(y
n
)g(y
0
)
y
n
y
0

f(x
n
)f(x
0
)
x
n
x
0
:
(i) Supposef is dierentiable atx
0
. Now, sincex
n
!x
0
we have, by continuity
of f at x
0
, y
n
!y
0
. Therefore,
f(x
n
)f(x
0
)
x
n
x
0
!f
0
(x
0
);
g(y
n
)g(y
0
)
y
n
y
0
!g
0
(y
0
):
Thus, (x
n
)!g
0
(y
0
)f
0
(x
0
) showing that gf is dierentiable at x
0
and
(gf)
0
(x
0
) =g
0
(y
0
)f
0
(x
0
):
(ii) Suppose g f is is dierentiable at x
0
and g
0
(y
0
) 6= 0. Then we have
(x
n
)! (gf)
0
(x
0
) and
f(x
n
)f(x
0
)
x
n
x
0
=
(x
n
)
g(yn)g(y
0
)
yny
0
!
(gf)
0
(x
0
)
g
0
(y
0
)
:
Hence, f is dierentiable at x
0
and f
0
(x
0
) =
(gf)
0
(x
0
)
g
0
(y
0
)
.
(iii) Proof of this part is analogous to the proof of (ii). Hence, we omit the
details.
Remark 2.6 The part (ii) and (iii) of Theorem 2.30 is not available in standard
books on Calculus. I found it useful while discussing derivative of logarithm function
in next section (Section ??). 
Exercise 2.17 Prove part (iii) of Theorem 2.30. J
We shall assume that the students are familiar with the following:
 For c2R, if f(x) =c; x2R, then f
0
(x) = 0 8x2R.
 If f(x) =x; x2R, then f
0
(x) = 1 8x2R.
 If f(x) = sinx; x2R, then f
0
(x) = cosx 8x2R.
 If f(x) = cosx then f
0
(x) = sinx.
From these, using theorems in the last subsection, we obtain the following:
 For n2N, if f(x) =x
n
; x2R, then f
0
(x) =nx
n1
8x2R.
 If f(x) = cosx = 1 2 sin
2
(x=2); x2R, then f
0
(x) = sinx 8x2R.
 Iff(x) = tanx forx2D :=fx2R : cosx6= 0g, thenf
0
(x) =sec
2
x 8x2D.
Example 2.24 The function e
x
is dierentiable for every x2R and
(e
x
)
0
=e
x
8x2R:
We note that for h6= 0,
e
x+h
e
x
h
e
x
=
e
x
h
(e
h
 1h) =
e
x
h
1
X
n=2
h
n
n!
:
Now, ifjhj 1, thenjhj
n
jhj
2
for all n2f2; 3;:::g. Thus,
jhj 1 =)




e
x+h
e
x
h
e
x




e
x
jhj
1
X
n=2
1
n!
=e
x
jhj(e 2):
From this we obtain that e
x
is dierentiable at x and its derivative is e
x
. 
Example 2.25 For a> 0, the function a
x
is dierentiable for every x2R and
(a
x
)
0
=a
x
lna 8x2R:
By the composition rule in Theorem 2.30,
(a
x
)
0
= (e
xlna
)
0
=e
xlna
lna =a
x
lna:
Example 2.26 The function lnx is dierentiable for every x> 0, and
(lnx)
0
=
1
x
; x> 0:
To see this, let f(x) = lnx and g(x) = e
x
. Then we have g(f(x)) = x for every
x > 0. Since gf is dierentiable, g is dierentiable, and g
0
(y) = e
y
6= 0 for
every y 2 R, by Theorem 2.30, f is dierentiable for every x > 0 and we have
g
0
(f(x))f
0
(x) = 1. Thus,
1 =e
lnx
(lnx)
0
=x(lnx)
0
so that (lnx)
0
= 1=x. 
Example 2.27 For a> 0, the function log
a
x is dierentiable for every x> 0, and
(log
a
x)
0
=
1
x lna
; x> 0:
We know that
log
a
x =
lnx
lna
:
Hence, (log
a
x)
0
=
1
x lna
for every x> 0. 
Example 2.28 For r2 R, let f(x) = x
r
for x > 0. Then f is dierentiable for
every x> 0 and
f
0
(x) =rx
r1
; x> 0:
By the composition rule in Theorem 2.30,
f
0
(x) = (e
rlnx
)
0
=e
rlnx
r
x
=
x
r
r
x
=rx
r1
:

Exercise 2.18 Prove the following.
(i) The function lnjxj is dierentiable for every x2R with x6= 0, and
(lnjxj)
0
=
1
x
; x6= 0:
(ii) For a> 0, the function log
a
jxj is dierentiable for every x2R with x6= 0,
and
(log
a
jxj)
0
=
1
x lna
; x6= 0: