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Chapter

5

CONTINUITY AND
DIFFERENTIABILITY
The whole of science is nothing more than a refinement
of everyday thinking
...
1 Introduction
This chapter is essentially a continuation of our study of
differentiation of functions in Class XI
...
In this chapter, we introduce the
very important concepts of continuity, differentiability and
relations between them
...
Further, we introduce a
new class of functions called exponential and logarithmic
functions
...
We illustrate certain geometrically obvious
conditions through differential calculus
...


5
...
Consider
the function
1, if x ≤ 0
f ( x) = 
 2, if x > 0
This function is of course defined at every
point of the real line
...
1
...
At the points near and to the
left of 0, i
...
, at points like – 0
...
01, – 0
...
At the points near
and to the right of 0, i
...
, at points like 0
...
01,

Fig 5
...
001, the value of the function is 2
...
In
particular the left and right hand limits do not coincide
...
Note that when we try to draw
the graph, we cannot draw it in one stroke, i
...
, without lifting pen from the plane of the
paper, we can not draw the graph of this function
...
This is one instance of function being not continuous at x = 0
...
Left and the right hand limits at x = 0
are both equal to 1
...
Again, we note that we
cannot draw the graph of the function without
lifting the pen
...

Naively, we may say that a function is
continuous at a fixed point if we can draw the
graph of the function around that point without
Fig 5
...

Mathematically, it may be phrased precisely as follows:
Definition 1 Suppose f is a real function on a subset of the real numbers and let c be
a point in the domain of f
...
Recall that
if the right hand and left hand limits at x = c coincide, then we say that the common
value is the limit of the function at x = c
...
If f is not continuous at c, we say f is discontinuous at c and c is called a point
of discontinuity of f
...

Solution First note that the function is defined at the given point x = 1 and its value is 5
...
Clearly
lim f ( x) = lim (2 x + 3) = 2 (1) + 3 = 5
x→1

Thus

x→1

lim f ( x ) = 5 = f (1)
x→ 1

Hence, f is continuous at x = 1
...

Solution First note that the function is defined at the given point x = 0 and its value is 0
...
Clearly
lim f ( x) = lim x2 = 0 2 = 0
x →0

Thus

x→ 0

lim f ( x ) = 0 = f (0)
x →0

Hence, f is continuous at x = 0
...

Solution By definition
 − x, if x < 0
f (x) = 
 x, if x ≥ 0
Clearly the function is defined at 0 and f(0) = 0
...
Hence, f is continuous at x = 0
...


150

MATHEMATICS

Solution The function is defined at x = 0 and its value at x = 0 is 1
...
Hence,
lim f ( x ) = lim ( x3 + 3) = 0 3 + 3 = 3
x →0
x →0

Since the limit of f at x = 0 does not coincide with f (0), the function is not continuous
at x = 0
...

Example 5 Check the points where the constant function f (x) = k is continuous
...
Let c be any real number
...

Example 6 Prove that the identity function on real numbers given by f (x) = x is
continuous at every real number
...
Also,
lim f ( x ) = lim x = c
x →c

x →c

Thus, lim f (x) = c = f(c) and hence the function is continuous at every real number
...

Definition 2 A real function f is said to be continuous if it is continuous at every point
in the domain of f
...
Suppose f is a function defined on a
closed interval [a, b], then for f to be continuous, it needs to be continuous at every
point in [a, b] including the end points a and b
...
As a consequence
x→a

x →b

of this definition, if f is defined only at one point, it is continuous there, i
...
, if the
domain of f is a singleton, f is a continuous function
...

Let c be a real number such that c < 0
...
Also
lim f ( x) = lim ( − x) = – c
x →c
x →c

(Why?)

Since lim f ( x ) = f ( c ) , f is continuous at all negative real numbers
...
Then f(c) = c
...
Hence, f
x →c

is continuous at all points
...

Solution Clearly f is defined at every real number c and its value at c is c3 + c2 – 1
...
This means
x →c

f is a continuous function
...

x

Solution Fix any non zero real number c, we have
lim f ( x) = lim
x →c

x→c

1 1
=
x c

1
Also, since for c ≠ 0, f (c ) = , we have lim f ( x) = f ( c) and hence, f is continuous
x →c
c
at every point in the domain of f
...


152

MATHEMATICS

We take this opportunity to explain the concept of infinity
...
To carry out this analysis we follow the usual trick of
x
finding the value of the function at real numbers close to 0
...
We tabulate this in the following (Table 5
...

Table 5
...
3

1 3
...


0
...
1 = 10–1

0
...
001 = 10–3 10– n

5

10

100 = 102

1000 = 103

10n

We observe that as x gets closer to 0 from the right, the value of f (x) shoots up
higher
...
In symbols, we write
lim f ( x) = + ∞

x →0 +

(to be read as: the right hand limit of f (x) at 0 is plus infinity)
...

Similarly, the left hand limit of f at 0 may be found
...

Table 5
...
3

– 1 – 3
...


– 0
...
2, we deduce that the
value of f (x) may be made smaller than any
given number by choosing a negative real
number very close to 0
...
Again, we wish to emphasise
that – ∞ is NOT a real number and hence the
left hand limit of f at 0 does not exist (as a real
number)
...
3 is a geometric representation
of the above mentioned facts
...
3

CONTINUITY AND DIFFERENTIABILITY

153

Example 10 Discuss the continuity of the function f defined by
 x + 2, if x ≤ 1
f (x) = 
 x − 2, if x > 1

Solution The function f is defined at all points of the real line
...
Therefore, lim f ( x) = lim ( x + 2) = c + 2
x→c

x→ c

Thus, f is continuous at all real numbers less than 1
...
Therefore,
lim f ( x) = lim (x – 2) = c – 2 = f (c)
x →c

x →c

Thus, f is continuous at all points x > 1
...
4
do not coincide, f is not continuous at x = 1
...
The graph of the function is given in Fig 5
...

Example 11 Find all the points of discontinuity of the function f defined by
 x + 2, if x < 1

f (x) =  0, if x = 1
 x − 2, if x > 1

Solution As in the previous example we find that f
is continuous at all real numbers x ≠ 1
...
Hence
x = 1 is the only point of discontinuity of f
...
5
...
5

154

MATHEMATICS

Example 12 Discuss the continuity of the function defined by
 x + 2, if x < 0
f (x) = 
 − x + 2, if x > 0

Solution Observe that the function is defined at all real numbers except at 0
...

Case 2 If c ∈ D2, then lim f ( x) = lim (– x + 2)
x →c

x →c

= – c + 2 = f (c) and hence f is continuous in D2
...
Graph of this
function is given in the Fig 5
...
Note that to graph
Fig 5
...

Example 13 Discuss the continuity of the function f given by
 x, if x ≥ 0

f (x) =  2
 x , if x < 0

Solution Clearly the function is defined at
every real number
...
7
...

Let

D1 = {x ∈ R : x < 0}, D2 = {0} and
D3 = {x ∈ R : x > 0}

Fig 5
...

Case 2 At any point in D3, we have f (x) = x and it is easy to see that it is continuous
there (see Example 6)
...
The value of the function at 0 is f(0) = 0
...
This means that f is
x →0

continuous at every point in its domain and hence, f is a continuous function
...

Solution Recall that a function p is a polynomial function if it is defined by
p(x) = a 0 + a 1 x +
...
Clearly this
function is defined for every real number
...
Since c is any real number, p is continuous at
every real number and hence p is a continuous function
...

Solution First observe that f is defined for all real numbers
...
8
...
Below we explore, if this is true
...
8

156

MATHEMATICS

Case 1 Let c be a real number which is not equal to any integer
...
e
...
Also f (c) = [c] and hence the function is continuous at all real
x →c

x →c

numbers not equal to integers
...
Then we can find a sufficiently small real number
r > 0 such that [c – r] = c – 1 whereas [c + r] = c
...

5
...
1 Algebra of continuous functions
In the previous class, after having understood the concept of limits, we learnt some
algebra of limits
...

Since continuity of a function at a point is entirely dictated by the limit of the function at
that point, it is reasonable to expect results analogous to the case of limits
...

Then
(1) f + g is continuous at x = c
...

(3) f
...

f 
(4)   is continuous at x = c, (provided g (c) ≠ 0)
...
Clearly it is defined at
x = c
...


x→c

(by definition of f + g)
(by the theorem on limits)
(as f and g are continuous)
(by definition of f + g)

Proofs for the remaining parts are similar and left as an exercise to the reader
...
e
...
g) defined by (λ
...
g(x) is also
continuous
...

(ii) As a special case of (4) above, if f is the constant function f (x) = λ, then the

λ
λ
λ
defined by (x) =
is also continuous wherever g (x) ≠ 0
...

g
The above theorem can be exploited to generate many continuous functions
...
The following examples
illustrate this:
function

Example 16 Prove that every rational function is continuous
...
The domain of f is all real numbers except
points at which q is zero
...

f (x) =

Example 17 Discuss the continuity of sine function
...

Now, observe that f (x) = sin x is defined for every real number
...
Put x = c + h
...
Therefore
lim f ( x) = lim sin x
x →c
x →c
= lim sin(c + h)
h →0
= lim [sin c cos h + cos c sin h]
h →0
= lim [sin c cos h ] + lim [cos c sin h ]
h →0
h→ 0
= sin c + 0 = sin c = f (c)
Thus lim f (x) = f (c) and hence f is a continuous function
...

Example 18 Prove that the function defined by f (x) = tan x is a continuous function
...
This is defined for all real numbers such
cos x

π

...
Thus tan x being a quotient of two continuous functions is
continuous wherever it is defined
...
Recall that if f and g are two real functions, then
(f o g) (x) = f (g (x))
that cos x ≠ 0, i
...
, x ≠ (2n +1)

is defined whenever the range of g is a subset of domain of f
...

Theorem 2 Suppose f and g are real valued functions such that (f o g) is defined at c
...

The following examples illustrate this theorem
...

Solution Observe that the function is defined for every real number
...
Since both g and h are continuous functions, by Theorem 2,
it can be deduced that f is a continuous function
...

Solution Define g by g (x) = 1 – x + | x | and h by h (x) = | x | for all real x
...
Hence g being a sum
of a polynomial function and the modulus function is continuous
...


CONTINUITY AND DIFFERENTIABILITY

159

EXERCISE 5
...
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5
...
Examine the continuity of the function f(x) = 2x2 – 1 at x = 3
...
Examine the following functions for continuity
...
Prove that the function f (x) = xn is continuous at x = n, where n is a positive
integer
...
Is the function f defined by
(c) f (x) =

 x, if x ≤ 1
f ( x) = 
 5, if x > 1
continuous at x = 0? At x = 1? At x = 2?
Find all points of discontinuity of f, where f is defined by

6
...


 x + 1, if x ≥ 1

f ( x) =  2
 x + 1, if x < 1


| x | +3, if x ≤ − 3

f ( x) =  −2 x, if − 3 < x < 3
 6 x + 2, if x ≥ 3


9
...


7
...


 x3 − 3, if x ≤ 2

f ( x) = 
 x2 + 1, if x > 2


 x10 − 1, if x ≤ 1

f ( x) = 
if x > 1
 x2 ,

13
...


 x + 5, if x ≤ 1
f ( x) = 
 x − 5, if x > 1
a continuous function?

160

MATHEMATICS

Discuss the continuity of the function f, where f is defined by
14
...


 −2, if x ≤ − 1

f ( x) =  2 x, if − 1 < x ≤ 1
 2, if x > 1


15
...
Find the relationship between a and b so that the function f defined by
 ax + 1, if x ≤ 3
f ( x) = 
bx + 3, if x > 3
is continuous at x = 3
...
For what value of λ is the function defined by

19
...

21
...

23
...
Here [x] denotes the greatest integer less than or equal to x
...
cos x
Discuss the continuity of the cosine, cosecant, secant and cotangent functions
...
Determine if f defined by
1
 2
 x sin , if x ≠ 0
f ( x) = 
x
 0,
if x = 0

is a continuous function?

CONTINUITY AND DIFFERENTIABILITY

161

25
...


26
...


 kx 2 , if x ≤ 2

f ( x) = 
if x > 2
3,


at x = 2

28
...
Find the values of a and b such that the function defined by
29
...

32
...

34
...

Show that the function defined by f (x) = cos (x2 ) is a continuous function
...

Examine that sin | x | is a continuous function
...


5
...
Differentiability
Recall the following facts from previous class
...
The derivative of f at c is
defined by

lim
h →0

f (c + h) − f (c)
h

162

MATHEMATICS

provided this limit exists
...
The
dx

function defined by

f (x + h) − f ( x)
h →0
h
wherever the limit exists is defined to be the derivative of f
...
The process of finding
dx
dx
derivative of a function is called differentiation
...

The following rules were established as a part of algebra of derivatives:
(1) (u ± v)′ = u′ ± v′
denoted by f ′ (x) or

(2) (uv)′ = u′v + uv′ (Leibnitz or product rule)
′
(3)  u  = u ′v − uv′ , wherever v ≠ 0 (Quotient rule)
...
3
n

f (x )

x

f ′(x )

nx n – 1

sin x

cos x

tan x

cos x

– sin x

sec2 x

Whenever we defined derivative, we had put a caution provided the limit exists
...

h
In other words, we say that a function f is differentiable at a point c in its domain if both
its answer
...
A function is said
h →0
h →0
h
h
to be differentiable in an interval [a, b] if it is differentiable at every point of [a, b]
...
Similarly, a function is said to be differentiable in an interval
(a, b) if it is differentiable at every point of (a, b)
...

Proof Since f is differentiable at c, we have
lim
x →c

f ( x) − f (c )
= f ′ ( c)
x−c

But for x ≠ c, we have
f (x) – f (c) =
Therefore

or

f ( x) − f (c )

...
( x − c) 
x →c
x →c 
x−c

f ( x) − f ( c) 
lim [ f ( x)] − lim [ f ( c)] = lim 

...
0 = 0
lim f ( x) = f (c)

or

x →c

Hence f is continuous at x = c
...

We remark that the converse of the above statement is not true
...
Consider the left
hand limit
lim–

h →0

f (0 + h ) − f (0) −h
=
= −1
h
h

The right hand limit
lim

h →0 +

f (0 + h ) − f (0) h
= =1
h
h

f (0 + h ) − f (0)
h
does not exist and hence f is not differentiable at 0
...

Since the above left and right hand limits at 0 are not equal, lim
h →0

5
...
1 Derivatives of composite functions
To study derivative of composite functions, we start with an illustrative example
...

d
d
(2 x + 1) 3 
f ( x) =


dx
dx
d
(8 x3 + 12 x 2 + 6 x + 1)
dx
= 24x2 + 24x + 6
= 6 (2x + 1)2
f (x) = (h o g) (x)
=

Now, observe that

where g(x) = 2x + 1 and h(x) = x3
...
Then f(x) = h(t) = t3
...
2 = 3t2
...
We may formalise this observation in the following
theorem called the chain rule
...
e
...
Suppose t = u (x) and if both

dt
dv
and
exist, we have
dx
dt

df dv dt
= ⋅
dx dt dx
We skip the proof of this theorem
...
Suppose
f is a real valued function which is a composite of three functions u, v and w ; i
...
,
f = (w o u) o v
...
Reader is invited to formulate chain
rule for composite of more functions
...

Solution Observe that the given function is a composite of two functions
...
Observe that

165

dv
dt
= cos t and
= 2 x exist
...
Thus
df
= cos t ⋅ 2 x = 2 x cos x2
dx
Alternatively, We can also directly proceed as follows:
y = sin (x2) ⇒
= cos x2

dy d
=
(sin x2)
dx dx

d 2
(x ) = 2x cos x2
dx

Example 22 Find the derivative of tan (2x + 3)
...
Then
(v o u) (x) = v(u(x)) = v(2x + 3) = tan (2x + 3) = f (x)
Thus f is a composite of two functions
...
Then

dv
= sec2 t and
dt

dt
= 2 exist
...

Solution The function f(x) = sin (cos (x2)) is a composition f (x) = (w o v o u) (x) of the
three functions u, v and w, where u(x) = x2, v(t) = cos t and w(s) = sin s
...
Hence by a generalisation of chain rule, we have

t = u(x) = x2 and s = v (t) = cos t
...
(– sin t)
...
cos (cos x2 )
dx ds dt dx

166

MATHEMATICS

Alternatively, we can proceed as follows:
y = sin (cos x2)
Therefore

dy d
d
=
sin (cos x2) = cos (cos x2)
(cos x2)
dx dx
dx
= cos (cos x2) (– sin x2)

d
(x2 )
dx

= – sin x2 cos (cos x2 ) (2x)
= – 2x sin x2 cos (cos x2)

EXERCISE 5
...

1
...
cos (sin x)

3
...
sec (tan ( x ))

sin (ax + b)
5
...
cos x3
...
cos ( x )
2 cot ( x2 )
9
...


f (x) = |x – 1|, x ∈ R
is not differentiable at x = 1
...
Prove that the greatest integer function defined by
f (x) = [x], 0 < x < 3
is not differentiable at x = 1 and x = 2
...
3
...

But it is not necessary that functions are always expressed in this form
...
In
the second case, it does not seem that there is an easy way to solve for y
...
When a
relationship between x and y is expressed in a way that it is easy to solve for y and
write y = f (x), we say that y is given as an explicit function of x
...
In this subsection, we learn to differentiate implicit
functions
...

dx

Solution One way is to solve for y and rewrite the above as
y=x –π
dy
=1
dx
Alternatively, directly differentiating the relationship w
...
t
...
r
...
, x
...

dx

Solution We differentiate the relationship directly with respect to x, i
...
,
dy d
d
+ (sin y) =
(cos x )
dx dx
dx
which implies using chain rule
dy
dy
+ cos y ⋅
= – sin x
dx
dx
This gives
where

dy
sin x
= −
dx
1 + cos y
y ≠ (2n + 1) π

168

MATHEMATICS

5
...
3 Derivatives of inverse trigonometric functions
We remark that inverse trigonometric functions are continuous functions, but we will
not prove this
...

Example 26 Find the derivative of f given by f (x) = sin–1 x assuming it exists
...
Then, x = sin y
...
r
...
x, we get
1 = cos y

which implies that

dy
dx

dy
1
1
=
=
cos y cos(sin−1 x)
dx

π π
Observe that this is defined only for cos y ≠ 0, i
...
, sin–1 x ≠ − , , i
...
, x ≠ – 1, 1,
2 2
i
...
, x ∈ (– 1, 1)
...

Recall that for x ∈ (– 1, 1), sin (sin–1 x) = x and hence
cos2 y = 1 – (sin y) 2 = 1 – (sin (sin–1 x))2 = 1 – x2
 π π
Also, since y ∈  − ,  , cos y is positive and hence cos y = 1 − x 2
 2 2
Thus, for x ∈ (– 1, 1),

dy
1
1
=
=
dx cos y
1 − x2
Example 27 Find the derivative of f given by f (x) = tan–1 x assuming it exists
...
Then, x = tan y
...
r
...
x, we get
1 = sec2 y

dy
dx

which implies that
dy
1
1
1
1
=
=
=
=
2
2
−1
2
dx sec y 1 + tan y 1 + (tan (tan x)) 1 + x 2

CONTINUITY AND DIFFERENTIABILITY

169

Finding of the derivatives of other inverse trigonometric functions is left as exercise
...
4):
Table 5
...
3
dy
in the following:
dx
1
...
xy + y2 = tan x + y

2
...
x2 + xy + y2 = 100

3
...
x3 + x2y + xy2 + y3 = 81

7
...
sin2 x + cos2 y = 1

 2x 
9
...
y = tan–1 
2 
3
3
 1− 3x 
11
...


 1− x 2 
y = sin −1 
 ,0 < x < 1
 1+ x 2 

13
...


1
1
y = sin −1 2 x 1 − x2 , −
< x<
2
2

15
...
4 Exponential and Logarithmic Functions
Till now we have learnt some aspects of different classes of functions like polynomial
functions, rational functions and trigonometric functions
...
It needs to be
emphasized that many statements made
in this section are motivational and precise
proofs of these are well beyond the scope
of this text
...
9 gives a sketch of
y = f1 (x) = x, y = f 2(x) = x2 , y = f3(x) = x3
and y = f4(x) = x4
...

Steeper the curve, faster is the rate of
growth
...
9
increment in the value of x (> 1), the
increment in the value of y = fn (x) increases as n increases for n = 1, 2, 3, 4
...

Essentially, this means that the graph of y = fn (x) leans more towards the y-axis as n
increases
...
If x increases from 1 to
2, f10 increases from 1 to 210 whereas f15 increases from 1 to 215
...

Upshot of the above discussion is that the growth of polynomial functions is dependent
on the degree of the polynomial function – higher the degree, greater is the growth
...
The answer is in affirmative and an example of such a function is
y = f (x) = 10x
...

For example, we can prove that 10x grows faster than f100 (x) = x100
...

Clearly f(x) is much greater than f100 (x)
...
But we will not attempt to give a proof of this here
...


CONTINUITY AND DIFFERENTIABILITY

171

Definition 3 The exponential function with positive base b > 1 is the function
y = f (x) = bx
The graph of y = 10x is given in the Fig 5
...

It is advised that the reader plots this graph for particular values of b like 2, 3 and 4
...

(2) Range of the exponential function is the set of all positive real numbers
...

(4) Exponential function is ever increasing; i
...
, as we move from left to right, the
graph rises above
...
In
other words, in the second quadrant, the graph approaches x-axis (but never
meets it)
...
In
the Appendix A
...
4 of Class XI, it was observed that the sum of the series

1 1
+ +
...
Using this e as the base we obtain an
extremely important exponential function y = ex
...

It would be interesting to know if the inverse of the exponential function exists and
has nice interpretation
...

1+

Definition 4 Let b > 1 be a real number
...

Logarithm of a to base b is denoted by logb a
...
Let us
work with a few explicit examples to get a feel for this
...
In terms of
logarithms, we may rewrite this as log2 8 = 3
...
Also, 625 = 54 = 252 is equivalent to saying log5 625 = 4 or
log25 625 = 2
...
This function, called the
logarithmic function, is defined by
logb : R+ → R
x → logb x = y if by = x

172

MATHEMATICS

As before if the base b = 10, we say it
is common logarithms and if b = e, then
we say it is natural logarithms
...
In this
chapter, log x denotes the logarithm
function to base e, i
...
, ln x will be written
as simply log x
...
10 gives the plots
of logarithm function to base 2, e and 10
...
10

(1) We cannot make a meaningful definition of logarithm of non-positive numbers
and hence the domain of log function is R+
...

(3) The point (1, 0) is always on the graph of the log function
...
e
...

(5) For x very near to zero, the value
of log x can be made lesser than
any given real number
...

(6) Fig 5
...
It is of interest to observe
that the two curves are the mirror
images of each other reflected in the line y = x
...
11

Two properties of ‘log’ functions are proved below:
(1) There is a standard change of base rule to obtain loga p in terms of log b p
...
This means a α = p, b β = p and b γ = a
...
But then
γ
log b p
log b a

(2) Another interesting property of the log function is its effect on products
...
Then bα = pq
...

But then bα = pq = bβ bγ = b β + γ
which implies α = β + γ, i
...
,
logb pq = logb p + logb q
A particularly interesting and important consequence of this is when p = q
...
In fact this is true for any real number n, but we will
not attempt to prove this
...

So the above equation is not true for non-positive real numbers
...
If
y > 0, we may take logarithm which gives us log y = log (elog x) = log x
...
Thus
y = x
...

One of the striking properties of the natural exponential function in differential
calculus is that it doesn’t change during the process of differentiation
...

Theorem 5*
(1) The derivative of ex w
...
t
...
e
...
r
...
, x is

d x
(e ) = ex
...
e
...

dx
x
x

* Please see supplementary material on Page 286
...
r
...
x:
(i) e –x
(ii) sin (log x), x > 0
(iii) cos–1 (ex)

ecos x

(iv)

Solution
(i) Let y = e– x
...
Using chain rule, we have
dy
d
cos (log x)
= cos (log x) ⋅ (log x) =
dx
dx
x
(iii) Let y = cos–1 (ex)
...
Using chain rule, we have
dy
= ecos x ⋅ ( − sin x) = − (sin x ) ecos x
dx

EXERCISE 5
...
r
...
x:
1
...
esin

−1

x

3
...
sin (tan–1 e–x)

5
...
ex + e x +
...


8
...


e

x

, x >0

cos x
, x >0
log x

10
...
5
...

u′(x) + v′(x)
...
This process of differentiation is
known as logarithms differentiation and is illustrated by the following examples:
Example 30 Differentiate

Solution Let y =

( x − 3) ( x2 + 4)
w
...
t
...

3 x2 + 4 x + 5

( x − 3) ( x2 + 4)
(3 x2 + 4 x + 5)

Taking logarithm on both sides, we have
1
[log (x – 3) + log (x2 + 4) – log (3x2 + 4x + 5)]
2
Now, differentiating both sides w
...
t
...
r
...
x, where a is a positive constant
...
Then
log y = x log a
Differentiating both sides w
...
t
...
log a = a x log a
...
r
...
x
...
Taking logarithm on both sides, we have
log y = sin x log x
Therefore

1 dy
d
d

...

dx

Solution Given that yx + xy + xx = ab
...
Taking logarithm on both sides, we have
log u = x log y
Differentiating both sides w
...
t
...
(1)

CONTINUITY AND DIFFERENTIABILITY

177

1 du
d
d
⋅
= x (log y ) + log y ( x)
u dx
dx
dx

1 dy
= x ⋅ + log y ⋅ 1
y dx
So

 x dy

 x dy

du
+ log y  = y x 
+ log y 
= u
dx
 y dx

 y dx



...
r
...
x, we have

1 dv
d
dy
⋅
= y (log x ) + log x
v dx
dx
dx
1
dy
= y ⋅ + log x ⋅
x
dx
So

dv
dy 
y
= v  + log x 
dx
x
dx 

dy 
y y
= x  + log x 
x
dx 
Again
w = xx
Taking logarithm on both sides, we have
log w = x log x
...
r
...
x, we have


...
e
...
(4)

178

MATHEMATICS

From (1), (2), (3), (4), we have
dy 
 x dy

y
yx 
+ log y  + x y  + log x  + xx (1 + log x) = 0
x
dx 
 y dx

or

dy
= – xx (1 + log x) – y
...
yx – 1 + xy
...
x y −1 + x x (1 + log x)]
dy
=
x
...
5
Differentiate the functions given in Exercises 1 to 11 w
...
t
...


(x − 1) ( x − 2)
( x − 3) ( x − 4) (x − 5)

1
...
cos 2x
...


3
...
xx – 2sin x
x



6
...
(log x)x + xlog x

8
...
xsin x + (sin x)cos x

10
...
(x + 3)
...
(x + 5)
2

x x cos x +

x

x2 + 1
x2 − 1

1

11
...

14
...

17
...

dx
xy + yx = 1
13
...
xy = e(x – y)
Find the derivative of the function given by f(x) = (1 + x) (1 + x2) (1 + x4 ) (1 + x8)
and hence find f ′(1)
...

(iii) by logarithmic differentiation
...
If u, v and w are functions of x, then show that
d
du
dv
dw
(u
...
w) =
v
...


...
v
dx
dx
dx
dx
in two ways - first by repeated application of product rule, second by logarithmic
differentiation
...
6 Derivatives of Functions in Parametric Forms
Sometimes the relation between two variables is neither explicit nor implicit, but some
link of a third variable with each of the two variables, separately, establishes a relation
between the first two variables
...
The third variable is called the parameter
...

In order to find derivative of function in such form, we have by chain rule
...

dx

Solution Given that
x = a cos θ, y = a sin θ
Therefore

dx
dy
= – a sin θ,
= a cos θ
dθ
dθ

Hence

dy
dy
d θ a cos θ = − cot θ
= dx =
− a sin θ
dx
dθ

180

MATHEMATICS

dy
, if x = at2, y = 2at
...

dx

Solution We have

dx
dy
= a(1 + cos θ),
= a (sin θ)
dθ
dθ
dy
dy
d θ = a sin θ = tan θ
=
dx a (1 + cos θ)
dx
2
dθ

Therefore

dy
is expressed in terms of parameter only
dx
without directly involving the main variables x and y
...

dx
Solution Let x = a cos3 θ, y = a sin3 θ
...


EXERCISE 5
...
x = 2at2, y = at4
3
...

dx
2
...
x = 4t, y =

4
t

5
...
x = a (θ – sin θ), y = a (1 + cos θ) 7
...
x = a sec θ, y = b tan θ

2
10
...


−1

−1

11
...
7 Second Order Derivative
Let

y = f (x)
...
(1)
dx
If f′(x) is differentiable, we may differentiate (1) again w
...
t
...
Then, the left hand

side becomes

d  dy 
  which is called the second order derivative of y w
...
t
...
The second order derivative of f (x) is denoted by f ″(x)
...
We remark that higher order derivatives may be
defined similarly
...

dx 2
Solution Given that y = x3 + tan x
...
sec x tan x = 6x + 2 sec2 x tan x

Example 39 If y = A sin x + B cos x, then prove that

d2y
+ y =0
...

2
dx
dx

Solution Given that y = 3e2x + 2e3 x
...

2
dx
dx

Solution We have y = sin– 1 x
...
 = 0
dx 
dx 

So

or

(1 − x 2 ) ⋅

d 2 y dy d
+
⋅
dx 2 dx dx

or

(1 − x 2 ) ⋅

d 2 y dy
2x
−
⋅
=0
2
dx
dx 2 1 − x 2

(

)

(1 − x2 ) = 0

d2y
dy
− x =0
2
dx
dx
Alternatively, Given that y = sin–1 x, we have
Hence

(1 − x2 )

y1 =
So
Hence

1
1− x

2

, i
...
, (1 − x 2 ) y2 = 1
1

2
(1 − x2 )
...
7
Find the second order derivatives of the functions given in Exercises 1 to 10
...
x2 + 3x + 2
4
...
x 20
3

5
...
x
...
ex sin 5x

7
...
tan–1 x
9
...
sin (log x)
d2y
+ y =0
11
...
If y = cos–1 x, Find

d2y
in terms of y alone
...
If y = 3 cos (log x) + 4 sin (log x), show that x2 y2 + xy1 + y = 0
14
...
If y = 500e + 600e , show that
dx 2
7x

– 7x

d 2 y  dy 
= 
16
...
If y = (tan–1 x)2, show that (x2 + 1)2 y2 + 2x (x2 + 1) y1 = 2

5
...
We
shall also learn the geometric interpretation of these theorems
...

Then there exists some c in (a, b) such that f ′(c) = 0
...
12 and 5
...


Fig 5
...
13

Observe what happens to the slope of the tangent to the curve at various points
between a and b
...

That is precisely the claim of the Rolle’s theorem as the slope of the tangent at any
point on the graph of y = f (x) is nothing but the derivative of f (x) at that point
...
Then there exists some c in (a, b) such that

f (b ) − f (a )
b− a
Observe that the Mean Value Theorem (MVT) is an extension of Rolle’s theorem
...
The graph of a function
y = f(x) is given in the Fig 5
...
We have already interpreted f′(c) as the slope of the
f ′(c ) =

f (b ) − f (a )
b−a
is the slope of the secant drawn between (a, f (a)) and (b, f(b))
...
In other words, there is a point c in
(a, b) such that the tangent at (c, f(c)) is parallel to the secant between (a, f(a)) and
(b, f (b))
...
From the Fig 5
...
14

Example 42 Verify Rolle’s theorem for the function y = x2 + 2, a = – 2 and b = 2
...

Also f(– 2) = f( 2) = 6 and hence the value of f(x) at – 2 and 2 coincide
...
Since f′ (x) = 2x, we
get c = 0
...

Example 43 Verify Mean Value Theorem for the function f (x) = x2 in the interval [2, 4]
...


186

Now,

MATHEMATICS

f(2) = 4 and f (4) = 16
...
But f′ (x) = 2x which
implies c = 3
...


EXERCISE 5
...
Verify Rolle’s theorem for the function f (x) = x2 + 2x – 8, x ∈ [– 4, 2]
...
Examine if Rolle’s theorem is applicable to any of the following functions
...

4
...

6
...

Verify Mean Value Theorem, if f (x) = x2 – 4x – 3 in the interval [a, b], where
a = 1 and b = 4
...
Find all c ∈ (1, 3) for which f′(c) = 0
...


Miscellaneous Examples
Example 44 Differentiate w
...
t
...
Therefore
3
1

1

−1 d
− −1 d
dy
1
 1
(3 x + 2) +  −  (2 x2 + 4) 2 ⋅ (2 x2 + 4)
= (3 x + 2) 2 ⋅
dx
2
dx
 2
dx

CONTINUITY AND DIFFERENTIABILITY
1

187

3

−
−
1
1
2
= (3 x + 2) 2 ⋅ (3) −   (2 x + 4) 2 ⋅ 4 x
 2
2

=

3
2 3x + 2

2x

−

3

(2x 2 + 4) 2

2
This is defined for all real numbers x > −
...
Therefore
1 
dy

sec2 x d
⋅ (sec2 x ) + 3  −
= e
dx
dx
 1 − x2 

d
1 


sec2 x 
⋅  2sec x (sec x ) + 3  −
= e



dx

1 − x2 
1 

sec2 x
+3 −
= 2sec x (sec x tan x) e
2 
 1− x 
1 

2
sec2 x
+ 3 −
= 2sec x tan x e
2 
 1− x 
Observe that the derivative of the given function is valid only in [ −1, 1] − {0} as
the derivative of cos– 1 x exists only in (– 1, 1) and the function itself is not
defined at 0
...

log 7
The function is defined for all real numbers x > 1
...
r
...
x
...
Observe that this function is defined for all real numbers
...

sin x 
(ii) Let f (x) = tan – 1 


...
e
...
We may rewrite this
function as
=

− 1  sin x 
f(x) = tan 

 1 + cos x 


 x
 x 

 2 sin  2  cos  2  
 


= tan 
x


2 cos 2



2

−1

 x  x
−1 
= tan  tan   =
  2  2
 x
Observe that we could cancel cos   in both numerator and denominator as it
2

is not equal to zero
...

2

 2x + 1 
(iii) Let f(x) = sin – 1 

...
Since the quantity in the middle is always positive,
1+ 4x

CONTINUITY AND DIFFERENTIABILITY

we need to find all x such that

2 x +1
1+ 4x

189

≤ 1 , i
...
, all x such that 2x + 1 ≤ 1 + 4x
...
Hence the function
2x
is defined at every real number
...

Solution The function y = (sin x) sin x is defined for all positive real numbers
...

= cos x log (sin x) + cos x
= (1 + log (sin x)) cos x

1
d
⋅ (sin x )
sin x dx

190

MATHEMATICS

Thus

dy
= y ((1 + log (sin x)) cos x) = (1 + log (sin x)) ( sin x)sin x cos x
dx
dy
, where
dx

Example 47 For a positive constant a find
1
t ,

a

 1
y =a
and x =  t + 
 t
Solution Observe that both y and x are defined for all real t ≠ 0
...
Thus for t ≠ ± 1,
dt
1

t+ 
1
dy
a t  1 − 2  log a
 t 
dy dt
=
=
a −1
dx dx
1
 1

a t +  ⋅  1 − 2 
dt
 t 
 t

=

a

t+

1
t

log a

 1
a t + 
 t
2 x w
...
t
...

Example 48 Differentiate sin

a −1

Solution Let u (x) = sin2 x and v (x) = e cos x
...
Clearly
dv dv / dx
du
dv
= 2 sin x cos x and
= e cos x (– sin x) = – (sin x) e cos x
dx
dx

CONTINUITY AND DIFFERENTIABILITY

2sin x cos x
2 cos x
du
= − cos x
=
cos x
dv
− sin x e
e

Thus

Miscellaneous Exercise on Chapter 5
Differentiate w
...
t
...

1
...
sin3 x + cos 6 x
3
...
sin–1(x

x ), 0 ≤ x ≤ 1

x
2 ,–22x + 7

cos −1
5
...
(log x)
8
...

6
...
xx + xa + a x + a a, for some fixed a > 0 and x > 0
9
...


xx

2

−3

(sin x – cos x)

,

2

x

+ ( x − 3) , for x > 3

dy
π
π
, if y = 12 (1 – cos t), x = 10 (t – sin t), − < t <
2
2
dx
dy
13
...
If x 1 + y + y 1 + x = 0 , for , – 1 < x < 1, prove that
12
...
If (x – a) 2 + (y – b) 2 = c2 , for some c > 0, prove that
3

  dy  2  2
1 +   
  dx  
d2y
dx2
is a constant independent of a and b
...
If cos y = x cos (a + y), with cos a ≠ ± 1, prove that

dy cos 2 ( a + y)
=

...
If x = a (cos t + t sin t) and y = a (sin t – t cos t), find

...
If f (x) = | x |3, show that f ″(x) exists for all real x and find it
...
Using mathematical induction prove that

d ( n)
x = nxn −1 for all positive
dx

integers n
...
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation,
obtain the sum formula for cosines
...
Does there exist a function which is continuous everywhere but not differentiable
at exactly two points? Justify your answer
...
If y = l
a

g ( x) h ( x)
f ′( x) g ′(x ) h′(x)
dy
m
n , prove that
= l
m
n
dx
b
c
a
b
c

2
23
...

dx 2
dx

Summary
A real valued function is continuous at a point in its domain if the limit of the
function at that point equals the value of the function at that point
...

Sum, difference, product and quotient of continuous functions are continuous
...
e
...

(f
...
g(x) is continuous
...



Every differentiable function is continuous, but the converse is not true
...
If f = v o u, t = u (x)
and if both

dt
dv
and
exist then
dx
dt

df dv dt
=
⋅
dx dt dx
Following are some of the standard derivatives (in appropriate domains):
d ( −1 )
1
sin x =
dx
1 − x2

d ( −1 )
1
cos x = −
dx
1 − x2

d ( −1 )
1
tan x =
dx
1 + x2

d ( −1 )
−1
cot x =
dx
1 + x2

d ( −1 )
1
sec x =
dx
x 1 − x2

d (
−1
cosec−1 x ) =
dx
x 1 − x2

d ( x) x
d
1
e =e
( log x ) =
dx
dx
x
Logarithmic differentiation is a powerful technique to differentiate functions
of the form f (x) = [u (x)]v (x)
...

Rolle’s Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable
on (a, b) such that f (a) = f (b), then there exists some c in (a, b) such that
f ′(c) = 0
...
Then there exists some c in (a, b) such that
f ′(c ) =

f (b ) − f (a )
b− a

—

—



---

Title: CONTINUITY AND DIFFERENTIABILITY
Description: NCERT

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