Unit– 2
Calculus
2. Rolle’s theorem
Important definitions-
Continuity- suppose that a function f(x) is defined in the interval I , then it is said to be continuous at x=a , if
Differentiability- A function f(x) is said to be differentiable at x=a if 
exists where ‘a’ belongs to I
Rolle’s theorem-
Suppose f(x) is a function defined on [a , b] and it satisfies the following conditions
1. f(x) is continuous in [a , b]
2. f(x) is differentiable in (a , b)
3. f(a) = f(b)
Then there exists atleast a point point c ϵ (a , b) , where a<b , such that f’(c) = 0
Proof: suppose y = f(x) is a function and A(a , f(a)) , B(b , f(b)) be two points on the curve f(x) and a,b are two end points. Now conditions for Rolle’s theorem-
1.f(x) is a continuous function in[a , b] , from the figure without breaks in between A&B on y = f(x).
2. f(x) is differentiable in (a , b), because joining A and B we get a line AB.
Slope of the line AB=0 then a point C at P also a tangent at P, or Q,R,S is parallel to x –axis.
Slope of the tangent at P or Q, R, S, will be 0, even the curve y = f(x) decreases or increases, that means f(x) is constant.
Derivative of f(x),
f’(c) = 0
That’s why, f’(c) = 0
3. the slope of the line AB is equal to zero, that means the line AB is parallel to x-axis.
So that, f(a) = f(b)
Example: Verify Rolle’s theorem for the function f(x) = x(x+3) 
in interval [-3, 0].
Sol. First we will differentiate the given function with respect to x, we get
f’(x) = (x²+3x) 
+ (2x + 3)
= 
This shows that f’(x) exists for all x, therefore f(x) is continuous for all x.
Now, f(-3) = 0 and f(0) = 0 , so that f(-3) = f(0).
Here f(x) satisfies all the conditions of Rolle’s theorem,
Then,
f’(x) = 0 , which gives
= 0
We get,
X = 3 and x = -2
Here we can see that clearly -3<-2<0 , therefore there exists -2 ∈ (-3,0) such that
f’(-2) = 0
that means the Rolle’s theorem is true for the given function.
Example: Verify Rolle’s theorem for the given functions below-
1. f(x) = x³ - 6x²+11x-6 in the interval [1,3]
2. f(x) = x²-4x+8 in the interval [1,3]
Sol. (1)
As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]
Also, f(1) = f(3) = 0
Now we find f’(x) = 0
3x² - 12x +11 = 0
We get, x = 2+ 
and 2 - 
Hence both of them lie in (1,3).
Hence the theorem holds good for the given function in interval [1,3]
(2) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]
Also, f(1) = 1 -4 +8 = 5 and f(3) = 9 – 12 + 8 = 5
Hence f(1) = f(3)
Now the first derivative of the function,
f’(x) = 0
2x – 4 = 0 , gives
X = 2
We can see that 1<2<3, hence there exists 2 between 1 and 3. And f’(2) = 0.
This means that the Rolle’s theorem holds good for the given function and given interval.
Example: Verify the Rolle’s theorem for sin x in the interval [
]
Sol. Suppose f(x) = sin x
We know that sin x is continuous for all x.
Now , f’(x) = cos x exists for all x in (
) and
f(
0
f(
0
thus f(x) satisfies all the conditions of Rolle’s theorem.
Now,
f’(x) = 0 that gives , cos x = 0
x = 
here we notice that both intervals lie in (
.
there exists, c = 
so that, f’(c) = 0
the Rolle’s theorem has been verified.
Lagranges’s mean value theorem-
Suppose that f(x) be a function of x such that,
1. if it is continuous in [a , b]
2. if it is differentiable in (a , b)
Then there atleast exists a value cϵ (a , b)
f’(c) = 
proof:
lets define a function g(x)
g(x) = f(x) – Ax ………………..(1)
here A is a constant which is to be determined,
so that, g(a) = g(b)
now,
g(a) = f(a) – Aa
g(b) = f(b) – Ab
so,
g(a) = g(b),
f(a) – Aa = f(b) – Ab ,
which gives
A = 
…………………..(2)
As right hand side of eq.(1) is continuous in [a,b] , so that g(x) is continuous.
And right hand side of eq.(1) is differential in (a,b) , so that g(x) is differentiable in (a,b).
And g(a) = g(b) , because of the choice of A.
Hence g(x) satisfies all the conditions of Rolle’s theorem.
So that,
There exists a value c such that a<c<b at which g’(c) = 0
Now, differentiate eq. (1) with respect to x, we get
g’(x) = f’(x) – A
here we know that, x = c,
g’(c) = f’(c) – A
as g’(c) = 0, then
f’(c) – A =0
so that,, f’(c) = A,
from equation (2) , we get
f’(c) = 
hence proved.
Example: Verify Lagrange’s mean value theorem for f(x) = (x-1)(x-2)(x-3) in [0,4].
Sol. As we see that the given function is a polynomial and we know that the polynomial is continuous in [0,4] and differentiable in (0,4).
f(x) = (x-1)(x-2)(x-3)
f(x) = x-6x²+11x-6
now at x = 0, we get
f(0) = -6 and
at x = 4, we get.
f(4) = 6
diff. the function w.r.t.x , we get
f’(x) = 3x²-6x+11
suppose x = c, we get
f’(c) = 3c²-6c+11
by Lagrange’s mean value theorem,
f’(c) = 
= 
= 
= 3
now we get,
3c²-6c+11 = 3
3c²-6c+8 = 0
On solving the quadratic equation, we get
C = 2 
Here we see that the value of c lies between 0 and 4
Therefore the given function is verified.
Example: Verify Lagrange’s mean value theorem for f(x) = log xin [1,e].
Sol. We already know that the function which is log x is continuous for all x>0.
So that this is the continuous function In [1,e]
Now,
f’(x) = 1/x
which is exists for all x in (1,e)
so that f(x) is differentiable in (1,e).
by Lagrange’s mean value theorem, we get
f’(c) = 
, let x = c,
then ,
f’(c) = 
we get,
c = e-1
e-1 will always lies between 1 and e .
hence the function is verified by Lagrange’s mean value theorem.
Cauchy’s mean value theorem-
Suppose we have two functions f(x) and g(x) of x, such that,
1. Both functions are continuous in [a,b]
2. Both functions are differentiable in (a,b)
3. g’(x) ≠ 0 for any x ϵ (a,b)
These three exists atleast , x = c ϵ (a,b) , at which
Proof: suppose ,we define a functions,
h(x) = f(x) – A.g(x) …………………….(1)
so that h(a) = h(b) and A is a constant to be determined.
Now,
h(a) = f(a) – Ag(a)
h(b) = f(b) – A.g(b)
so that,
f(a) – Ag(a) = f(b) – A.g(b) , which gives
A = 
…………………………….(2)
Now , h(x) is continuous in [a,b] as RHS of eq. (1) is continuous in [a,b] and h(x) is diff. in (a,b) as RHS of eq. (1) is diff. in (a,b)
Also,
h(a) = h(b)
Therefore all the conditions of Rolle’s theorem are satistfied then there exists a Value x = cϵ (a,b)
so that h’(c) = 0
Differentiate eq.(1) w.r.t. x , we get
h’(x) = f’(x) – A.g’(x)
At x = c
h’(c) = f’(c) – A.g’(c)
0 = f’(c) – A.g’(c)
A = 
So that , we get
where a<c<b
Hence the Cauchy’s mean value theorem is proved.
Example: Verify Cauchy’s mean value theorem for the function f(x) = sin x and g(x) = cos x in [ 0 , π/2]
Sol. It is given tha,
f(x) = sin x and g(x) = cos x
Now,
f’(x) = cos x and g’(x) = - sin x
We know that both the functions are continuous in [ 0 , π/2] and differentiable in ( 0 , π/2 )
Also, g’(x) = -sin x ≠ 0 for all x ϵ( 0 , π/2 )
By Cauchy’s mean value theorem, we get
for some c: 0< c < 
That means
which gives,
Cot c = 1
C = 
Now we see that 
lies between 0 and 
Example: Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]
Sol. We are given, f(x) = x⁴ and g(x) = x
Derivative of these fucntions ,
f’(x) = 4x³ and g’(x) = 2x
put these values in Cauchy’s formula, we get
2c² = 
c² = 
c = 
now put the values of a = 1 and b = 2 ,we get
c = 
=
= 
(approx)
Hence the Cauchy’s theorem is verified.
Taylor’s Theorem-
If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-
+ …….+ 
+ ……..
Which is called Taylor’s theorem.
If we put x = a, we get-
+ …….+ 
+ …….. (1)
Maclaurin’s Theorem-
If we put a = 0 and h = x then equation(1) becomes-
+ …….
Which is called Maclaurin’s theorem.
Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)
We get-
+ …….
Example: Express the polynomial 
in powers of (x-2).
Sol. Here we have,
f(x) = 
differentiating the function w.r.t.x-
f’(x) = 
f’’(x) = 12x + 14
f’’’(x) = 12
f’’’’(x)=0
now using Taylor’s theorem-
+ ……. (1)
Here we have, a = 2,
Put x = 2 in the derivatives of f(x), we get-
f(2) = 
f’(2) = 
f’’(2) = 12(2)+14 = 38
f’’’(2) = 12 and f’’’’(2) = 0
now put a = 2 and substitute the above values in equation(1), we get-
Example: Expand sin x in powers of 
Sol. Let f(x) = sin x
Then,
= 
By using Taylor’s theorem-
+ ……. (1)
Here f(x) = sin x and a = π/2
f’(x) = cos x , f’’(x) = - sin x , f’’’(x) = - cos x and so on.
Putting x = π/2 , we get
f(x) = sin x = 
= 1
f’(x) = cos x = 
= 0
f’’(x) = -sin x = 
= -1
f’’’(x) = -cos x = 
= 0
from equation (1) put a = 
and substitute these values, we get-
+ …….
= 
………………………..
2.4. Indeterminate forms & L’Hospital’s rule
Let we have two functions f(x) and g(x) and-
Then-
is an expression of the form 
, in that case we can say that f(x)/g(x) is an indeterminate for of the type 
at x = a.
Now, Let we have two functions f(x) and g(x) and-
Then-
is an expression of the form 
, in that case we can say that f(x)/g(x) is an indeterminate for of the type 
at x = a.
Some other indeterminate forms are 
L’Hospital’s rule for 
form-
Working steps-
1. Check that the limits f(x)/g(x) is an indeterminate form of type 
.
(note- we can not apply L’Hospital rule if it is not in indeterminate form)
2. Differentiate f and g separately.
3. Find the limits of the derivatives .if the limit is finite , 
then it is equal to the limit of f(x)/g(x).
Example-1: Evaluate 
Sol. Here we notice that it is an indeterminate form of 
.
So that , we can apply L’Hospital rule-
Example-2: Evaluate 
.
Sol. Let f(x) = 
and g(x) = 
.
Here we see that this is the indeterminate form of 0/0 at x = 0.
Now by using L’Hospital rule, we get-
= 
= 
= 
= 1
Note- Suppose we get an indeterminate form even after finding first derivative, then in that case , we use the other form of L’Hospital’s rule.
If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞ , -∞), then
Example-3: Evaluate 
Sol. Let f(x) = 
, then
And
= 0
= 0
But if we use L’Hospital rule again, then we get-
Example-4: Evaluate 
Sol. We can see that this is an indeterminate form of type 0/0.
Apply L’Hospital’s rule, we get
But this is again an indeterminate form, so that we will again apply L’Hospital’s rule-
We get
= 
L’Hospital’s rule for 
form-
Let f and g are two differentiable functions on an open interval containing x = a, except possibly at x = a and that
If 
has a finite limit, or if it is 
, then
Theorem- If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞ , -∞), then
Example-5: Find 
, n>0.
Sol. Let f(x) = log x and g(x) = 
These two functions satisfied the theorem that we have discussed above-
So that,
Example-6: Evaluate 
Sol. Apply L’Hospital rule as we can see that this is the form of 
= 
Note- In some cases like above example, we can not apply L’Hospital’s rule.
Other types of indeterminate forms-
Example-7: Evaluate 
Sol. Here we find that-
So that this limit is the form of 0
.
Now,
Change 
to obtain the limit-
Now this is the form of 0/0,
Apply L’Hospital’s rule-
If f(x) is a single valued function defined in a region R then
Maxima is a maximum point 
if and only if 
Minima is a minimum point 
if and only if 
Maxima and Minima of a function of two independent variables
Let 
be a defined function of two independent variables.
Then the point 
is said to be a maximum point of 
if
Or 
= 
For all positive and negative values of h and k.
Similarly the point 
is said to be a minimum point of 
if
Or 
= 
For all positive and negative values of h and k.
Saddle point: Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.
A point is a saddle point of a function of two variables if
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at the point.
Stationary Value
The value 
is said to be a stationary value of 
if
i.e. the function is a stationary at (a , b).
Rule to find the maximum and minimum values of 
4. (a) If 
(b) If 
(c) If 
(d) If 
Example1 Find out the maxima and minima of the function
Given 
…(i)
Partially differentiating (i) with respect to x we get
….(ii)
Partially differentiating (i) with respect to y we get
….(iii)
Now, form the equations 
Using (ii) and (iii) we get
using above two equations
Squaring both side we get
Or 
This show that 
Also we get 
Thus we get the pair of value as 
Now, we calculate
Putting above values in
At point (0,0) we get
So, the point (0,0) is a saddle point.
At point 
we get
So the point 
is the minimum point where 
In case 
So the point 
is the maximum point where 
Example2 Find the maximum and minimum point of the function
Partially differentiating given equation with respect to and x and y then equate them to zero
On solving above we get 
Also 
Thus we get the pair of values (0,0), (
,0) and (0,
Now, we calculate
At the point (0,0)
So function has saddle point at (0,0).
At the point (
So the function has maxima at this point (
.
At the point (0,
So the function has minima at this point (0,
.
At the point (
So the function has an saddle point at (
Example-3: There is rectangular box which is open at the top, is to have a volume of 32 c.c.
Find the dimensions of the box requiring least (minimum) material to construct it.
Sol. Here it is given that-
Volume (V) = 32 c.c.
Suppose ‘l’ , ‘b’, ‘h’ are the length, breadth, height of the rectangular box respectively and its surface area is ‘S’.
As we know that-
Volume (V) = l b h = 32
b (breadth) = 32/lh
And the surface area of the rectangular box which is open at the top is-
S = 2 (l + b) h + l b …………… (1)
On putting the value of ‘b’ in (1), we get
S = 2 
S = 
……………… (2)
Differentiate partially equation (2) with respect to l and h respectively, we get-
…………… (3) and 
…………….. (4)
For Max. and Min. S, we get
And 
The values of ‘l’, ‘h’ and ‘b’ will be-
L = 4, h = 2 and b = 4
Now-
And
So that-
, then S is minimum for l = 4, b = 4, and h = 2
Unit– 2
Calculus
2. Rolle’s theorem
Important definitions-
Continuity- suppose that a function f(x) is defined in the interval I , then it is said to be continuous at x=a , if
Differentiability- A function f(x) is said to be differentiable at x=a if 
exists where ‘a’ belongs to I
Rolle’s theorem-
Suppose f(x) is a function defined on [a , b] and it satisfies the following conditions
1. f(x) is continuous in [a , b]
2. f(x) is differentiable in (a , b)
3. f(a) = f(b)
Then there exists atleast a point point c ϵ (a , b) , where a<b , such that f’(c) = 0
Proof: suppose y = f(x) is a function and A(a , f(a)) , B(b , f(b)) be two points on the curve f(x) and a,b are two end points. Now conditions for Rolle’s theorem-
1.f(x) is a continuous function in[a , b] , from the figure without breaks in between A&B on y = f(x).
2. f(x) is differentiable in (a , b), because joining A and B we get a line AB.
Slope of the line AB=0 then a point C at P also a tangent at P, or Q,R,S is parallel to x –axis.
Slope of the tangent at P or Q, R, S, will be 0, even the curve y = f(x) decreases or increases, that means f(x) is constant.
Derivative of f(x),
f’(c) = 0
That’s why, f’(c) = 0
3. the slope of the line AB is equal to zero, that means the line AB is parallel to x-axis.
So that, f(a) = f(b)
Example: Verify Rolle’s theorem for the function f(x) = x(x+3) 
in interval [-3, 0].
Sol. First we will differentiate the given function with respect to x, we get
f’(x) = (x²+3x) 
+ (2x + 3)
= 
This shows that f’(x) exists for all x, therefore f(x) is continuous for all x.
Now, f(-3) = 0 and f(0) = 0 , so that f(-3) = f(0).
Here f(x) satisfies all the conditions of Rolle’s theorem,
Then,
f’(x) = 0 , which gives
= 0
We get,
X = 3 and x = -2
Here we can see that clearly -3<-2<0 , therefore there exists -2 ∈ (-3,0) such that
f’(-2) = 0
that means the Rolle’s theorem is true for the given function.
Example: Verify Rolle’s theorem for the given functions below-
1. f(x) = x³ - 6x²+11x-6 in the interval [1,3]
2. f(x) = x²-4x+8 in the interval [1,3]
Sol. (1)
As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]
Also, f(1) = f(3) = 0
Now we find f’(x) = 0
3x² - 12x +11 = 0
We get, x = 2+ 
and 2 - 
Hence both of them lie in (1,3).
Hence the theorem holds good for the given function in interval [1,3]
(2) As we know that every polynomial is continuous and differentiable for all points, so that the given function is continuous and differentiable in the interval [1,3]
Also, f(1) = 1 -4 +8 = 5 and f(3) = 9 – 12 + 8 = 5
Hence f(1) = f(3)
Now the first derivative of the function,
f’(x) = 0
2x – 4 = 0 , gives
X = 2
We can see that 1<2<3, hence there exists 2 between 1 and 3. And f’(2) = 0.
This means that the Rolle’s theorem holds good for the given function and given interval.
Example: Verify the Rolle’s theorem for sin x in the interval [
]
Sol. Suppose f(x) = sin x
We know that sin x is continuous for all x.
Now , f’(x) = cos x exists for all x in (
) and
f(
0
f(
0
thus f(x) satisfies all the conditions of Rolle’s theorem.
Now,
f’(x) = 0 that gives , cos x = 0
x = 
here we notice that both intervals lie in (
.
there exists, c = 
so that, f’(c) = 0
the Rolle’s theorem has been verified.
Lagranges’s mean value theorem-
Suppose that f(x) be a function of x such that,
1. if it is continuous in [a , b]
2. if it is differentiable in (a , b)
Then there atleast exists a value cϵ (a , b)
f’(c) = 
proof:
lets define a function g(x)
g(x) = f(x) – Ax ………………..(1)
here A is a constant which is to be determined,
so that, g(a) = g(b)
now,
g(a) = f(a) – Aa
g(b) = f(b) – Ab
so,
g(a) = g(b),
f(a) – Aa = f(b) – Ab ,
which gives
A = 
…………………..(2)
As right hand side of eq.(1) is continuous in [a,b] , so that g(x) is continuous.
And right hand side of eq.(1) is differential in (a,b) , so that g(x) is differentiable in (a,b).
And g(a) = g(b) , because of the choice of A.
Hence g(x) satisfies all the conditions of Rolle’s theorem.
So that,
There exists a value c such that a<c<b at which g’(c) = 0
Now, differentiate eq. (1) with respect to x, we get
g’(x) = f’(x) – A
here we know that, x = c,
g’(c) = f’(c) – A
as g’(c) = 0, then
f’(c) – A =0
so that,, f’(c) = A,
from equation (2) , we get
f’(c) = 
hence proved.
Example: Verify Lagrange’s mean value theorem for f(x) = (x-1)(x-2)(x-3) in [0,4].
Sol. As we see that the given function is a polynomial and we know that the polynomial is continuous in [0,4] and differentiable in (0,4).
f(x) = (x-1)(x-2)(x-3)
f(x) = x-6x²+11x-6
now at x = 0, we get
f(0) = -6 and
at x = 4, we get.
f(4) = 6
diff. the function w.r.t.x , we get
f’(x) = 3x²-6x+11
suppose x = c, we get
f’(c) = 3c²-6c+11
by Lagrange’s mean value theorem,
f’(c) = 
= 
= 
= 3
now we get,
3c²-6c+11 = 3
3c²-6c+8 = 0
On solving the quadratic equation, we get
C = 2 
Here we see that the value of c lies between 0 and 4
Therefore the given function is verified.
Example: Verify Lagrange’s mean value theorem for f(x) = log xin [1,e].
Sol. We already know that the function which is log x is continuous for all x>0.
So that this is the continuous function In [1,e]
Now,
f’(x) = 1/x
which is exists for all x in (1,e)
so that f(x) is differentiable in (1,e).
by Lagrange’s mean value theorem, we get
f’(c) = 
, let x = c,
then ,
f’(c) = 
we get,
c = e-1
e-1 will always lies between 1 and e .
hence the function is verified by Lagrange’s mean value theorem.
Cauchy’s mean value theorem-
Suppose we have two functions f(x) and g(x) of x, such that,
1. Both functions are continuous in [a,b]
2. Both functions are differentiable in (a,b)
3. g’(x) ≠ 0 for any x ϵ (a,b)
These three exists atleast , x = c ϵ (a,b) , at which
Proof: suppose ,we define a functions,
h(x) = f(x) – A.g(x) …………………….(1)
so that h(a) = h(b) and A is a constant to be determined.
Now,
h(a) = f(a) – Ag(a)
h(b) = f(b) – A.g(b)
so that,
f(a) – Ag(a) = f(b) – A.g(b) , which gives
A = 
…………………………….(2)
Now , h(x) is continuous in [a,b] as RHS of eq. (1) is continuous in [a,b] and h(x) is diff. in (a,b) as RHS of eq. (1) is diff. in (a,b)
Also,
h(a) = h(b)
Therefore all the conditions of Rolle’s theorem are satistfied then there exists a Value x = cϵ (a,b)
so that h’(c) = 0
Differentiate eq.(1) w.r.t. x , we get
h’(x) = f’(x) – A.g’(x)
At x = c
h’(c) = f’(c) – A.g’(c)
0 = f’(c) – A.g’(c)
A = 
So that , we get
where a<c<b
Hence the Cauchy’s mean value theorem is proved.
Example: Verify Cauchy’s mean value theorem for the function f(x) = sin x and g(x) = cos x in [ 0 , π/2]
Sol. It is given tha,
f(x) = sin x and g(x) = cos x
Now,
f’(x) = cos x and g’(x) = - sin x
We know that both the functions are continuous in [ 0 , π/2] and differentiable in ( 0 , π/2 )
Also, g’(x) = -sin x ≠ 0 for all x ϵ( 0 , π/2 )
By Cauchy’s mean value theorem, we get
for some c: 0< c < 
That means
which gives,
Cot c = 1
C = 
Now we see that 
lies between 0 and 
Example: Verify Cauchy’s mean value theorem for the function f(x) = x⁴ and g(x) = x² in the interval [1,2]
Sol. We are given, f(x) = x⁴ and g(x) = x
Derivative of these fucntions ,
f’(x) = 4x³ and g’(x) = 2x
put these values in Cauchy’s formula, we get
2c² = 
c² = 
c = 
now put the values of a = 1 and b = 2 ,we get
c = 
=
= 
(approx)
Hence the Cauchy’s theorem is verified.
Taylor’s Theorem-
If f(x + h) is a function of h which can be expanded in the ascending powers of h and is differentiable by any number of times with respect to h, then-
+ …….+ 
+ ……..
Which is called Taylor’s theorem.
If we put x = a, we get-
+ …….+ 
+ …….. (1)
Maclaurin’s Theorem-
If we put a = 0 and h = x then equation(1) becomes-
+ …….
Which is called Maclaurin’s theorem.
Note – if we put h = x - a then there will be the expansion of F(x) in powers of (x – a)
We get-
+ …….
Example: Express the polynomial 
in powers of (x-2).
Sol. Here we have,
f(x) = 
differentiating the function w.r.t.x-
f’(x) = 
f’’(x) = 12x + 14
f’’’(x) = 12
f’’’’(x)=0
now using Taylor’s theorem-
+ ……. (1)
Here we have, a = 2,
Put x = 2 in the derivatives of f(x), we get-
f(2) = 
f’(2) = 
f’’(2) = 12(2)+14 = 38
f’’’(2) = 12 and f’’’’(2) = 0
now put a = 2 and substitute the above values in equation(1), we get-
Example: Expand sin x in powers of 
Sol. Let f(x) = sin x
Then,
= 
By using Taylor’s theorem-
+ ……. (1)
Here f(x) = sin x and a = π/2
f’(x) = cos x , f’’(x) = - sin x , f’’’(x) = - cos x and so on.
Putting x = π/2 , we get
f(x) = sin x = 
= 1
f’(x) = cos x = 
= 0
f’’(x) = -sin x = 
= -1
f’’’(x) = -cos x = 
= 0
from equation (1) put a = 
and substitute these values, we get-
+ …….
= 
………………………..
2.4. Indeterminate forms & L’Hospital’s rule
Let we have two functions f(x) and g(x) and-
Then-
is an expression of the form 
, in that case we can say that f(x)/g(x) is an indeterminate for of the type 
at x = a.
Now, Let we have two functions f(x) and g(x) and-
Then-
is an expression of the form 
, in that case we can say that f(x)/g(x) is an indeterminate for of the type 
at x = a.
Some other indeterminate forms are 
L’Hospital’s rule for 
form-
Working steps-
1. Check that the limits f(x)/g(x) is an indeterminate form of type 
.
(note- we can not apply L’Hospital rule if it is not in indeterminate form)
2. Differentiate f and g separately.
3. Find the limits of the derivatives .if the limit is finite , 
then it is equal to the limit of f(x)/g(x).
Example-1: Evaluate 
Sol. Here we notice that it is an indeterminate form of 
.
So that , we can apply L’Hospital rule-
Example-2: Evaluate 
.
Sol. Let f(x) = 
and g(x) = 
.
Here we see that this is the indeterminate form of 0/0 at x = 0.
Now by using L’Hospital rule, we get-
= 
= 
= 
= 1
Note- Suppose we get an indeterminate form even after finding first derivative, then in that case , we use the other form of L’Hospital’s rule.
If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞ , -∞), then
Example-3: Evaluate 
Sol. Let f(x) = 
, then
And
= 0
= 0
But if we use L’Hospital rule again, then we get-
Example-4: Evaluate 
Sol. We can see that this is an indeterminate form of type 0/0.
Apply L’Hospital’s rule, we get
But this is again an indeterminate form, so that we will again apply L’Hospital’s rule-
We get
= 
L’Hospital’s rule for 
form-
Let f and g are two differentiable functions on an open interval containing x = a, except possibly at x = a and that
If 
has a finite limit, or if it is 
, then
Theorem- If we have f(x) and g(x) are two functions such that 
.
If 
exist or (∞ , -∞), then
Example-5: Find 
, n>0.
Sol. Let f(x) = log x and g(x) = 
These two functions satisfied the theorem that we have discussed above-
So that,
Example-6: Evaluate 
Sol. Apply L’Hospital rule as we can see that this is the form of 
= 
Note- In some cases like above example, we can not apply L’Hospital’s rule.
Other types of indeterminate forms-
Example-7: Evaluate 
Sol. Here we find that-
So that this limit is the form of 0
.
Now,
Change 
to obtain the limit-
Now this is the form of 0/0,
Apply L’Hospital’s rule-
If f(x) is a single valued function defined in a region R then
Maxima is a maximum point 
if and only if 
Minima is a minimum point 
if and only if 
Maxima and Minima of a function of two independent variables
Let 
be a defined function of two independent variables.
Then the point 
is said to be a maximum point of 
if
Or 
= 
For all positive and negative values of h and k.
Similarly the point 
is said to be a minimum point of 
if
Or 
= 
For all positive and negative values of h and k.
Saddle point: Critical points of a function of two variables are those points at which both partial derivatives of the function are zero. A critical point of a function of a single variable is either a local maximum, a local minimum, or neither. With functions of two variables there is a fourth possibility - a saddle point.
A point is a saddle point of a function of two variables if
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at the point.
Stationary Value
The value 
is said to be a stationary value of 
if
i.e. the function is a stationary at (a , b).
Rule to find the maximum and minimum values of 
4. (a) If 
(b) If 
(c) If 
(d) If 
Example1 Find out the maxima and minima of the function
Given 
…(i)
Partially differentiating (i) with respect to x we get
….(ii)
Partially differentiating (i) with respect to y we get
….(iii)
Now, form the equations 
Using (ii) and (iii) we get
using above two equations
Squaring both side we get
Or 
This show that 
Also we get 
Thus we get the pair of value as 
Now, we calculate
Putting above values in
At point (0,0) we get
So, the point (0,0) is a saddle point.
At point 
we get
So the point 
is the minimum point where 
In case 
So the point 
is the maximum point where 
Example2 Find the maximum and minimum point of the function
Partially differentiating given equation with respect to and x and y then equate them to zero
On solving above we get 
Also 
Thus we get the pair of values (0,0), (
,0) and (0,
Now, we calculate
At the point (0,0)
So function has saddle point at (0,0).
At the point (
So the function has maxima at this point (
.
At the point (0,
So the function has minima at this point (0,
.
At the point (
So the function has an saddle point at (
Example-3: There is rectangular box which is open at the top, is to have a volume of 32 c.c.
Find the dimensions of the box requiring least (minimum) material to construct it.
Sol. Here it is given that-
Volume (V) = 32 c.c.
Suppose ‘l’ , ‘b’, ‘h’ are the length, breadth, height of the rectangular box respectively and its surface area is ‘S’.
As we know that-
Volume (V) = l b h = 32
b (breadth) = 32/lh
And the surface area of the rectangular box which is open at the top is-
S = 2 (l + b) h + l b …………… (1)
On putting the value of ‘b’ in (1), we get
S = 2 
S = 
……………… (2)
Differentiate partially equation (2) with respect to l and h respectively, we get-
…………… (3) and 
…………….. (4)
For Max. and Min. S, we get
And 
The values of ‘l’, ‘h’ and ‘b’ will be-
L = 4, h = 2 and b = 4
Now-
And
So that-
, then S is minimum for l = 4, b = 4, and h = 2
