NCERT Textbook: Application of Derivatives

# NCERT Textbook: Application of Derivatives | NCERT Textbooks (Class 6 to Class 12) - CTET & State TET PDF Download

``` Page 1

v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature.” — WHITEHEAD v
6.1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions.
In this chapter, we will study applications of the derivative in various disciplines, e.g., in
engineering, science, social science, and many  other fields. For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs. W e will also use derivative to find intervals
on which a function is increasing or decreasing. Finally, we use the derivative to find
approximate value of certain quantities.
6.2 Rate of Change of Quantities
Recall that by the derivative
ds
dt
, we mean the rate of change of distance s with
respect to the time t. In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule ( ) y f x = , then
dy
dx
(or f'(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
?
?
?
=
0
(or f'(x
0
)) represents the  rate of change
of y with respect to x at
0
x x = .
Further, if two variables x and y are varying with respect to another variable t, i.e.,
if ( ) x f t = and ( ) y g t = , then by Chain Rule
dy
dx
=
dy dx
dt dt
,  if
0
dx
dt
?
Chapter 6
APPLICATION OF
DERIV ATIVES
Reprint 2024-25
Page 2

v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature.” — WHITEHEAD v
6.1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions.
In this chapter, we will study applications of the derivative in various disciplines, e.g., in
engineering, science, social science, and many  other fields. For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs. W e will also use derivative to find intervals
on which a function is increasing or decreasing. Finally, we use the derivative to find
approximate value of certain quantities.
6.2 Rate of Change of Quantities
Recall that by the derivative
ds
dt
, we mean the rate of change of distance s with
respect to the time t. In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule ( ) y f x = , then
dy
dx
(or f'(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
?
?
?
=
0
(or f'(x
0
)) represents the  rate of change
of y with respect to x at
0
x x = .
Further, if two variables x and y are varying with respect to another variable t, i.e.,
if ( ) x f t = and ( ) y g t = , then by Chain Rule
dy
dx
=
dy dx
dt dt
,  if
0
dx
dt
?
Chapter 6
APPLICATION OF
DERIV ATIVES
Reprint 2024-25
MATHEMATICS 148
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t.
Let us consider some examples.
Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm.
Solution The area A of a circle with radius r is given by A = pr
2
. Therefore, the rate
of change of the area A with respect to its radius r is given by
2
A
( ) 2
d d
r r
dr dr
= p = p
.
When r = 5 cm,
A
10
d
dr
= p
. Thus, the area of the circle is changing at the rate of
10p cm
2
/s.
Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second. How fast is the surface area increasing when the length of an edge is 10
centimetres ?
Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube. Then, V = x
3
and S = 6x
2
, where x is a function of time t.
Now
V d
dt
= 9cm
3
/s (Given)
Therefore 9 =
3 3
V
( ) ( )
d d d dx
x x
dt dt dx dt
= = ·
(By Chain Rule)
=
2
3
dx
x
dt
·
or
dx
dt
=
2
3
x
... (1)
Now
dS
dt
=
2 2
(6 ) (6 )
d d dx
x x
dt dx dt
= ·
(By Chain Rule)
=
2
3 36
12x
x x
? ?
· =
? ?
? ?
(Using (1))
Hence, when x = 10 cm,
2
3.6 cm /s
dS
dt
=
Reprint 2024-25
Page 3

v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature.” — WHITEHEAD v
6.1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions.
In this chapter, we will study applications of the derivative in various disciplines, e.g., in
engineering, science, social science, and many  other fields. For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs. W e will also use derivative to find intervals
on which a function is increasing or decreasing. Finally, we use the derivative to find
approximate value of certain quantities.
6.2 Rate of Change of Quantities
Recall that by the derivative
ds
dt
, we mean the rate of change of distance s with
respect to the time t. In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule ( ) y f x = , then
dy
dx
(or f'(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
?
?
?
=
0
(or f'(x
0
)) represents the  rate of change
of y with respect to x at
0
x x = .
Further, if two variables x and y are varying with respect to another variable t, i.e.,
if ( ) x f t = and ( ) y g t = , then by Chain Rule
dy
dx
=
dy dx
dt dt
,  if
0
dx
dt
?
Chapter 6
APPLICATION OF
DERIV ATIVES
Reprint 2024-25
MATHEMATICS 148
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t.
Let us consider some examples.
Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm.
Solution The area A of a circle with radius r is given by A = pr
2
. Therefore, the rate
of change of the area A with respect to its radius r is given by
2
A
( ) 2
d d
r r
dr dr
= p = p
.
When r = 5 cm,
A
10
d
dr
= p
. Thus, the area of the circle is changing at the rate of
10p cm
2
/s.
Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second. How fast is the surface area increasing when the length of an edge is 10
centimetres ?
Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube. Then, V = x
3
and S = 6x
2
, where x is a function of time t.
Now
V d
dt
= 9cm
3
/s (Given)
Therefore 9 =
3 3
V
( ) ( )
d d d dx
x x
dt dt dx dt
= = ·
(By Chain Rule)
=
2
3
dx
x
dt
·
or
dx
dt
=
2
3
x
... (1)
Now
dS
dt
=
2 2
(6 ) (6 )
d d dx
x x
dt dx dt
= ·
(By Chain Rule)
=
2
3 36
12x
x x
? ?
· =
? ?
? ?
(Using (1))
Hence, when x = 10 cm,
2
3.6 cm /s
dS
dt
=
Reprint 2024-25
APPLICATION OF DERIVATIVES 149
Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing?
Solution The area A of a circle with radius r is given by A = pr
2
. Therefore, the rate
of change of area A with respect to time t is
A d
dt
=
2 2
( ) ( )
d d dr
r r
dt dr dt
p = p ·
= 2p r
dr
dt
(By Chain Rule)
It is given that
dr
dt
= 4cm/s
Therefore, when r = 10 cm,
A d
dt
= 2p (10) (4) = 80p
Thus, the enclosed area is increasing at the rate of 80p cm
2
/s, when r = 10 cm.
A
Note
dy
dx
is positive if y increases as x increases and is negative if y decreases
as x increases.
Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and
the width y is increasing at the rate of 2cm/minute. When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle.
Solution Since the length x is decreasing and the width y is increasing with respect to
time, we have
3 cm/min
dx
dt
= - and 2 cm/min
dy
dt
=
(a) The perimeter P of a rectangle is given by
P = 2 (x + y)
Therefore
P d
dt
=
2 2 3 2 2
dx
dt
dy
dt
+
?
?
?
?
?
?
= - + = - ( ) cm/min
(b) The area A of the rectangle is given by
A = x . y
Therefore
A d
dt
=
dx dy
y x
dt dt
· + ·
= – 3(6) + 10(2) (as x = 10 cm and y = 6 cm)
= 2 cm
2
/min
Reprint 2024-25
Page 4

v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature.” — WHITEHEAD v
6.1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions.
In this chapter, we will study applications of the derivative in various disciplines, e.g., in
engineering, science, social science, and many  other fields. For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs. W e will also use derivative to find intervals
on which a function is increasing or decreasing. Finally, we use the derivative to find
approximate value of certain quantities.
6.2 Rate of Change of Quantities
Recall that by the derivative
ds
dt
, we mean the rate of change of distance s with
respect to the time t. In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule ( ) y f x = , then
dy
dx
(or f'(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
?
?
?
=
0
(or f'(x
0
)) represents the  rate of change
of y with respect to x at
0
x x = .
Further, if two variables x and y are varying with respect to another variable t, i.e.,
if ( ) x f t = and ( ) y g t = , then by Chain Rule
dy
dx
=
dy dx
dt dt
,  if
0
dx
dt
?
Chapter 6
APPLICATION OF
DERIV ATIVES
Reprint 2024-25
MATHEMATICS 148
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t.
Let us consider some examples.
Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm.
Solution The area A of a circle with radius r is given by A = pr
2
. Therefore, the rate
of change of the area A with respect to its radius r is given by
2
A
( ) 2
d d
r r
dr dr
= p = p
.
When r = 5 cm,
A
10
d
dr
= p
. Thus, the area of the circle is changing at the rate of
10p cm
2
/s.
Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second. How fast is the surface area increasing when the length of an edge is 10
centimetres ?
Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube. Then, V = x
3
and S = 6x
2
, where x is a function of time t.
Now
V d
dt
= 9cm
3
/s (Given)
Therefore 9 =
3 3
V
( ) ( )
d d d dx
x x
dt dt dx dt
= = ·
(By Chain Rule)
=
2
3
dx
x
dt
·
or
dx
dt
=
2
3
x
... (1)
Now
dS
dt
=
2 2
(6 ) (6 )
d d dx
x x
dt dx dt
= ·
(By Chain Rule)
=
2
3 36
12x
x x
? ?
· =
? ?
? ?
(Using (1))
Hence, when x = 10 cm,
2
3.6 cm /s
dS
dt
=
Reprint 2024-25
APPLICATION OF DERIVATIVES 149
Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing?
Solution The area A of a circle with radius r is given by A = pr
2
. Therefore, the rate
of change of area A with respect to time t is
A d
dt
=
2 2
( ) ( )
d d dr
r r
dt dr dt
p = p ·
= 2p r
dr
dt
(By Chain Rule)
It is given that
dr
dt
= 4cm/s
Therefore, when r = 10 cm,
A d
dt
= 2p (10) (4) = 80p
Thus, the enclosed area is increasing at the rate of 80p cm
2
/s, when r = 10 cm.
A
Note
dy
dx
is positive if y increases as x increases and is negative if y decreases
as x increases.
Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and
the width y is increasing at the rate of 2cm/minute. When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle.
Solution Since the length x is decreasing and the width y is increasing with respect to
time, we have
3 cm/min
dx
dt
= - and 2 cm/min
dy
dt
=
(a) The perimeter P of a rectangle is given by
P = 2 (x + y)
Therefore
P d
dt
=
2 2 3 2 2
dx
dt
dy
dt
+
?
?
?
?
?
?
= - + = - ( ) cm/min
(b) The area A of the rectangle is given by
A = x . y
Therefore
A d
dt
=
dx dy
y x
dt dt
· + ·
= – 3(6) + 10(2) (as x = 10 cm and y = 6 cm)
= 2 cm
2
/min
Reprint 2024-25
MATHEMATICS 150
Example 5 The total cost C(x) in Rupees, associated with the production of x units of
an item is given by
C (x) = 0.005 x
3
– 0.02 x
2
+ 30x + 5000
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output.
Solution Since marginal cost is the rate of change of total cost with respect to the
output, we have
Marginal cost (MC) =
2
0.005(3 ) 0.02(2 ) 30
dC
x x
dx
= - +
When x = 3, MC =
2
0.015(3 ) 0.04(3) 30 - +
= 0.135 – 0.12 + 30 = 30.015
Hence, the required marginal cost is ` 30.02 (nearly).
Example 6 The total revenue in Rupees received from the sale of x units of a product
is given by R(x) = 3x
2
+ 36x + 5. Find the marginal revenue, when x = 5, where by
marginal revenue we mean the rate of change of total revenue with respect to the
number of items sold at an instant.
Solution Since marginal revenue is the rate of change of total revenue with respect to
the number of units sold, we have
Marginal Revenue (MR) =
R
6 36
d
x
dx
= +
When x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is ` 66.
EXERCISE 6.1
1. Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm (b) r = 4 cm
2. The volume of a cube is increasing at the rate of 8 cm
3
/s. How fast is the
surface area increasing when the length of an edge is 12 cm?
3. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate
at which the area of the circle is increasing when the radius is 10 cm.
4. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the
volume of the cube increasing when the edge is 10 cm long?
5. A stone is dropped into a quiet lake and waves move in circles at the speed of
5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is
the enclosed area increasing?
Reprint 2024-25
Page 5

v With the Calculus as a key, Mathematics can be successfully applied
to the explanation of the course of Nature.” — WHITEHEAD v
6.1 Introduction
In Chapter 5, we have learnt how to find derivative of composite functions, inverse
trigonometric functions, implicit functions, exponential functions and logarithmic functions.
In this chapter, we will study applications of the derivative in various disciplines, e.g., in
engineering, science, social science, and many  other fields. For instance, we will learn
how the derivative can be used (i) to determine rate of change of quantities, (ii) to find
the equations of tangent and normal to a curve at a point, (iii) to find turning points on
the graph of a function which in turn will help us to locate points at which largest or
smallest value (locally) of a function occurs. W e will also use derivative to find intervals
on which a function is increasing or decreasing. Finally, we use the derivative to find
approximate value of certain quantities.
6.2 Rate of Change of Quantities
Recall that by the derivative
ds
dt
, we mean the rate of change of distance s with
respect to the time t. In a similar fashion, whenever one quantity y varies with another
quantity x, satisfying some rule ( ) y f x = , then
dy
dx
(or f'(x)) represents the rate of
change of y with respect to x and
dy
dx
x x
?
?
?
=
0
(or f'(x
0
)) represents the  rate of change
of y with respect to x at
0
x x = .
Further, if two variables x and y are varying with respect to another variable t, i.e.,
if ( ) x f t = and ( ) y g t = , then by Chain Rule
dy
dx
=
dy dx
dt dt
,  if
0
dx
dt
?
Chapter 6
APPLICATION OF
DERIV ATIVES
Reprint 2024-25
MATHEMATICS 148
Thus, the rate of change of y with respect to x can be calculated using the rate of
change of y and that of x both with respect to t.
Let us consider some examples.
Example 1 Find the rate of change of the area of a circle per second with respect to
its radius r when r = 5 cm.
Solution The area A of a circle with radius r is given by A = pr
2
. Therefore, the rate
of change of the area A with respect to its radius r is given by
2
A
( ) 2
d d
r r
dr dr
= p = p
.
When r = 5 cm,
A
10
d
dr
= p
. Thus, the area of the circle is changing at the rate of
10p cm
2
/s.
Example 2 The volume of a cube is increasing at a rate of 9 cubic centimetres per
second. How fast is the surface area increasing when the length of an edge is 10
centimetres ?
Solution Let x be the length of a side, V be the volume and S be the surface area of
the cube. Then, V = x
3
and S = 6x
2
, where x is a function of time t.
Now
V d
dt
= 9cm
3
/s (Given)
Therefore 9 =
3 3
V
( ) ( )
d d d dx
x x
dt dt dx dt
= = ·
(By Chain Rule)
=
2
3
dx
x
dt
·
or
dx
dt
=
2
3
x
... (1)
Now
dS
dt
=
2 2
(6 ) (6 )
d d dx
x x
dt dx dt
= ·
(By Chain Rule)
=
2
3 36
12x
x x
? ?
· =
? ?
? ?
(Using (1))
Hence, when x = 10 cm,
2
3.6 cm /s
dS
dt
=
Reprint 2024-25
APPLICATION OF DERIVATIVES 149
Example 3 A stone is dropped into a quiet lake and waves move in circles at a speed
of 4cm per second. At the instant, when the radius of the circular wave is 10 cm, how
fast is the enclosed area increasing?
Solution The area A of a circle with radius r is given by A = pr
2
. Therefore, the rate
of change of area A with respect to time t is
A d
dt
=
2 2
( ) ( )
d d dr
r r
dt dr dt
p = p ·
= 2p r
dr
dt
(By Chain Rule)
It is given that
dr
dt
= 4cm/s
Therefore, when r = 10 cm,
A d
dt
= 2p (10) (4) = 80p
Thus, the enclosed area is increasing at the rate of 80p cm
2
/s, when r = 10 cm.
A
Note
dy
dx
is positive if y increases as x increases and is negative if y decreases
as x increases.
Example 4 The length x of a rectangle is decreasing at the rate of 3 cm/minute and
the width y is increasing at the rate of 2cm/minute. When x =10cm and y = 6cm, find
the rates of change of (a) the perimeter and (b) the area of the rectangle.
Solution Since the length x is decreasing and the width y is increasing with respect to
time, we have
3 cm/min
dx
dt
= - and 2 cm/min
dy
dt
=
(a) The perimeter P of a rectangle is given by
P = 2 (x + y)
Therefore
P d
dt
=
2 2 3 2 2
dx
dt
dy
dt
+
?
?
?
?
?
?
= - + = - ( ) cm/min
(b) The area A of the rectangle is given by
A = x . y
Therefore
A d
dt
=
dx dy
y x
dt dt
· + ·
= – 3(6) + 10(2) (as x = 10 cm and y = 6 cm)
= 2 cm
2
/min
Reprint 2024-25
MATHEMATICS 150
Example 5 The total cost C(x) in Rupees, associated with the production of x units of
an item is given by
C (x) = 0.005 x
3
– 0.02 x
2
+ 30x + 5000
Find the marginal cost when 3 units are produced, where by marginal cost we
mean the instantaneous rate of change of total cost at any level of output.
Solution Since marginal cost is the rate of change of total cost with respect to the
output, we have
Marginal cost (MC) =
2
0.005(3 ) 0.02(2 ) 30
dC
x x
dx
= - +
When x = 3, MC =
2
0.015(3 ) 0.04(3) 30 - +
= 0.135 – 0.12 + 30 = 30.015
Hence, the required marginal cost is ` 30.02 (nearly).
Example 6 The total revenue in Rupees received from the sale of x units of a product
is given by R(x) = 3x
2
+ 36x + 5. Find the marginal revenue, when x = 5, where by
marginal revenue we mean the rate of change of total revenue with respect to the
number of items sold at an instant.
Solution Since marginal revenue is the rate of change of total revenue with respect to
the number of units sold, we have
Marginal Revenue (MR) =
R
6 36
d
x
dx
= +
When x = 5, MR = 6(5) + 36 = 66
Hence, the required marginal revenue is ` 66.
EXERCISE 6.1
1. Find the rate of change of the area of a circle with respect to its radius r when
(a) r = 3 cm (b) r = 4 cm
2. The volume of a cube is increasing at the rate of 8 cm
3
/s. How fast is the
surface area increasing when the length of an edge is 12 cm?
3. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate
at which the area of the circle is increasing when the radius is 10 cm.
4. An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the
volume of the cube increasing when the edge is 10 cm long?
5. A stone is dropped into a quiet lake and waves move in circles at the speed of
5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is
the enclosed area increasing?
Reprint 2024-25
APPLICATION OF DERIVATIVES 151
6. The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of
increase of its circumference?
7. The length x of a rectangle is decreasing at the rate of 5 cm/minute and the
width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find
the rates of change  of (a) the perimeter, and (b) the area of the rectangle.
8. A balloon, which always remains spherical on inflation, is being inflated by pumping
in 900 cubic centimetres of gas per second. Find the rate at which the radius of
the balloon increases when the radius is 15 cm.
9. A balloon, which always remains spherical has a variable radius. Find the rate at
which its volume is increasing with the radius when the later is 10 cm.
10. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled
along the ground, away from the wall, at the rate of 2cm/s. How fast is its height
on the wall decreasing when the foot of the ladder is 4 m away from the wall ?
11. A particle moves along the curve 6y = x
3
+2. Find the points on the curve at
which the y-coordinate is changing 8 times as fast as the x-coordinate.
12. The radius of an air bubble is increasing at the rate of
1
2
cm/s. At what rate is the
volume of the bubble increasing when the radius is 1 cm?
13. A balloon, which always remains spherical, has a variable diameter
3
(2 1)
2
x+
.
Find the rate of change of its volume with respect to x.
14. Sand is pouring from a pipe at the rate of 12 cm
3
/s. The falling sand forms a cone
on the ground in such a way that the height of the cone is always one-sixth of the
radius of the base. How fast is the height of the sand cone increasing when the
height is 4 cm?
15. The total cost C (x) in Rupees associated with the production of x units of an
item is given by
C (x) = 0.007x
3
– 0.003x
2
+ 15x + 4000.
Find the marginal cost when 17 units are produced.
16. The total revenue in Rupees received from the sale of x units of a product is
given by
R (x) = 13x
2
+ 26x + 15.
Find the marginal revenue when  x = 7.
Choose the correct answer for questions 17 and 18.
17. The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10p (B) 12p (C) 8p (D) 11p
Reprint 2024-25
```

## NCERT Textbooks (Class 6 to Class 12)

681 docs|672 tests

## FAQs on NCERT Textbook: Application of Derivatives - NCERT Textbooks (Class 6 to Class 12) - CTET & State TET

 1. What is the concept of derivatives in mathematics?
Ans. Derivatives are a fundamental concept in calculus that measures the rate at which a function changes. It represents the slope of a function at a particular point and provides information about its behavior, such as whether it is increasing or decreasing. Derivatives are used to solve various real-world problems involving rates of change, optimization, and approximation.
 2. How can derivatives be applied in real-life situations?
Ans. Derivatives have numerous applications in various fields. For example, in physics, derivatives are used to calculate velocity and acceleration of objects in motion. In economics, derivatives help determine the marginal cost and revenue functions, which are crucial for decision-making. They are also used in engineering to analyze rates of change in electrical circuits and the behavior of systems. Overall, derivatives provide a mathematical tool to model and understand change in different real-life scenarios.
 3. What are the different techniques for finding derivatives of functions?
Ans. There are several techniques for finding derivatives of functions. The most common techniques include the power rule, product rule, quotient rule, chain rule, and trigonometric rules. The power rule is used to find the derivative of functions with a variable raised to a constant power. The product rule helps in finding the derivative of the product of two functions. The quotient rule is used to find the derivative of the quotient of two functions. The chain rule is employed when the function involves composition of multiple functions. Trigonometric rules are specific rules to find the derivatives of trigonometric functions.
 4. How can derivatives be used to optimize functions?
Ans. Derivatives play a crucial role in optimization problems. To optimize a function, we need to find the maximum or minimum values of that function. By taking the derivative of the function and equating it to zero, we can find the critical points where the function may have extreme values. We can then analyze these critical points using the first and second derivative tests to determine whether they correspond to a maximum or minimum value. This information helps us to optimize functions in various fields such as economics, engineering, and physics.
 5. Can derivatives be used to approximate values of a function?
Ans. Yes, derivatives can be used to approximate values of a function. The derivative of a function at a particular point represents the instantaneous rate of change at that point. By using this information, we can estimate the value of the function at nearby points. This technique is known as linear approximation or tangent line approximation. It involves finding the equation of the tangent line at a given point and using it to estimate the function's value. This method is particularly useful when dealing with complex functions or when an exact evaluation is difficult to obtain.

## NCERT Textbooks (Class 6 to Class 12)

681 docs|672 tests

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