## 1. Study applications of the derivative

Chapter 6

Applications of Derivatives:

Derivatives  have various important applications in Mathematics such as

• To find the Rate of Change of a Quantities,
• To find the equation of Tangent and Normal to a Curve, and
• 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,
• To find intervals on which a function is increasing or decreasing.
• To find the Approximation Value,
• To find the Minimum and Maximum Values of algebraic expressions.

Application of Derivatives in Real Life

• To calculate the profit and loss in business using graphs.
• To check the temperature variation.
• To determine the speed or distance covered such as miles per hour, kilometre per hour etc.
• Derivatives are used to derive many equations in Physics.
• Infinite series representation of functions,
• Optimization problems

## 2. Rate of Change of Quantities

Rate of Change of Quantities :

, we mean the rate of change of distance s with respect to the time t.

the rate of change of y with respect to x.= f′ (x)

x=x0 (or f′ (x0 )) represents the rate of change

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.

,if

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.

Example :

Find the rate of change of the area of a circle per second with respect to its radius r when r = 5 cm

The area A of a circle with radius r is given by A = π r 2 . Therefore, the rate of change of the area A with respect to its radius r is given by

When r = 5 cm.

Thus, the area of the circle is changing at the rate of 10π cm2 /s.

Example :

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

Example :  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) = = 0.005(3x2) - 0.02(2x ) + 30

When x = 3, MC = 0.015(32)- 0.04(3) + 30 = 0.135 – 0.12 + 30 = 30.015.

Hence, the required marginal cost is ` 30.02 (nearly).

## 3. Increasing and Decreasing Functions

Increasing and Decreasing Functions:

Definition :  Let I be an interval contained in the domain of a real valued function f.

Then f is said to be

1. increasing on I if x1 < x2 in I f(x1 ) < f(x2 ) for all x1 , x2 I.

(ii) decreasing on I, if x 1 , x 2 in I f(x 1 ) < f(x 2 ) for all x 1 , x 2 I.

1. constant on I, if f(x) = c for all x I, where c is a constant.

(iv) decreasing on I if x1 < x2 in I f (x1 ) ≥ f(x2 ) for all x1 , x2 I.

1. strictly decreasing on I if x1 < x2 in I f(x1 ) > f(x2 ) for all x1 , x2 I.

Definition :

Let x 0 be a point in the domain of definition of a real valued function f. Then f is said to be increasing, decreasing at x0 if there exists an open interval I containing x 0 such that f is increasing, decreasing, respectively, in I

Example :

Show that the function given by f(x) = 7x – 3 is increasing on R.

Solution:

Let x1 and x Î R. Then x1 < x2 7x1 < 7x2 7x1 – 3 < 7x2 – 3 f(x1 ) < f(x2 ) Thus, by Definition ,

it follows that f is strictly increasing on R.

Theorem 1 :

Let f be continuous on [a, b] and differentiable on the open interval (a,b).

Then (a) f is increasing in [a,b] if f ′(x) > 0 for each x (a, b)

(b) f is decreasing in [a,b] if f ′(x) < 0 for each x (a, b)

(c) f is a constant function in [a,b] if f ′(x) = 0 for each x (a, b)

Proof :

1. Let x1 , x2 [a, b] be such that x1 < x2 . Then, by Mean Value Theorem ,

there exists a point c between x1 and x2

such that f(x2 ) – f(x1 ) = f ′(c) (x2 – x1 )

i.e. f(x2 ) – f(x1 ) > 0 (as f ′(c) > 0 (given))

i.e. f(x2 ) > f(x1 )

Thus, we have x1< x2 =>f (x1 )<f (x2 ), for all x1,x2Î [ a,b ]

Hence, f is an increasing function in [a,b].

The proofs of part (b) and (c) are similar

Example :

Find the intervals in which the function f given by f(x) = x 2 – 4x + 6 is

1. increasing (b) decreasing

Solution :

We have f (x) = x 2 – 4x + 6

or f ′(x) = 2x – 4

Therefore, f ′(x) = 0 gives x = 2.

Now the point x = 2 divides the real line into two disjoint intervals namely, (– ∞, 2) and (2, ∞)

In the interval (– ∞, 2), f ′(x) = 2x – 4 < 0.

Therefore, f is decreasing in this interval. Also, in the interval (2,  ∞) , f  ′(x )> 0

and so the function f is increasing in this interval.

Example :

Find the intervals in which the function f given by f (x) = 4x 3 – 6x 2 – 72x + 30 is

1. increasing (b) decreasing.

Solution:  We have f(x) = 4x 3 – 6x 2 – 72x + 30

or f ′(x) = 12x 2 – 12x – 72

= 12(x 2 – x – 6) = 12(x – 3) (x + 2)

Therefore, f ′(x) = 0 gives x = – 2, 3.

The points x = – 2 and x = 3 divides the real line into three disjoint intervals, namely,

(– ∞, – 2), (– 2, 3)

and (3, ∞).

In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3), f ′(x) is negative. Consequently, the function f is increasing in the intervals (– ∞, – 2) and (3, ∞) while the function is decreasing in the interval (– 2, 3).

However, f is neither increasing nor decreasing in R.

## 4. Tangents and Normals

Tangents and Normals:

The equation of a straight line passing through a given point (x 1 , y1 ) having finite slope m is given by

y – y1 = m (x – x1 )

The slope of the tangent to the curve y = f(x) at the point (x1 , y1 ) is given by

So the equation of the tangent at (x1 , y1 ) to the curve y = f (x) is given by

y – y0 = f ′(x1 )(x – x1 )

Note: If a tangent line to the curve y = f (x) makes an angle θ with x-axis in the positive direction, then slope of the tangent =dy/ dx = tan θ .

Particular cases:

[1] If slope of the tangent line is zero, then tan θ = 0 and so θ = 0 which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at the point (x0 , y0 ) is given by y = y0 .

Example ;

Find the slope of the tangent to the curve y = x 3 – x at x = 2.

Solution :

The slope of the tangent at x = 2 is given by

## 5. Approximations

Approximations:

To find a very small change or variation of a quantity, we can use derivatives to give the approximate value of it. The approximate value is represented by delta â–³.

Suppose change in the value of x, dx = x then,

dy/dx = â–³x = x.

Since the change in x, dx ≈ x therefore, dy ≈ y.

Example :

Approximate 25.5

using differential.

Solution:

Let us consider y = x

, where x = 25 and âˆ†x = 0.5. Then,

âˆ†y = f(x+âˆ†x) - f(x)= x+â–³x-x

• âˆ†y =25.5-25
• âˆ†y =25.55
• 25.5=â–³y+5

Since dy is approximately equal to âˆ†y, therefore

dy=dydxâ–³x=12x(0.5)=0.05

Therefore, the approximate value of 25.5=5+0.05=5.05

Example :

Find the approximate value of the function f(3.02), where f(x) is given as 3x2+5x+3.

Solution:

Given that, f(x) = 3x2+5x+3

Assume x = 3, and âˆ†x = 0.02.

Hence, we can write the given function as:

f (3. 02) = f (x + âˆ†x) = 3(x + âˆ†x)2 + 5(x + âˆ†x) + 3

We know that,

âˆ†y = f (x + âˆ†x) – f (x).

The above expression can be written as

f (x + âˆ†x) = f (x) + âˆ†y

As, dx = âˆ†x, it can be approximately written as f (x) + f ′(x) âˆ†x

Hence, f (3.02) ≈ (3x2 + 5x + 3) + (6x + 5) âˆ†x

Now, substitute the values of x and âˆ†x, we get

= (3(3)2 + 5(3) + 3) + (6(3) + 5) (0.02)

Now, simplify it to get the approximate value

= (27 + 15 + 3) + (18 + 5) (0.02)

= 45 + 0.46

= 45.46

Therefore, the approximate value of f(3.02) is 45.46.

## 6. Maxima and Minima

Maxima and Minima

Let x1,x D f(a) > f(x1) and f(a) > f(x2)

A function is said to have a maximum value at a point 'x=a' in its domain D if f(a) ≥ f(x), x D.

f(a) is called the maximum value of f. "x=a" is called the point of maximum value of f.

Let x1,x D f(a) < f(x1) and f(a) < f(x2)

A function is said to have a minimum value at a point 'a' in its domain D if f(a) ≤ f(x), x D.

f(a) is called the minimum value of f. "x=a" is called the point of minimum value of f.

Extreme value of function f:

Function f(x) is said to have an extreme value in its domain D if there exists a point a  D such

that f (a) is either the maximum value or the minimum value of f in D.

The number f (a) is called an extreme value of f in D, and point 'a' is called an extreme point.

Consider f(x) = 2x, x R

The ordered pairs are (-2,8),(-1,2),(0,0),(1,2) and (2,8).

From the graph, we have f(x) = 0 if x = 0 and f(x) ≥ 0 x R

The minimum value of f is 0, and the point of minimum value of f is x = 0.

From the graph of the function, f has no maximum value, and hence, no point of maximum value of f in R.

Suppose the domain of f is restricted to [- 2, 1].

The maximum value of the function, f(-2) = 2(-2)2 = 8, which is at x = -2

Note: A function has an extreme value at a point even if it is not differentiable at that point.

Every monotonic function assumes its maximum/minimum value at the end points of the domain of definition of the function.

A more general result is

Every continuous function on a closed interval has a maximum and a minimum

value.

N.B.: Every increasing or decreasing function assumes its maximum or minimum value at the end points of the domain of definition of the function.

Or

Every continuous function on a closed interval has a maximum and a minimum value.

Similarly, the function has maximum value in some neighbourhood of points B and D which are at the top of their respective hills. For this reason, the points A and C may be regarded as points of local minimum value (or relative minimum value) and points B and D may be regarded as points of local maximum value (or relative maximum value) for the function. The local maximum value and local minimum value of the function are referred to as local maxima and local minima, respectively, of the function.

The graph of the function has minimum values in some neighbourhood(nbd) of points Q and S, and has maximum values in some neighbourhood(nbd) of points P, R and T.

The x-coordinates of points P, R and T are called the points of local maximum, and the y-coordinates of points P, R and T, are the local maximum values of f(x).

Similarly, the x-coordinates of points Q and S are called the points of local minimum, and the y-coordinates of points Q and S, are called the local minimum values of f(x).

The local maximum value of a function is said to be the local maxima of that function.

The local minimum value of a function is said to be the local minima of that function.

• When x = a, if f(x) ≤ f(a) for every x in the domain, then f(x) has an Absolute Maximum value and the point a is the point of the maximum value of f.
• When x = a, if f(x) ≤ f(a) for every x in some open interval (p, q) then f(x) has a Relative Maximum value.
• When x= a, if f(x) ≥ f(a) for every x in the domain then f(x) has an Absolute Minimum value and the point a is the point of the minimum value of f.
• When x = a, if f(x) ≥ f(a) for every x in some open interval (p, q) then f(x) has a Relative Minimum value.

Let the point of local maximum value of f in the graph be x = a. The function f is increasing in

the interval (a - h, a) and decreasing in the interval (a , a+h ), where h > 0.

DefinitionLet f be a real valued function and let c be an interior point in the domain

of f. Then

(a) c is called a point of local maxima if there is an h > 0 such that

f (c) f (x), for all x in (c – h, c + h), x c

The value f (c) is called the local maximum value of f.

(b) c is called a point of local minima if there is an h > 0 such that

f (c) f (x), for all x in (c – h, c + h)

The value f (c) is called the local minimum value of f .

Geometrically, the above definition states that if x = c is a point of local maxima of f,

then the graph of f around c will be as shown in Fig 6.14(a). Note that the function f is

increasing (i.e., f (x) > 0) in the interval (c – h, c) and decreasing (i.e., f (x) < 0) in the

interval (c, c + h).

This suggests that f (c) must be zero.

If f is increasing, then f '(x) > 0, and if f is decreasing, then f '(x) < 0.

If f is neither decreasing nor increasing, then f '(x) = 0. i.e. f '(a) = 0.

Let the point of local minimum value of f in the graph be x = a.

Function f is decreasing in the interval (a - h, a) and increasing in the interval (a , a+h ), where h > 0. we have f '(a) = 0

Monotonicity

Functions are said to be monotonic if they are either increasing or decreasing in their entire domain. f(x) = ex, f(x) = nx, f(x) = 2x + 3 are some examples.

Functions which are increasing and decreasing in their domain are said to be non-monotonic

For example: f(x) = sin x , f(x) = x2

Monotonicity Of A function At A Point

A function is said to be monotonically decreasing at x = a if f(x) satisfy;

f(x + h) < f(a) for a small positive h

• f'(x) will be positive if the function is increasing
• f'(x) will be negative if the function is decreasing
• f'(x) will be zero when the function is at its maxima or minima

Theorem:

Let f be a real valued function defined on an open interval I. Suppose point a is any arbitrary point in I. If f has a local maxima or a local minima at x = a, then either f '(a) = 0, or f is not differentiable at a.

However, the converse need not be true, i.e. a point at which the derivative vanishes need not be the point of local maxima or local minima.

Every continuous function on a closed interval has a maximum and a minimum value.

Point of Inflection:

For continuous function f(x), if f'(x0) = 0 or f’”(x0) does not exist at points where f'(x0) exists and if f”(x) changes sign when passing through x = xthen x0 is called the point of inflection.

(a) If f”(x) < 0, x ∈ (a, b) then the curve y = f(x) in concave downward

(b) if f” (x) > 0, x ∈ (a, b) then the curve y = f(x) is concave upwards in (a, b)

For example: f(x) = sin x

Solution: f'(x) = cos x

f”(x) = sinx = 0 x = nπ, n ∈ z

Critical point:

A point c in the domain of a function f at which either f (c) = 0 or f is not

differentiable is called a critical point of f.

Theorem  (First Derivative Test) : Let f be a function defined on an open interval I.

Let f be continuous at a critical point c in I. Then

(i) If f (x) changes sign from positive to negative as x increases through c, i.e., if  f (x) > 0 at every point sufficiently close to and to the left of c, and f (x) < 0 at every point sufficiently close to and to the right of c, then c is a point of local maxima.

(ii) If f (x) changes sign from negative to positive as x increases through c, i.e., if f (x) < 0 at every point sufficiently close to and to the left of c, and f (x) > 0 at every point sufficiently close to and to the right of c, then c is a point of local minima.

(iii) If f (x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection .

Example : Find all points of local maxima and local minima of the function f

given by

f (x) = x3 – 3x + 3.

Solution We have

f (x) = x3 – 3x + 3.

or f (x) = 3x2 – 3 = 3(x – 1) (x + 1)

or f (x) = 0 at x = 1 and x = – 1

Thus, x = ± 1 are the only critical points which could possibly be the points of local

maxima and/or local minima of f . Let us first examine the point x = 1.

Example ; Find all the points of local maxima and local minima of the function f

given by

f (x) = 2x3 – 6x2 + 6x +5.

Solution We have

f (x) = 2x3 – 6x2 + 6x + 5

or f (x) = 6x2 – 12x + 6 = 6(x – 1)2

or f (x) = 0 at x = 1

Thus, x = 1 is the only critical point of f . We shall now examine this point for local maxima and/or local minima of f. Observe that f (x) 0, for all x R and in particular f (x) > 0, for values close to 1 and to the left and to the right of 1. Therefore, by first derivative test, the point x = 1 is neither a point of local maxima nor a point of local minima. Hence x = 1 is a point of inflexion.

Theorem  (Second Derivative Test):  Let f be a function defined on an interval I and c I. Let f be twice differentiable at c. Then

(i) x = c is a point of local maxima if f (c) = 0 and f (c) < 0 The value f (c) is local maximum value of f .

(ii) x = c is a point of local minima if f (c) 0 and f (c) > 0 In this case, f (c) is local minimum value of f .

(iii) The test fails if f (c) = 0 and f (c) = 0. In this case, we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion.

Example : Find local maximum and local minimum values of the function f given by

f (x) = 3x4 + 4x3 – 12x2 + 12

Solution We have

f (x) = 3x4 + 4x3 – 12x2 + 12

or f (x) = 12x3 + 12x2 – 24x = 12x (x – 1) (x + 2)

or f (x) = 0 at x = 0, x = 1 and x = – 2.

Now f (x) = 36x2 + 24x – 24 = 12(3x2 + 2x – 2)

f ‘’(0)=-24 <0

f ‘’(1)= 36 >0

f ‘’(-2)= 72 >0

Therefore, by second derivative test, x = 0 is a point of local maxima and local maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = – 2 are the points of local minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20, respectively.

Example : If length of three sides of a trapezium other than base are equal to 10cm, then find the area of the trapezium when it is maximum.

Solution The required trapezium is as given in Figure 6.19.

Draw perpendiculars DP and

CQ on AB. Let AP = x cm. Note that APD ~ BQC. Therefore, QB = x cm. Also, by

Pythagoras theorem, DP = QC = √(100-x 2). Let A be the area of the trapezium. Then

AA(x) = (sum of parallel sides) (height)

Theorem:  Let f be a continuous function on an interval I = [a, b]. Then f has the

absolute maximum value and f attains it at least once in I. Also, f has the absolute

minimum value and attains it at least once in I.

Theorem:  Let f be a differentiable function on a closed interval I and let c be any

interior point of I. Then

(i) f (c) = 0 if f attains its absolute maximum value at c.

(ii) f (c) = 0 if f attains its absolute minimum value at c.

In view of the above results, we have the following working rule for finding absolute maximum and/or absolute minimum values of a function in a given closed interval [a, b].

Working Rule

Step 1: Find all critical points of f in the interval, i.e., find points x where either f (x) 0 or f is not differentiable.

Step 2: Take the end points of the interval.

Step 3: At all these points (listed in Step 1 and 2), calculate the values of f .

Step 4: Identify the maximum and minimum values of f out of the values calculated in

Step 3: This maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f .

Example: Find the absolute maximum and minimum values of a function f given by

f (x) = 2x3 – 15x2 + 36x +1 on the interval [1, 5].

Solution We have  f (x) = 2x3 – 15x2 + 36x +1

or f (x) = 6x2 – 30x + 36 = 6(x – 3) (x – 2)

Note that f (x) = 0 gives x = 2 and x = 3.

We shall now evaluate the value of f at these points and at the end points of the

interval [1, 5], i.e., at x = 1, x = 2, x = 3 and at x = 5. So

f (1) = 2(13) – 15(12) + 36(1) + 1 = 24

f (2) = 2(23) – 15(22) + 36(2) + 1 = 29

f (3) = 2(33) – 15(32) + 36(3) + 1 = 28

f (5) = 2(53) – 15(52) + 36(5) + 1 = 56

Thus, we conclude that absolute maximum value of f on [1, 5] is 56, occurring at x =5, and absolute minimum value of f on [1, 5] is 24 which occurs at x = 1.