Class 12 Applications of Derivatives Maths Chapter 6 Notes By PW

Applying differentiation to practical problems can be challenging when you need to connect mathematical formulas with changing quantities, motion or the behaviour of functions. You must understand how derivatives indicate increasing and decreasing trends, how critical points help identify maxima and minima, and how to form equations for optimisation problems.

A clear understanding of these concepts requires both formula revision and regular problem-solving practice. PW supports your Class 12 learning through live classes, recorded lectures, and tests. These resources provide opportunities to revisit concepts, practise questions and assess your preparation alongside your board studies.

Key Concepts Covered in Applications of Derivatives Class 12

Applications of Derivatives is broadly divided into three main areas: rate of change of quantities, increasing and decreasing functions, and maxima and minima. Each concept uses differentiation to analyse how a function behaves and solve mathematical or practical problems.

Rate of Change of Quantities

The rate of change of a quantity tells you how quickly it changes with respect to another variable, such as time.

If y depends on x, its rate of change with respect to x is: dy/dx = f′(x)

When a quantity changes with time t, its rate of change is represented by: dy/dt

For example, if the radius of a circle increases at 2 cm/s, then: dr/dt = 2 cm/s

Here, dr/dt represents the rate at which the radius changes with time.

Rate of Change of the Area of a Circle

The area of a circle is: A = πr²

Differentiating both sides with respect to time: dA/dt = 2πr · (dr/dt)

Given dr/dt = 2 cm/s, substituting the value:

dA/dt = 2πr × 2

dA/dt = 4πr cm²/s

The rate of change of area depends on the radius at that particular instant.

Rate of Change of the Circumference of a Circle

The circumference of a circle is: C = 2πr

Differentiating with respect to time: dC/dt = 2π · (dr/dt)

Substituting dr/dt = 2: dC/dt = 4π cm/s

Remember: Always include the correct units. The rate of change of area has units of area per unit time, while the rate of change of circumference has units of length per unit time.

Positive and Negative Rates of Change

The sign of the derivative indicates the direction in which a quantity changes:

• If dy/dt > 0, the quantity is increasing.

• If dy/dt < 0, the quantity is decreasing.

• If dy/dt = 0, the instantaneous rate of change is zero.

Steps to Solve Related-Rate Problems

Related-rate questions involve two or more quantities connected through a mathematical relationship.

Follow these steps:

1. Identify the quantities that are changing.

2. Write the formula connecting those quantities.

3. Differentiate the equation with respect to time.

4. Substitute the given rate of change.

5. Substitute the specified values of the variables.

6. Calculate the required rate and include the correct units.

7. Interpret the sign of the answer.

Important applications:

• Cube: Use the volume or surface-area formula to establish the relationship between the edge and the required quantity.

• Isosceles triangle: Draw a perpendicular from the vertex to the base. It bisects the base, allowing you to use the Pythagorean theorem to relate the height, base and equal sides.

• Cone: Use similar triangles to establish the relationship between the radius and height before differentiating the volume formula.

For problems involving multiple changing quantities, establish their relationship before differentiating. Substituting numerical values too early can make the calculation more difficult.

Applications of Derivatives in Motion

Derivatives help describe the motion of a particle by connecting displacement, velocity and acceleration.

If x represents displacement and t represents time, velocity is: v = dx/dt

Acceleration is the rate of change of velocity with respect to time: a = dv/dt

Since velocity is the first derivative of displacement: a = d²x/dt²

When Is a Particle at Rest?

A particle is at rest when its instantaneous velocity is zero.

Therefore: dx/dt = 0

If you need to find when a particle starts from rest or stops moving, solve the velocity equation and examine the valid time values.

For example, if: v(t) = 3t² − 12t

For a particle at rest:

3t² − 12t = 0

3t(t − 4) = 0

Thus: t = 0 or t = 4

The particle is at rest at t = 0 and t = 4. The first value represents the initial instant, while the second represents a later instant.

Remember: At rest means velocity is zero, not necessarily acceleration.

Increasing and Decreasing Functions

The derivative of a function indicates its slope and helps determine whether the function is increasing or decreasing over an interval.

A function f(x) is increasing on an interval if: x₁ < x₂ ⇒ f(x₁) ≤ f(x₂)

It is strictly increasing if: x₁ < x₂ ⇒ f(x₁) < f(x₂)

Similarly, a function is decreasing if: x₁ < x₂ ⇒ f(x₁) ≥ f(x₂)

It is strictly decreasing if: x₁ < x₂ ⇒ f(x₁) > f(x₂)

Derivative Test for Increasing and Decreasing Functions

For a differentiable function on an interval, the following conditions help determine its behaviour:

The derivative test is applied over the entire interval. A derivative equal to zero at an isolated point does not necessarily mean that the function is constant or has an extremum.

Steps to Find Increasing and Decreasing Intervals

1. Find the first derivative f′(x).

2. Solve f′(x) = 0 and identify points where the derivative is undefined, if they lie in the domain.

3. Divide the domain into intervals using these critical values.

4. Choose a test value from each interval.

5. Determine the sign of f′(x) in each interval.

6. State the increasing and decreasing intervals.

Example:

Consider: f′(x) = 6(x − 1)(x − 2)

To find the critical values: 6(x − 1)(x − 2) = 0

Therefore: x = 1, x = 2

These values divide the number line into three intervals:

• (−∞, 1)

• (1, 2)

• (2, ∞)

Now check the sign of the derivative:

Therefore:

• The function is increasing on (−∞, 1) and (2, ∞).

• The function is decreasing on (1, 2).

Parameter-Based Questions

Some questions ask you to find the value of a parameter for which a function is increasing or decreasing on a specified interval.

For an increasing function, check: f′(x) ≥ 0

For a decreasing function, check: f′(x) ≤ 0

Useful mathematical properties include:

For trigonometric expressions: For an expression of the form A sin x + B cos x, its range is:

−√(A² + B²) ≤ A sin x + B cos x ≤ √(A² + B²)

For exponential expressions:

eˣ > 0 for all real x

For algebraic expressions:

x² ≥ 0

1 + x² ≥ 1

These properties can simplify inequalities involving derivatives and help determine the required parameter values.

Maxima and Minima

Maxima and minima refer to the greatest and least values of a function over a specified domain or interval.

• A local maximum is a value greater than or equal to the nearby function values.

• A local minimum is a value less than or equal to the nearby function values.

• An absolute maximum is the greatest value over the entire specified domain.

• An absolute minimum is the least value over the entire specified domain.

Examples:

1. For f(x) = x², the minimum value is 0 at x = 0. It has no maximum over the real numbers.

2. For f(x) = |x + 2| − 1, the minimum value is −1 at x = −2.

3. For f(x) = 3x² + 6x + 8:
   f(x) = 3(x + 1)² + 5
   Since (x + 1)² ≥ 0, the minimum value is:
   5 at x = −1

4. For f(x) = sin x:
   −1 ≤ sin x ≤ 1
   Therefore, its maximum value is 1 and its minimum value is −1.

Maximum and Minimum of Trigonometric Expressions

Trigonometric identities can simplify expressions and help determine their ranges.

For example: sin x cos x = ½ sin 2x

Since: −1 ≤ sin 2x ≤ 1

We get: −½ ≤ sin x cos x ≤ ½

Thus, the maximum value is ½, and the minimum value is −½.

Point of Extrema and Extreme Value

The point at which a function attains a maximum or minimum is called the point of extrema. The corresponding function value is the maximum or minimum value.

For example, consider: f(x) = (x − 2)² + 3

The function attains its minimum at x = 2, so:

• Point of minimum: x = 2

• Minimum value: f(2) = 3

First-Derivative Test

The first-derivative test identifies local maxima and minima by examining the sign of f′(x) on either side of a critical point.

First, find the critical points by solving: f′(x) = 0

Then check whether the derivative changes sign around each point.

Memory Tip:

• Positive to negative: The function increases and then decreases, giving a local maximum.

• Negative to positive: The function decreases and then increases, giving a local minimum.

Second-Derivative Test

The second-derivative test uses the value of f″(x) at a critical point to identify the nature of the extremum.

Suppose f′(c) = 0.

• If f″(c) > 0, the function has a local minimum at x = c.

• If f″(c) < 0, the function has a local maximum at x = c.

• If f″(c) = 0, the test is inconclusive. Use the first-derivative test or another method.

Example:

Consider: f(x) = x² − 4x + 5

First derivative: f′(x) = 2x − 4

At a critical point: 2x − 4 = 0 ⇒ x = 2

Second derivative: f″(x) = 2

Since f″(2) = 2 > 0, the function has a local minimum at x = 2.

The minimum value is: f(2) = 2² − 4(2) + 5 = 1

Absolute Maxima and Minima on a Closed Interval

For a continuous function on a closed interval [a, b], the absolute maximum and minimum can be found by comparing its values at the critical points and endpoints.

Follow these steps:

1. Find f′(x).

2. Identify all critical points inside (a, b).

3. Calculate the function value at each critical point.

4. Calculate the function values at both endpoints, a and b.

5. Compare all the values.

6. The greatest value is the absolute maximum, and the least value is the absolute minimum.

This method is important because a local maximum or minimum is not necessarily the absolute maximum or minimum on the interval.

Optimisation Word Problems

Optimisation problems involve finding the maximum or minimum value of a quantity under given conditions. These questions often use practical situations involving area, volume, cost, revenue or profit.

The main challenge is to form the correct mathematical function before applying differentiation.

General Steps to Solve Optimisation Problems

1. Identify the quantity that needs to be maximised or minimised.

2. Define the variables involved.

3. Establish the relationship between the variables using the given conditions.

4. Express the required quantity in terms of a single variable.

5. Differentiate the function.

6. Set the first derivative equal to zero to find critical points.

7. Check the valid domain and reject physically impossible values.

8. Use the first- or second-derivative test to verify the result.

9. Substitute the valid value into the original function to obtain the required maximum or minimum.

Important Optimisation Examples

1. Profit Problems

Profit is calculated as: Profit = Revenue − Cost

If R(x) represents revenue and C(x) represents cost, then: P(x) = R(x) − C(x)

To find the maximum profit, differentiate: P′(x) = R′(x) − C′(x)

Then solve: P′(x) = 0

Verify the critical point using the derivative tests.

2. Open Box Made From a Square Sheet

Suppose a square sheet has side length a. Squares of side length x are cut from each corner and the sides are folded to form an open box.

The dimensions of the box are:

• Length: a − 2x

• Breadth: a − 2x

• Height: x

Therefore, its volume is: V = x(a − 2x)²

The physically valid range is: 0 < x < a/2

To find the maximum volume, differentiate V with respect to x, equate the derivative to zero and verify the valid critical point.

3. Maximum Volume of an Open Cylindrical Tumbler

For an open cylindrical tumbler with a fixed total surface area S, the surface area is:

S = πr² + 2πrh

where r is the radius and h is the height.

The volume is: V = πr²h

Using the surface-area equation: h = (S − πr²) / (2πr)

Substituting this into the volume equation: V = Sr/2 − πr³/2

Differentiating with respect to r: dV/dr = S/2 − 3πr²/2

For maximum volume, dV/dr = 0: S = 3πr²

Using the surface-area equation, this gives: h = r

Thus, for an open cylindrical tumbler with fixed surface area, the volume is maximum when its height equals its radius.

Applications of Derivatives connect differentiation with practical problems involving changing quantities, motion, function behaviour and optimisation. Understanding the derivative tests, practising related-rate questions and learning to identify maxima and minima can help you approach different question types with greater clarity.

For Class 12 preparation, regular revision and question practice are important alongside conceptual understanding. PW supports CBSE Class 12 preparation through live classes, recorded lectures, and tests. You can use live sessions to learn concepts, recorded lectures to revisit difficult topics and practice resources to assess your understanding.