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Mathematics Grade XII

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Chapter 6: Application of Derivatives

Class 12 • Mathematics • Chapter 6

Application of Derivatives

Using the slope of a curve to find rates, peaks, valleys and the best possible answer.

CBSE / NCERT Aligned

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Chapter Roadmap

Derivative as a Rate of Change • Increasing and Decreasing Functions • Tangents and Normals • Maxima and Minima • First and Second Derivative Tests • Absolute Extrema on an Interval

Turning Slopes into Decisions

Once you can find a derivative, a whole world of practical questions opens up. How fast is something changing right now? Where is a quantity rising and where is it falling? What is the largest profit or the smallest cost possible? Each of these is answered by reading the derivative, because the derivative is the instantaneous rate of change and the slope of the tangent line.

In this chapter the derivative becomes a decision-making tool. You will compute rates of change, decide where functions increase or decrease, write the equations of tangents and normals, and locate maximum and minimum values using the first and second derivative tests. The same ideas drive optimisation in business, engineering and science.

Foundation

The sign of f′(x) tells you the behaviour: f′(x) > 0 means f is increasing, f′(x) < 0 means f is decreasing, and f′(x) = 0 marks a stationary point that may be a maximum, a minimum or neither.

Key Terms You Must Know

Term

Meaning

Example

Rate of change

How fast one quantity changes with respect to another, given by a derivative.

dV/dt for volume over time

Increasing function

Values rise as x rises; here f′(x) > 0.

f(x) = x³ rises throughout

Tangent

The line touching a curve with the curve’s slope at that point.

slope = f′ at the point

Normal

The line at a point perpendicular to the tangent.

slope = −1/f′

Stationary point

A point where f′(x) = 0.

peaks and valleys

Local maximum/minimum

A highest or lowest value in a small neighbourhood.

top of a hill or bottom of a dip

Core Concepts, Step by Step

1. The Derivative as a Rate of Change

If a quantity y depends on x, then dy/dx is the rate at which y changes per unit change in x. When the variable is time, derivatives become speeds and growth rates: dV/dt is how fast a volume grows, dA/dt how fast an area spreads. Related-rate problems link several changing quantities through one equation and then differentiate it with respect to time.

2. Increasing and Decreasing Functions

Where the derivative is positive the function climbs; where it is negative the function falls. To find these intervals, set f′(x) = 0 to locate the turning points, then test the sign of f′ on each side. This converts a question about the whole graph into simple sign checks at a few points.

3. Tangents and Normals

The tangent at a point has slope f′ evaluated there, and its equation follows from the point-slope form. The normal is perpendicular to the tangent, so its slope is the negative reciprocal, −1/f′. Tangents and normals describe how a curve behaves locally and appear in optics, design and physics.

The tangent shares the curve’s slope; the normal is perpendicular to it

Parabola y equals x squared at point (1,1) with tangent line slope 2 and perpendicular normal line

4. Maxima and Minima

Peaks and valleys of a smooth curve occur where the tangent is horizontal, that is where f′(x) = 0. These stationary points are the candidates for local maxima and minima. Not every stationary point is an extremum, so each candidate must be tested, which is what the next two tests do.

A cubic with a local maximum and a local minimum, each with a horizontal tangent

Cubic curve y equals x cubed minus 3x showing a local maximum at minus one comma two and a local minimum at one comma minus two

5. First and Second Derivative Tests

The first derivative test checks how the sign of f′ changes across a stationary point: from positive to negative gives a local maximum, from negative to positive gives a local minimum. The second derivative test is often quicker: at a stationary point, f″ > 0 means a minimum and f″ < 0 means a maximum; if f″ = 0 the test is inconclusive and you fall back to the first test.

6. Absolute Extrema and Optimisation

On a closed interval the largest and smallest values of a continuous function occur either at a stationary point inside or at an end of the interval. So we list the function’s values at all stationary points and at both endpoints and pick the biggest and smallest. This is the heart of optimisation: shaping a real problem into a function and finding its best value.

Key Results with Proofs

Result 1: Stationary Point at a Local Extremum

Statement. If f has a local maximum or minimum at an interior point c and f is differentiable there, then f′(c) = 0.

Proof

At a smooth peak the tangent cannot tilt either way, so it must be flat.

Suppose c is a local maximum, so f(c) is at least as large as nearby values. – assume a local maximum

From the left the slope cannot be negative without values rising past f(c); from the right it cannot be positive. – squeeze the one-sided slopes

The left limit gives f′(c) ≥ 0 and the right limit gives f′(c) ≤ 0. – two inequalities

Both can hold only if f′(c) = 0. – combine

This is why we always begin by solving f′(x) = 0 to find candidates.

Result 2: The First Derivative Test

Statement. At a stationary point c, if f′ changes from positive to negative then c is a local maximum; if from negative to positive then a local minimum.

Proof

The behaviour of the function follows directly from the sign of its slope on each side.

To the left of a maximum the function is rising, so f′ > 0. – left of a peak

To the right it is falling, so f′ < 0. – right of a peak

A change of sign from + to − therefore marks a local maximum. – read the sign change

The opposite change, − to +, marks a local minimum. – mirror case

If f′ does not change sign, the point is neither a maximum nor a minimum.

Result 3: The Second Derivative Test

Statement. At a stationary point c where f′(c) = 0: if f″(c) > 0 then c is a local minimum, and if f″(c) < 0 then c is a local maximum.

Proof

The sign of the second derivative reveals which way the curve bends at the stationary point.

f″ measures how the slope is changing. – meaning of the second derivative

If f″(c) > 0 the slope is increasing through 0, going from negative to positive, a valley. – positive second derivative

If f″(c) < 0 the slope is decreasing through 0, going from positive to negative, a peak. – negative second derivative

If f″(c) = 0 the test gives no answer and the first derivative test is used instead. – inconclusive case

For f(x) = x³ the point x = 0 has f″ = 0 and is not an extremum, showing why the test can be inconclusive.

Worked Examples

Example 1

Question: The side of a square increases at 2 cm/s. How fast is its area growing when the side is 10 cm?

▶ Show full working

Differentiate A = s² with respect to time.

A = s², so dA/dt = 2s (ds/dt). – relate the rates

Substitute s = 10 and ds/dt = 2: dA/dt = 2(10)(2). – put in numbers

= 40 cm²/s. – result

Answer: The area grows at 40 cm²/s.

Example 2

Question: Find where f(x) = x² − 4x is increasing and where it is decreasing.

▶ Show full working

Use the sign of f′.

f′(x) = 2x − 4; set it to 0 to get x = 2. – find the turning point

For x > 2, f′ > 0 so f is increasing; for x < 2, f′ < 0 so f is decreasing. – test the sign on each side

Answer: Increasing on x > 2, decreasing on x < 2.

Example 3

Question: Find the equation of the tangent to y = x² at the point (2, 4).

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Slope is f′ at the point.

f′(x) = 2x, so the slope at x = 2 is 4. – slope

Point-slope form: y − 4 = 4(x − 2), giving y = 4x − 4. – write the line

Answer: y = 4x − 4.

Example 4

Question: Find the equation of the normal to y = x² at the point (2, 4).

▶ Show full working

Normal slope is the negative reciprocal.

Tangent slope is 4, so the normal slope is −1/4. – perpendicular slope

y − 4 = −(1/4)(x − 2). – point-slope form

Answer: y − 4 = −(1/4)(x − 2).

Example 5

Question: Find the local maximum and minimum of f(x) = x³ − 3x.

▶ Show full working

Solve f′ = 0, then use the second derivative test.

f′(x) = 3x² − 3 = 0 gives x = 1 and x = −1. – stationary points

f″(x) = 6x. At x = 1, f″ > 0, a minimum with value 1 − 3 = −2. – minimum

At x = −1, f″ < 0, a maximum with value −1 + 3 = 2. – maximum

Answer: Local maximum 2 at x = −1; local minimum −2 at x = 1.

Example 6

Question: Use the second derivative test on f(x) = x² − 4x + 3.

▶ Show full working

Find the stationary point and check f″.

f′(x) = 2x − 4 = 0 gives x = 2. – stationary point

f″(x) = 2 > 0, so x = 2 is a minimum. – second derivative test

Value: 4 − 8 + 3 = −1. – minimum value

Answer: Minimum value −1 at x = 2.

Example 7

Question: Find the absolute maximum and minimum of f(x) = x³ − 3x on [0, 2].

▶ Show full working

Check stationary points inside and both endpoints.

f′ = 3x² − 3 = 0 gives x = 1 (inside the interval). – stationary point in range

Values: f(0) = 0, f(1) = −2, f(2) = 8 − 6 = 2. – evaluate at all candidates

Largest is 2 at x = 2; smallest is −2 at x = 1. – pick extremes

Answer: Absolute maximum 2 at x = 2; absolute minimum −2 at x = 1.

Example 8

Question: A rectangle has perimeter 20 m. What dimensions give the greatest area?

▶ Show full working

Express area as a function of one side.

If one side is x, the other is 10 − x, so area A = x(10 − x) = 10x − x². – build the function

A′ = 10 − 2x = 0 gives x = 5; A″ = −2 < 0 confirms a maximum. – optimise

Both sides are 5 m and the area is 25 m². – interpret

Answer: A 5 m by 5 m square, area 25 m².

Example 9

Question: A sphere’s volume is V = (4/3)πr³. Find dV/dr at r = 2.

▶ Show full working

Differentiate with respect to r.

dV/dr = 4πr². – derivative

At r = 2: 4π(4) = 16π. – substitute

Answer: dV/dr = 16π.

Example 10

Question: A cost function is C(x) = x² + 5x + 10. Find the marginal cost at x = 10.

▶ Show full working

Marginal cost is C′(x).

C′(x) = 2x + 5. – differentiate

At x = 10: 2(10) + 5 = 25. – substitute

Answer: Marginal cost = 25.

Example 11

Question: Show that f(x) = x³ + x is increasing for all real x.

▶ Show full working

Check the sign of f′ everywhere.

f′(x) = 3x² + 1. – derivative

Since 3x² ≥ 0, we have f′(x) ≥ 1 > 0 for every x. – always positive

Answer: f is increasing everywhere.

Example 12

Question: Where does the curve y = x² − 6x + 5 have a horizontal tangent?

▶ Show full working

Horizontal tangent means f′ = 0.

f′(x) = 2x − 6 = 0 gives x = 3. – solve

Then y = 9 − 18 + 5 = −4, so the point is (3, −4). – find the point

Answer: At (3, −4).

Where You Meet This in Real Life

Business

Companies maximise profit and minimise cost by setting the derivative of the relevant function to zero.

Engineering design

The strongest beam, the least material packaging and the smoothest cam profile are all optimisation problems solved with derivatives.

Physics of motion

Speed and acceleration are derivatives, and the highest point of a thrown ball is where the velocity is momentarily zero.

Medicine and biology

Models find the dose or time at which a drug concentration peaks by locating where its rate of change is zero.

Logistics

Delivery routes and inventory levels are tuned to minimise total cost using the same maximum and minimum methods.

Practice Sets A–D

Practice Set A – Basics

A1. Find f′(x) for f(x) = x² − 6x.

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Power rule.

f′(x) = 2x − 6.

Answer: 2x − 6.

A2. What does f′(x) > 0 tell you about f?

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Sign meaning.

The function is increasing there.

Answer: It is increasing.

A3. Find the slope of the tangent to y = x² at x = 3.

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Use f′.

f′(x) = 2x, so at x = 3 the slope is 6.

Answer: 6.

A4. At a stationary point, what is the value of f′?

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Definition.

f′ = 0 there.

Answer: Zero.

Practice Set B – Conceptual

B1. Why does a smooth local maximum have f′ = 0?

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Think about the tangent.

At a smooth peak the tangent is horizontal, so its slope, which is f′, is 0.

Answer: Because the tangent is horizontal at a smooth peak.

B2. How does the second derivative test distinguish a maximum from a minimum?

▶ Reveal full working

Sign of f″.

At a stationary point, f″ > 0 means a minimum and f″ < 0 means a maximum.

Answer: By the sign of f″ at the stationary point.

B3. Why must we also check endpoints for absolute extrema on a closed interval?

▶ Reveal full working

Extremes can hide at the ends.

The largest or smallest value may occur at an end rather than at a stationary point inside.

Answer: Because extreme values can occur at the endpoints.

B4. What is the slope of the normal if the tangent slope is 2?

▶ Reveal full working

Negative reciprocal.

Normal slope = −1/2.

Answer: −1/2.

Practice Set C – Application / Numerical

C1. Find the interval on which f(x) = 6 + 12x + 3x² − 2x³ is increasing.

▶ Reveal full working

Solve f′ > 0.

f′(x) = 12 + 6x − 6x² = −6(x² − x − 2) = −6(x − 2)(x + 1). – factor

This is positive between the roots, so f increases on −1 < x < 2. – sign chart

Answer: Increasing on (−1, 2).

C2. Find the local extrema of f(x) = 2x³ − 6x.

▶ Reveal full working

Solve f′ = 0 and use f″.

f′ = 6x² − 6 = 0 gives x = ±1. – stationary points

f″ = 12x: at x = 1 minimum f(1) = −4; at x = −1 maximum f(−1) = 4. – test

Answer: Max 4 at x = −1, min −4 at x = 1.

C3. A balloon’s volume V = (4/3)πr³ grows so r increases at 0.5 cm/s. Find dV/dt at r = 3.

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Chain rule in time.

dV/dt = 4πr² (dr/dt). – relate rates

= 4π(9)(0.5) = 18π cm³/s. – substitute

Answer: 18π cm³/s.

C4. Find the absolute maximum of f(x) = x² on [−3, 2].

▶ Reveal full working

Check the endpoints and any stationary point.

f′ = 2x = 0 at x = 0, where f = 0. – stationary point

Endpoints: f(−3) = 9, f(2) = 4; the largest is 9. – compare

Answer: Absolute maximum 9 at x = −3.

Practice Set D – HOTS / Multi-step

D1. A rectangular box with a square base and no top must hold 32 m³. Minimise the material (surface area).

▶ Reveal full working

Express surface area in one variable.

Base side x, height h with x²h = 32, so h = 32/x². – constraint

Surface S = x² + 4xh = x² + 128/x; S′ = 2x − 128/x² = 0 gives x³ = 64, x = 4. – optimise

Then h = 32/16 = 2; the minimum-material box has base 4 m and height 2 m. – interpret

Answer: Base 4 m, height 2 m.

D2. Find the point on y = x² closest to the point (0, 3) by minimising the squared distance.

▶ Reveal full working

Minimise D(x) = x² + (x² − 3)².

D(x) = x² + (x² − 3)²; D′(x) = 2x + 2(x² − 3)(2x) = 2x(2x² − 5). – differentiate

Setting D′ = 0: x = 0 or x² = 5/2. – solve

Checking values, x² = 5/2 gives the closest points, at (±√(5/2), 5/2). – choose the minimum

Answer: The points (±√(5/2), 5/2).

D3. Show that of all rectangles with a fixed perimeter, the square has the greatest area.

▶ Reveal full working

Set perimeter 2(x + y) = constant and maximise area.

With perimeter fixed, y = p − x where p is half the perimeter; area A = x(p − x). – one variable

A′ = p − 2x = 0 gives x = p/2, so y = p/2 = x. – optimise

Equal sides mean a square, and A″ = −2 < 0 confirms it is the maximum. – conclude

Answer: The square gives the maximum area.

D4. Verify the second derivative test is inconclusive for f(x) = x⁴ at x = 0, and find the nature of the point.

▶ Reveal full working

Compute f′, f″ at 0, then reason.

f′ = 4x³ = 0 at x = 0; f″ = 12x² = 0 at x = 0, so the test is inconclusive. – test fails

But f′ changes from negative to positive across 0, so it is a local minimum. – first derivative test

Indeed f(0) = 0 is the smallest value of x⁴. – confirm

Answer: x = 0 is a local minimum (test inconclusive, first test confirms).

Chapter Summary

Everything in One Glance

Rate of Change

The derivative gives instantaneous rates; related-rate problems differentiate one linking equation in time.

Increasing/Decreasing

f′ > 0 rises, f′ < 0 falls; turning points are where f′ = 0.

Tangent & Normal

Tangent slope is f′; the normal slope is −1/f′.

Maxima & Minima

Stationary points solve f′ = 0 and are the candidates for extrema.

The Two Tests

First test reads the sign change of f′; second test reads the sign of f″.

Optimisation

On a closed interval compare stationary values and endpoint values to find absolute extrema.

Are You Exam-Ready?

8-Point Exam Quick-Check

Find f′(x) for f(x) = x³ − 3x².

On which interval is f(x) = x² decreasing?

Find the tangent slope to y = x³ at x = 2.

What is the normal slope if the tangent slope is 5?

Find the stationary points of f(x) = x³ − 12x.

Use the second derivative test on f(x) = x² + 2x.

A circle’s area is A = πr²; find dA/dr at r = 4.

Find the maximum area of a rectangle with perimeter 16.

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Class 12 Mathematics Chapter 6: Application of Derivatives, Complete Notes and Practice

These free Class 12 Maths Application of Derivatives notes follow the NCERT 2026 to 27 syllabus and cover rates of change, increasing and decreasing functions, tangents and normals, maxima and minima, the first and second derivative tests and optimisation, with twelve worked examples and sixteen graded practice questions. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.