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CBSE Class 12 Mathematics★ 8-10 marks · CBSE Class 12 Mathematics Chapter 6

Application of Derivatives (Rate of Change, Tangents, Maxima/Minima)

The derivative is not just an abstract concept — it tells you how fast something is CHANGING, the slope of a TANGENT, and where a function reaches its MAXIMUM or MINIMUM. CBSE Class 12 Mathematics Chapter 6.

⏱ 22 min readHard difficulty✓ Free · NCERT-aligned

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By the end of this chapter you'll be able to…

1Compute rates of change and apply to related rates problems (one variable changing w.r.t. another)
2Find the equation of the tangent and normal to a curve at a given point
3Determine intervals where a function is increasing or decreasing using f'(x)
4Find local maxima and minima using both the first and second derivative tests
5Solve optimisation word problems: maximise/minimise area, volume, cost, profit

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Why this chapter matters

Application of Derivatives is the most practically powerful chapter in Class 12 Mathematics. Optimisation problems (find dimensions/quantity that maximise/minimise something) test mathematical reasoning, not just computation. The tangent/normal equation and monotonicity interval problems are reliable board exam questions.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

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Application of Derivatives

"The derivative is the most practical mathematical tool ever invented. It answers: how fast? Which way? What's the best we can do?"

1. Chapter Overview

The derivative is the SWISS ARMY KNIFE of applied mathematics. This chapter covers: Rate of Change (how fast one quantity changes with respect to another), Tangents and Normals (the slope of the curve at a point), Increasing and Decreasing Functions, and Maxima and Minima — finding the HIGHEST and LOWEST values a function attains (optimisation).

2. Rate of Change

If y = f(x), dy/dx = rate of change of y with respect to x
Applications: velocity (ds/dt), acceleration (d²s/dt²), marginal cost/revenue in economics (dC/dq, dR/dq), related rates (one rate linked to another through a relationship)

3. Tangents and Normals

Tangent: Slope = f'(x₀) at point (x₀, y₀). Equation: y — y₀ = f'(x₀)(x — x₀)
Normal: Perpendicular to tangent. Slope = -1/f'(x₀). Equation: y — y₀ = -1/f'(x₀) × (x — x₀)

4. Increasing and Decreasing Functions

f'(x) > 0 ⇒ f is STRICTLY INCREASING
f'(x) < 0 ⇒ f is STRICTLY DECREASING
f'(x) = 0 ⇒ STATIONARY POINT (could be maxima, minima, or inflection)

5. Maxima and Minima (Optimisation)

First Derivative Test

Find f'(x) = 0 → critical points
Check sign of f'(x) to left and right:
- - → — : MAXIMUM
- — → + : MINIMUM
- No sign change: INFLECTION POINT

Second Derivative Test

f''(x) < 0 ⇒ LOCAL MAXIMUM (curve concave down)
f''(x) > 0 ⇒ LOCAL MINIMUM (curve concave up)
f''(x) = 0 ⇒ TEST FAILS (use first derivative test or higher-order derivative)

Absolute (Global) Maxima/Minima on [a,b]

Check: (a) critical points in (a,b), (b) endpoints f(a), f(b). The LARGEST is absolute maximum. The SMALLEST is absolute minimum.

Applications

'Find the dimensions of a box that MAXIMISE volume for a given surface area.'
'At what quantity is profit MAXIMISED?' (MR = MC). This is the foundation of optimisation — one of the most practical skills in mathematics.

6. Exam Focus

Rate of change — related rates problems
Tangent and normal equations
Increasing/decreasing intervals. Sign analysis of f'(x).
Maxima and minima — first derivative test, second derivative test. Absolute extrema on [a,b].
Optimisation word problems — dimensions, profit maximisation, cost minimisation.

7. Conclusion

The derivative answers the most PRACTICAL questions:

HOW FAST is the population growing? (rate of change)
WHAT IS THE SLOPE at this point? (tangent)
IS IT RISING OR FALLING? (increasing/decreasing)
WHAT IS THE BEST POSSIBLE VALUE? (maxima/minima — OPTIMISATION)

'Calculus is the mathematics of optimisation. And optimisation — making the best of what you have — is the central problem of economics, engineering, and life.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Tangent and Normal Equations

At point (x₀, y₀) on curve y = f(x): TANGENT slope = f'(x₀). Tangent equation: y − y₀ = f'(x₀)(x − x₀). NORMAL (perpendicular to tangent): slope = −1/f'(x₀) (undefined if f'(x₀) = 0 → vertical normal). Normal equation: y − y₀ = −(1/f'(x₀))(x − x₀). SPECIAL CASE: If f'(x₀) = 0 → tangent is horizontal (y = y₀), normal is vertical (x = x₀).

For parametric curves x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt) = g'(t)/f'(t). Substitute the parameter value at the given point to get the slope.

Increasing/Decreasing on an Interval

f is STRICTLY INCREASING on I if f'(x) > 0 for all x ∈ I. f is STRICTLY DECREASING on I if f'(x) < 0 for all x ∈ I. PROCEDURE: (1) Find f'(x). (2) Set f'(x) = 0 → find critical points. (3) Test sign of f'(x) in each interval between critical points. (4) State: 'f is increasing on [...] and decreasing on [...].'

CRITICAL POINTS are where f'(x) = 0 or f'(x) doesn't exist. Between consecutive critical points, f'(x) maintains one sign.

First and Second Derivative Tests

FIRST DERIVATIVE TEST: At critical point c: (i) f'(x) changes + → − : LOCAL MAXIMUM at c. (ii) f'(x) changes − → + : LOCAL MINIMUM at c. (iii) f'(x) doesn't change sign: NEITHER (inflection point). SECOND DERIVATIVE TEST: At critical point c (where f'(c) = 0): (i) f''(c) < 0: LOCAL MAXIMUM. (ii) f''(c) > 0: LOCAL MINIMUM. (iii) f''(c) = 0: TEST INCONCLUSIVE (use first derivative test).

ABSOLUTE (GLOBAL) EXTREMA on CLOSED INTERVAL [a,b]: evaluate f at ALL critical points in (a,b) AND at both endpoints f(a), f(b). The largest value is the absolute maximum; smallest is the absolute minimum. DO NOT forget the endpoints!

Optimisation Strategy

STEP 1: Identify the quantity to be maximised/minimised (objective). STEP 2: Identify the constraint. STEP 3: Express the objective as a function of ONE variable (eliminate others using the constraint). STEP 4: Find the derivative; set to zero (or undefined). STEP 5: Apply second derivative test to confirm max/min. STEP 6: State the answer with units. CLASSIC PROBLEMS: Box from rectangle sheet; cylinder inscribed in sphere; fence with minimum material; profit/cost optimisation.

The exam problem always gives both an objective AND a constraint. If you have two variables and no constraint, you're missing something. The constraint eliminates one variable.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Forgetting to check endpoints for absolute extrema on a closed interval

✓ For ABSOLUTE extrema on [a,b]: the maximum/minimum could be at a critical point OR at an endpoint. You MUST evaluate f at a, f at b, AND f at all critical points in (a,b). The largest of all these values is the absolute maximum. Forgetting endpoints gives the wrong answer.

WATCH OUT

✗ Using the second derivative test when f''(c) = 0

✓ When f''(c) = 0, the second derivative test is INCONCLUSIVE — it cannot tell you whether c is a maximum, minimum, or inflection point. You MUST then use the FIRST DERIVATIVE TEST: check the sign of f'(x) on either side of c.

NCERT exercises (with solutions)

Every NCERT exercise from this chapter — what it covers and how many questions to expect.

Ex s 6.1–6.6

Exercises 6.1–6.6

Rate of change problems; tangent and normal equations; monotonicity intervals; local/absolute extrema; optimisation word problems (box, fence, cylinder, profit)

Questions

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· tangent-normal

Find the equations of the tangent and normal to the curve y = x² − 3x − 2 at the point (2, −4).

▶ Show solution

f(x) = x² − 3x − 2. f'(x) = 2x − 3. At x = 2: slope of tangent = f'(2) = 2(2) − 3 = 1. VERIFY POINT: f(2) = 4 − 6 − 2 = −4 ✓ (2, −4) is on the curve. TANGENT (slope 1, through (2,−4)): y − (−4) = 1(x − 2) → y + 4 = x − 2 → y = x − 6. NORMAL (slope = −1/1 = −1, through (2,−4)): y − (−4) = −1(x − 2) → y + 4 = −x + 2 → y = −x − 2.

Q2MEDIUM· optimisation

A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window that will admit maximum light through the whole opening.

▶ Show solution

Let width = 2r (so the semicircle has radius r) and height of rectangle = h. PERIMETER: The perimeter consists of: bottom (2r) + two sides (2h) + semicircular arc (πr) = 10. Constraint: 2r + 2h + πr = 10 → 2h = 10 − 2r − πr → h = (10 − 2r − πr)/2 = 5 − r − πr/2. AREA (objective, to maximise): A = 2r·h + πr²/2 (rectangle + semicircle). Substitute h: A(r) = 2r(5 − r − πr/2) + πr²/2 = 10r − 2r² − πr² + πr²/2 = 10r − 2r² − πr²/2. dA/dr = 10 − 4r − πr = 10 − r(4+π). Set dA/dr = 0: r = 10/(4+π). d²A/dr² = −(4+π) < 0 → MAXIMUM. h = 5 − r − πr/2 = 5 − r(1 + π/2) = 5 − [10/(4+π)] × (2+π)/2 = 5 − 5(2+π)/(4+π) = [5(4+π) − 5(2+π)]/(4+π) = 10/(4+π) = r. RESULT: Width = 2r = 20/(4+π) m, Height = h = 10/(4+π) m = r (height equals radius).

Q3HARD· long-answer

Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.

▶ Show solution

Let the cone have height h and base radius r, inscribed in a sphere of radius R. The cone's apex is at the top of the sphere, and the centre of the base circle is at distance (h−R) from the sphere's centre (or R−h if h < R... let's set it up carefully). SETUP: Place sphere centre at O. Let cone have apex at top of sphere, height h. The centre of the base is at distance (h−R) from O. The base radius r satisfies: r² + (h−R)² = R² (Pythagoras in sphere). So r² = R² − (h−R)² = R² − h² + 2hR − R² = 2hR − h². VOLUME: V = (1/3)πr²h = (1/3)π(2hR−h²)h = (π/3)(2Rh²−h³). OPTIMISE: dV/dh = (π/3)(4Rh − 3h²) = (πh/3)(4R − 3h). Set dV/dh = 0: h = 0 (trivial) or h = 4R/3. d²V/dh² = (π/3)(4R − 6h). At h = 4R/3: d²V/dh² = (π/3)(4R − 8R) = (π/3)(−4R) < 0. MAXIMUM at h = 4R/3. MAXIMUM VOLUME: r² = 2(4R/3)R − (4R/3)² = 8R²/3 − 16R²/9 = 24R²/9 − 16R²/9 = 8R²/9. V_max = (π/3)(8R²/9)(4R/3) = 32πR³/81. SPHERE VOLUME: V_sphere = (4/3)πR³ = 108πR³/81. RATIO: V_max/V_sphere = (32πR³/81)/(108πR³/81) = 32/108 = 8/27. The maximum cone volume = (8/27) × sphere volume. ✓

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Tangent slope = f'(x₀). Normal slope = −1/f'(x₀).
•Increasing: f'(x) > 0. Decreasing: f'(x) < 0. Critical point: f'(x) = 0 or undefined.
•First derivative test: +→− = max; −→+ = min.
•Second derivative test: f''(c) < 0 = max; f''(c) > 0 = min; f''(c) = 0 = inconclusive.
•Absolute extrema on [a,b]: evaluate f at all critical points AND both endpoints.
•Optimisation: identify objective, identify constraint, eliminate variable, differentiate, solve, verify.
•Classic results: largest rectangle in circle = square. Largest cylinder in sphere: h = 2R/√3.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 8-10 marks

Question type	Marks each	Typical count	What it tests
Tangent/Normal or Rate of Change	3	1	Tangent and normal equations; rate of change (area, volume, related rates)
Monotonicity or Extrema	3-4	1	Find increasing/decreasing intervals; find local max/min by both derivative tests
Optimisation Word Problem	5-6	1	Box, cylinder, rectangle, profit/cost optimisation — find dimensions/quantities for max/min

Prep strategy

Optimisation problems: the CONSTRAINT is always given (total perimeter, total surface area, total cost). Use it to eliminate one variable so the objective function has only one variable. Then differentiate, set to zero, apply second derivative test.
For monotonicity: find critical points, draw a sign chart for f'(x), and state the intervals explicitly using inequality/bracket notation.
Tangent problems: always VERIFY the given point is actually on the curve by substituting into the equation. Then compute slope, then write equation.

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Profit Maximisation in Business

Every business decision about pricing, quantity, and resource allocation is an application of derivatives optimisation. The rule 'Marginal Revenue = Marginal Cost' (MR = MC) is literally f'(Revenue) = f'(Cost) — the profit function P = R − C has P'(q) = R'(q) − C'(q) = 0 at the optimal quantity. Every Economics textbook derives this from the same second derivative test used in this chapter.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Optimisation problems always end with 'verify it's a maximum/minimum.' Apply the second derivative test and write: 'Since d²V/dr² = [value] < 0, V is maximum.' This verification line earns 1 mark and shows rigour.
For related rates: identify ALL variables changing with time, write an equation relating them, differentiate BOTH SIDES w.r.t. t (using chain rule), substitute given values and rates to find the unknown rate.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Study the BRACHISTOCHRONE PROBLEM — Bernoulli's 1696 challenge: find the curve along which a bead slides from one point to another in minimum time under gravity. The answer (a cycloid) was found independently by Newton, Leibniz, and Bernoulli using calculus of variations — the extension of derivatives to optimise over FUNCTIONS rather than numbers
The KARUSH-KUHN-TUCKER (KKT) CONDITIONS are the generalisation of 'set derivative to zero' for constrained optimisation in multiple variables — the foundation of machine learning, linear programming, and operations research

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 12 Board (Mathematics)High

JEE Main (AOD — Application of Derivatives)High

CUET (Mathematics)High

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Questions students ask

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An INFLECTION POINT is a point where the curve changes its CONCAVITY (from concave up to concave down, or vice versa). It is where f''(x) = 0 AND f'' changes sign. At an inflection point, f' does NOT change sign (unlike at a max or min). A maximum or minimum is where f'(x) = 0 AND f' changes sign. Key difference: at a local max/min, f' changes sign. At an inflection point, f'' changes sign but f' does not.

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