Application of Derivatives Class 12 Notes

These application of derivatives class 12 notes compress NCERT Chapter 6 of Mathematics Part I into the ideas that matter for revision: what the derivative measures, the chain-rule shortcut for related rates, and the sign-table test for increasing and decreasing functions.

Work through the two original worked examples, drill the five-entry mistake table, then finish with the night-before recap. Every formula below carries its symbols and units, so you can revise without flipping back to the textbook.

Application of Derivatives Class 12 Notes: The Derivative’s Four Jobs

Chapter 5 taught you how to differentiate; Chapter 6 asks what the derivative is good for. The introduction (NCERT, p. 148) lists the jobs, and the table below maps them.

The turning-point idea appears in that same introduction (NCERT, p. 148): points where \( f'(x) \) changes sign give the local largest or smallest values of a function. Everything in this chapter hangs off that one sentence.

Rate of Change of Quantities: dy/dx as a Speedometer

If \( y = f(x) \), then \( \frac{dy}{dx} \) — the same thing as \( f'(x) \) — is the rate of change of y with respect to x. The value \( f'(x_0) \) is that rate at the single instant \( x = x_0 \) (NCERT, p. 148).

Read the sign before you read the number. \( \frac{dy}{dx} \) is positive when y grows as x grows, and negative when y falls as x grows (NCERT, p. 149). A negative rate is not an error; it is the whole answer.

When both x and y depend on time t, the chain rule connects the three rates (NCERT, p. 148):

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt},\quad \frac{dx}{dt} \neq 0 \]

This one formula powers every related-rates problem: differentiate the known geometric formula, then substitute the given rate.

The solved examples to study are Examples 1-6 (NCERT, pp. 148-151): circle area against radius, cube volume and surface, circular waves, rectangle perimeter and area, then cost and revenue. Fifteen of the eighteen problems in Exercise 6.1 are the same species.

Example 5 (NCERT, p. 150) introduces marginal cost \( MC = \frac{dC}{dx} \): the instantaneous rate of change of total cost C with output x. Example 6 (NCERT, pp. 150-151) introduces marginal revenue \( MR = \frac{dR}{dx} \): the instantaneous rate of change of total revenue R with units sold x. Both are quoted in ₹ per unit.

Increasing and Decreasing Functions: Reading the Sign of f'(x)

Definition 1 (NCERT, p. 152) settles whether an interval is rising or falling. For \( x_1, x_2 \) in an interval I:

Definition 2 (NCERT, p. 153) moves the same idea to a single point: f is increasing at \( x_0 \) if it is increasing inside some open interval around \( x_0 \). A point inherits the behaviour of its neighbourhood.

Theorem 1 (NCERT, p. 153) is the workhorse. If f is continuous on [a, b] and differentiable on (a, b), then \( f'(x) \geq 0 \) throughout (a, b) makes f increasing on [a, b]; \( f'(x) \leq 0 \) makes it decreasing; \( f'(x) = 0 \) makes it constant.

Why it works: the Mean Value Theorem of Chapter 5 gives \( f(x_2) – f(x_1) = f'(c)(x_2 – x_1) \), so the sign of \( f’ \) decides the sign of the difference.

A remark on p. 154 extends the rule: strict signs \( f’ \gt 0 \) or \( f’ \lt 0 \) on an interval excluding the endpoints, together with continuity, give the same conclusions. Example 8 (p. 154) proves \( f(x) = x^3 – 3x^2 + 4x \) is increasing on R by writing \( f'(x) = 3(x-1)^2 + 1 \gt 0 \).

Examples 7-9 (NCERT, pp. 153-155) show the two proof styles: Example 7 applies Definition 1 directly to \( f(x) = 7x – 3 \); Example 9 shows \( \cos x \) decreases on \( (0, \pi) \), increases on \( (\pi, 2\pi) \), and is therefore neither on \( (0, 2\pi) \).

Definitions to Memorise: Chapter 6 Vocabulary in One Table

One memorisable table. The running example is \( f(x) = x^2 \), which is decreasing for \( x \lt 0 \) and increasing for \( x \gt 0 \) (NCERT, p. 152).

The Formula Box: Every Symbol and Its Meaning

Every rate formula in one table, with each symbol and its unit. Everything here comes from the solved examples of Section 6.2 (NCERT, pp. 148-151) and Theorem 1 (NCERT, p. 153).

The Sign-Table Method: Finding Intervals Step by Step

The sign-table method turns the question “is f increasing or decreasing?” into a fixed five-step routine. The textbook develops it in Examples 10-13 (NCERT, pp. 155-158).

1. Differentiate: compute \( f'(x) \).
2. Solve: find every x for which \( f'(x) = 0 \).
3. Split: use those solutions to divide the domain into disjoint intervals.
4. Test: pick one value from each interval and record the sign of \( f'(x) \).
5. Conclude: \( f’ \gt 0 \) means increasing there, \( f’ \lt 0 \) means decreasing.

The figures that accompany these examples draw the split on the real line: Fig 6.3 (p. 155) and Fig 6.4 (p. 156) for polynomial functions, Fig 6.5 (p. 157) and Fig 6.6 (p. 158) for trigonometric functions on closed intervals. The examiner expects the same picture, usually written as a table.

Endpoint note from Example 12 (NCERT, p. 157): because the function is continuous at the split points, the conclusion extends to closed intervals — that is why the answer can read \( [0, \pi/6] \) increasing and \( [\pi/6, \pi/2] \) decreasing.

Practise Examples 10-13 (quadratic, cubic, \( \sin 3x \), \( \sin x + \cos x \)) and then Exercise 6.2, especially Q4, Q5 and Q6.

Two Worked Examples with Original Numbers

Fresh numbers, exam method. In both problems, name the rule before substituting — naming the rule is the step that earns the mark.

Worked Example A: Rate of change of a circle’s area

The radius of a circle is increasing at 2 cm/s. Find the rate at which the area is increasing when r = 6 cm.

Worked Example B: Intervals for f(x) = x^3 – 3x^2 – 9x + 2

Find the intervals where \( f(x) = x^3 – 3x^2 – 9x + 2 \) is increasing and where it is decreasing. Method: first derivative test with a sign table.

The graph rises on the outside intervals and dips between the two turning points. Note that f is neither increasing nor decreasing on all of R — monotonicity is always reported interval by interval.

Where Maxima and Minima Fit: Mapping Exercise 6.3

Exercise 6.3 (NCERT, pp. 175-177) is the optimisation workout: 29 questions built on the same turning-point idea. Use this map to target your practice.

These are direction markers for practice, not an answer key. The pattern to internalise: every question in the exercise is a turning-point problem of the kind promised in the 6.1 introduction (NCERT, p. 148), and on a closed interval the comparison always includes the endpoints.

Figure Walkthrough: Reading Fig 6.1 and Fig 6.2

No image is reproduced here; open the textbook at pp. 152-153 and read the two figures with this guide.

Fig 6.1 (p. 152) is the parabola \( f(x) = x^2 \) with two value tables. Reading the left table — \( x = -2, -\frac{3}{2}, -1, -\frac{1}{2}, 0 \) gives heights \( 4, \frac{9}{4}, 1, \frac{1}{4}, 0 \) — the height falls as you move toward the origin, so f is decreasing for \( x \lt 0 \).

Reading the right table — \( 0, \frac{1}{2}, 1, \frac{3}{2}, 2 \) gives heights \( 0, \frac{1}{4}, 1, \frac{9}{4}, 4 \) — the height rises, so f is increasing for \( x \gt 0 \). The turn happens at x = 0, exactly where \( f'(x) = 2x \) changes sign.

Fig 6.2 (p. 153) puts three curves side by side: (i) a strictly increasing graph that climbs at every step, (ii) a strictly decreasing graph that falls at every step, and (iii) a graph that is neither — it rises in places and falls in others. Together they are the visual translation of Definition 1.

Both figures are printed in the official textbook. Cross-check this description against the original in the official NCERT index for Mathematics Part I.

Common Mistakes in Chapter 6 and Their Corrections

Five slips cost the most marks in this chapter. The correction column is the rule to remember; the last column is your self-check.

Exam Notes: What Examiners Look For in Chapter 6

The patterns below are observed directly in the NCERT exercises, with the step that earns the mark named for each type.

• Proof questions — show that f is increasing on R (Exercise 6.2 Q1, Q2, Q18) — need one of two weapons: the Definition 1 chain \( x_1 \lt x_2 \Rightarrow f(x_1) \lt f(x_2) \), or a derivative that is always positive, exactly as Example 8 (p. 154) writes \( f'(x) = 3(x-1)^2 + 1 \gt 0 \). The proof step is the mark.
• Interval questions (Exercise 6.2 Q4, Q5, Q6, Q19) expect a sign table. The five visible parts are f’, the roots, the interval split, one sign per interval, and the conclusion written as intervals; when the function is continuous at the split points, extend the conclusion to closed intervals, as Example 12 (p. 157) does.
• Related-rates word problems dominate Exercise 6.1. The mark-earning steps are: write the given rate with its correct sign, apply the chain rule before substituting, and state the final unit — \( \text{cm}^2/\text{s} \), \( \text{cm/min} \), or ₹ per unit.
• Objective questions close both halves: Exercise 6.1 Q17 and Q18, then Exercise 6.3 Q27-Q29. They reward one-line derivatives of standard formulas — Q17 settles with \( dA/dr = 2\pi r \), Q18 with \( dR/dx = 6x + 36 \).

Night-Before Revision Recap for Chapter 6

If you can tick every bullet below and finish the five self-checks, Chapter 6 is under control.

• Rate of change: \( \frac{dy}{dx} \) is the rate of change of y with respect to x; \( f'(x_0) \) pins it to one instant.
• Related rates: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \) — differentiate the known formula with the chain rule, then substitute.
• Economics: \( MC = \frac{dC}{dx} \), \( MR = \frac{dR}{dx} \), both in ₹ per unit.
• Theorem 1: \( f’ \geq 0 \Rightarrow \) increasing, \( f’ \leq 0 \Rightarrow \) decreasing, \( f’ = 0 \Rightarrow \) constant.
• Interval routine: solve \( f'(x) = 0 \), split the real line, test one point per interval, conclude with intervals.

Memory device: think of the derivative as the speedometer of any quantity, and the sign of \( f’ \) as the road slope — plus means uphill (increasing), minus means downhill (decreasing), zero means flat (constant).

Revision self-check — five quick tasks:

1. State Definition 1 for an increasing function without looking.
2. Write the chain-rule formula for \( \frac{dy}{dx} \) when both x and y depend on t.
3. Differentiate \( A = \pi r^2 \) and write both \( \frac{dA}{dr} \) and \( \frac{dA}{dt} \).
4. Redo Worked Example B in under three minutes, sign table included.
5. Name all the candidates for the absolute maximum of f on [a, b].

These notes sit inside the wider Class 12 Mathematics notes collection. Chapter 6 assumes fluency with the prerequisite continuity and differentiability chapter, and the next chapter, Integrals, reverses the derivative — the rates you compute here are exactly what integration undoes.

For the complete picture across subjects, browse the Class 12 CBSE notes and the main CBSE notes index.

Frequently Asked Questions About Application of Derivatives

The five questions students search most, answered directly.

What is the rate of change of the area of a circle with respect to its radius?

\( \frac{dA}{dr} = 2\pi r \): the area \( A = \pi r^2 \) differentiated with respect to r (NCERT, p. 148). The rate is not constant — it grows with r, so at r = 5 cm it is \( 10\pi \) cm² per cm. If the question instead gives a time rate \( dr/dt \), use the chain rule \( \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \).

How do you check whether a function is increasing or decreasing on an interval?

Use Theorem 1 (NCERT, p. 153): if \( f'(x) \geq 0 \) throughout the open interval, f is increasing there; if \( f'(x) \leq 0 \), it is decreasing. The practical routine is the sign-table method — solve \( f'(x) = 0 \), split the domain into intervals, test one point per interval, and read the sign.

What is the difference between marginal cost and marginal revenue?

Marginal cost \( MC = \frac{dC}{dx} \) is the instantaneous rate of change of total cost C with respect to output x — the extra cost of producing one more unit. Marginal revenue \( MR = \frac{dR}{dx} \) is the same idea for total revenue R against units sold x. Both are measured in ₹ per unit (NCERT, pp. 150-151).

Why do we split the real line at the points where f'(x) = 0?

Because the sign of \( f’ \) is what decides increasing or decreasing, and a continuous derivative can only change sign where it is zero (or fails to exist). Inside each interval between the roots the sign is constant, so one test value per interval tells you the behaviour everywhere — that is the logic of Examples 10-13 (NCERT, pp. 155-158).

How do you find the absolute maximum and minimum of a function on a closed interval?

Find every turning point where \( f'(x) = 0 \) inside the interval, compute f at each turning point AND at both endpoints, then compare. The largest value is the absolute maximum and the smallest is the absolute minimum — the comparison pattern behind Exercise 6.3 questions like Q5, Q7, Q10 and Q12 (NCERT, pp. 175-176).

Reference: NCERT Class 12 Mathematics textbook, Part I, chapter Application of Derivatives.