Chapter 6 · Application of Derivatives · Go for it

Application of Derivatives, the quick way

The derivative tells you how fast something changes. Use it for rates, for where a graph rises or falls, and for the highest and lowest points.

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The chapter in one minute

What the chapter does, and what the exam asks.

The derivative $\frac{dy}{dx}$ measures how fast $y$ changes when $x$ changes. This chapter uses it for three jobs.

Rates: how fast an area, a volume or a cost is changing. Increasing and decreasing: where a graph goes up and where it goes down. Maxima and minima: the highest and lowest points, including the best answer to a real problem, like the biggest box.

Your edition of the book has only these three topics, with no tangents or approximations. They are exactly the topics in the CBSE syllabus.

Most exam questions on this chapter are one of ten kinds, and each kind has a fixed recipe. You will learn all ten below.

10question types

13formula cards

3topics: rates, rising or falling, highs and lows

1habit that saves marks: differentiate first, then put in the numbers

Formulas

The formulas you must know

Learn these cards. Almost everything in the chapter is built from them.

Circle

\begin{aligned} A &= \pi r^2 \\ \frac{dA}{dr} &= 2\pi r \end{aligned}

Its circumference is $C = 2\pi r$ ⁠, so $\frac{dC}{dr} = 2\pi$ ⁠.

Sphere

\begin{aligned} V &= \frac{4}{3}\pi r^3 \\ \frac{dV}{dr} &= 4\pi r^2 \end{aligned}

Its surface area is $S = 4\pi r^2$ ⁠.

Cube of edge

x

\begin{aligned} V &= x^3 \\ \frac{dV}{dx} &= 3x^2 \end{aligned}

Its surface area is $S = 6x^2$ ⁠, so $\frac{dS}{dx} = 12x$ ⁠.

Cone

V = \frac{1}{3}\pi r^2 h

$r$ is the base radius, $h$ the height. A cylinder has $V = \pi r^2 h$ ⁠.

Rates in time (chain rule)

\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}

Differentiate with respect to time $t$ ⁠, then put in the numbers.

Marginal cost and revenue

\begin{aligned} \text{MC} &= \frac{dC}{dx} \\ \text{MR} &= \frac{dR}{dx} \end{aligned}

$x$ is the number of units. "Marginal" means the rate of change.

Increasing (Definition 1)

\begin{aligned} &x_1 < x_2 \\ &\Rightarrow f(x_1) \le f(x_2) \end{aligned}

For all $x_1, x_2$ in the interval. Strictly increasing: $f(x_1) < f(x_2)$ ⁠. Decreasing: $f(x_1) \ge f(x_2)$ ⁠.

Theorem 1

\begin{aligned} &f'(x) \ge 0 \text{ on } (a, b) \\ &\Rightarrow f \text{ increasing on } [a, b] \end{aligned}

$f$ must be continuous on $[a, b]$ and differentiable on $(a, b)$ ⁠. With $f'(x) \le 0$ ⁠, $f$ is decreasing. With $f'(x) > 0$ ⁠, it is strictly increasing.

Critical point

c

\begin{aligned} &f'(c) = 0, \text{ or} \\ &f \text{ not differentiable at } c \end{aligned}

Local maxima and minima can only be at critical points (Theorem 2).

First derivative test

\begin{aligned} &f' \text{ changes } + \text{ to } - \\ &\Rightarrow \text{local maxima at } c \end{aligned}

$-$ to $+$ ⁠: local minima. No change of sign: neither (a point of inflexion).

Second derivative test

\begin{aligned} &f'(c) = 0,\ f''(c) < 0 \\ &\Rightarrow \text{local maxima at } c \end{aligned}

$f'(c) = 0$ and $f''(c) > 0$ ⁠: local minima. If $f''(c) = 0$ ⁠, the test fails: use the first derivative test.

Closed interval

[a, b]

\begin{aligned} &\text{compare } f(a),\ f(b) \\ &\text{and } f(c) \text{ at critical } c \end{aligned}

The largest is the absolute maximum value, the smallest the absolute minimum value.

Bounds

-1 \le \sin x \le 1

The same holds for $\cos x$ ⁠. And a square is never negative: $(x - a)^2 \ge 0$ ⁠.

Question types

The question types and how to crack each one

For each type: how to spot it, what to do, and one real NCERT question worked the exam way.

1. Rate of change with respect to a length

Spot it: "the rate of change of the area (or volume) with respect to the radius". No time is mentioned.

Write the formula that links the two quantities (a formula card).
Differentiate with respect to the second quantity.
Put in the given value, and write the units, such as cm³ per cm.

Exercise 6.1 Q9. A spherical balloon has a variable radius $r$ ⁠. Find the rate of change of its volume with respect to $r$ when $r = 10$ cm.

\begin{aligned} V &= \frac{4}{3}\pi r^3 \\ \frac{dV}{dr} &= 4\pi r^2 \end{aligned}

At $r = 10$ ⁠,

\frac{dV}{dr} = 4\pi(10)^2 = 400\pi

∴ $\frac{dV}{dr} = 400\pi$ cm³/cm

See a full exam answer · Go deeper: see it visually · Basics: area and volume formulas

In past papers (1): click to see

Sample 2023-24 Q38(i)

2. Rates in time: how fast is it changing now?

Spot it: words like "per second", "how fast", "at the rate of", and one moment, such as "when the radius is 10 cm".

Name the quantities and write the given rate as a derivative. Growing is $+$ ⁠, shrinking is $-$ ⁠.
Write the formula that links them, and differentiate it with respect to $t$ (chain rule).
Only now put in the numbers for that moment.
Answer in words, with units, and say "increasing" or "decreasing".

Exercise 6.1 Q3. The radius of a circle is increasing at 3 cm/s. Find how fast the area is increasing when the radius is 10 cm.

Given $\frac{dr}{dt} = 3$ cm/s.

\begin{aligned} A &= \pi r^2 \\ \frac{dA}{dt} &= 2\pi r\,\frac{dr}{dt} \end{aligned}

At $r = 10$ ⁠,

\frac{dA}{dt} = 2\pi(10)(3) = 60\pi

∴ the area is increasing at $60\pi$ cm²/s

See a full exam answer · Go deeper: see it visually · Basics: related rates

In past papers (8): click to see

Board 2026 Q22 · Board 2025 Q26 · Board 2024 Q36 · Board 2023 Q37 · Sample 2026-27 Q27 · Sample 2025-26 Q27 · Sample 2024-25 Q26 · Sample 2022-23 Q22

3. Marginal cost and marginal revenue

Spot it: the words "marginal cost" or "marginal revenue", and a formula in $x$ ⁠, the number of units.

Marginal cost is $C'(x)$ ⁠. Marginal revenue is $R'(x)$ ⁠.
Differentiate term by term.
Put in the number of units. The answer is in rupees.

Exercise 6.1 Q16. The total revenue in rupees from selling $x$ units is $R(x) = 13x^2 + 26x + 15$ ⁠. Find the marginal revenue when $x = 7$ ⁠.

R'(x) = 26x + 26

\begin{aligned} R'(7) &= 26(7) + 26 \\ &= 208 \end{aligned}

∴ the marginal revenue is ₹208

See a full exam answer · Go deeper: see it visually · Basics: rate of change

In past papers (1): click to see

Sample 2024-25 Q22

4. Show that a function is increasing everywhere

Spot it: "Show that (or prove that) $f$ is increasing on $\R$ ⁠."

Find $f'(x)$ ⁠. Write it so its sign is clear: a number times a square, or a square plus a positive number.
Show $f'(x) \ge 0$ for every real $x$ ⁠.
Conclude with Theorem 1: $f$ is increasing on $\R$ ⁠.

Exercise 6.2 Q18. Prove that $f(x) = x^3 - 3x^2 + 3x - 100$ is increasing in $\R$ ⁠.

\begin{aligned} f'(x) &= 3x^2 - 6x + 3 \\ &= 3(x - 1)^2 \end{aligned}

Since $(x - 1)^2 \ge 0$ ⁠, $f'(x) \ge 0$ for all real $x$ ⁠.

∴ by Theorem 1, $f$ is increasing on $\R$

See a full exam answer · Go deeper: see it visually · Basics: sign of a product

In past papers (6): click to see

Board 2024 Q22 · Board 2023 Q16 · Board 2023 Q25 · Sample 2024-25 Q3 · Sample 2024-25 Q27 · Sample 2023-24 Q38(ii)

5. Find where a function increases and decreases

Spot it: "Find the intervals in which $f$ is increasing or decreasing", or "strictly increasing or strictly decreasing".

Find $f'(x)$ and factorise it.
Solve $f'(x) = 0$ ⁠. These points cut the number line into intervals.
Find the sign of $f'(x)$ in each interval: test one number in it.
Where $f'(x) > 0$ ⁠, $f$ is strictly increasing (so also increasing). Where $f'(x) < 0$ ⁠, it is strictly decreasing.

Exercise 6.2 Q5. Find the intervals in which $f(x) = 2x^3 - 3x^2 - 36x + 7$ is increasing and decreasing.

\begin{aligned} f'(x) &= 6x^2 - 6x - 36 \\ &= 6(x - 3)(x + 2) \end{aligned}

$f'(x) = 0$ gives $x = -2$ or $x = 3$ ⁠.

In $(-\infty, -2)$ ⁠, $f'(x) > 0$ ⁠. In $(-2, 3)$ ⁠, $f'(x) < 0$ ⁠. In $(3, \infty)$ ⁠, $f'(x) > 0$ ⁠.

∴ $f$ is increasing in $(-\infty, -2)$ and $(3, \infty)$ ⁠, and decreasing in $(-2, 3)$

See a full exam answer · Go deeper: see it visually · Basics: interval notation

In past papers (7): click to see

Board 2026 Q36(iii)(b) · Board 2025 Q24 · Sample 2026-27 Q9 · Sample 2025-26 Q9 · Sample 2023-24 Q22 · Sample 2022-23 Q36(iii)(a) · Sample 2020-21 Q32

6. Maximum and minimum values you can see without calculus

Spot it: a square plus a number, a modulus plus a number, or sin or cos plus a number.

Use a bound: a square or a modulus is at least 0, and $\sin x$ lies between $-1$ and 1.
Find an $x$ where the bound is reached. That gives the minimum (or maximum) value.
If the values grow without limit, there is no maximum value. If they fall without limit, there is no minimum value.

Exercise 6.3 Q1 (i). Find the maximum and minimum values, if any, of $f(x) = (2x - 1)^2 + 3$ ⁠.

For all $x$ ⁠, $(2x - 1)^2 \ge 0$ ⁠, so

\begin{aligned} f(x) &= (2x - 1)^2 + 3 \\ &\ge 3 \end{aligned}

and $f\left(\frac12\right) = 3$ ⁠. Also $f(x) \to \infty$ as $x \to \infty$ ⁠.

∴ minimum value 3; no maximum value

See a full exam answer · Go deeper: see it visually

In past papers (1): click to see

Sample 2023-24 Q23(a)

7. Local maxima and minima: the second derivative test

Spot it: "Find the local maxima and local minima" or "the local maximum value".

Find $f'(x)$ ⁠, factorise it, and solve $f'(x) = 0$ ⁠.
Find $f''(x)$ and its sign at each point: negative gives local maxima, positive gives local minima.
If $f''(c) = 0$ ⁠, the test fails: use the first derivative test instead.
Put each point into $f$ to get the local values.

Exercise 6.3 Q3 (v). Find the local maxima and minima of $f(x) = x^3 - 6x^2 + 9x + 15$ ⁠, and the local values.

\begin{aligned} f'(x) &= 3x^2 - 12x + 9 \\ &= 3(x - 1)(x - 3) \end{aligned}

$f'(x) = 0$ gives $x = 1$ or $x = 3$ ⁠. Also $f''(x) = 6x - 12$ ⁠.

$f''(1) = -6 < 0$ ⁠, so $x = 1$ is a point of local maxima.

$f''(3) = 6 > 0$ ⁠, so $x = 3$ is a point of local minima.

∴ local maximum value $f(1) = 19$ ⁠; local minimum value $f(3) = 15$

See a full exam answer · Go deeper: see it visually

In past papers (4): click to see

Sample 2026-27 Q37 · Sample 2025-26 Q37 III A · Sample 2023-24 Q23(b) · Sample 2022-23 Q36(iii)(b)

8. When the second derivative test fails: the first derivative test

Spot it: $f''(c) = 0$ at a critical point, or $f$ has a corner there (like $|x|$ at 0), or the question says "first derivative test".

Find the critical points: where $f'(x) = 0$ ⁠, or where $f$ is not differentiable.
Find the sign of $f'(x)$ just to the left and just to the right of each one. A squared factor never changes sign.
$+$ then $-$ ⁠: local maxima. $-$ then $+$ ⁠: local minima. No change of sign: neither, a point of inflexion.

Example 21. Find all the points of local maxima and local minima of $f(x) = 2x^3 - 6x^2 + 6x + 5$ ⁠.

\begin{aligned} f'(x) &= 6x^2 - 12x + 6 \\ &= 6(x - 1)^2 \\ f''(x) &= 12(x - 1) \end{aligned}

$f'(x) = 0$ gives $x = 1$ ⁠, and $f''(1) = 0$ ⁠, so the second derivative test fails.

For $x \ne 1$ ⁠, $f'(x) = 6(x - 1)^2 > 0$ ⁠. So $f'$ does not change sign at $x = 1$ ⁠.

∴ no local maxima or minima; $x = 1$ is a point of inflexion

See a full exam answer · Go deeper: see it visually · Basics: sign of a product

In past papers (2): click to see

Board 2024 Q22 · Sample 2023-24 Q19

9. Absolute maximum and minimum on a closed interval

Spot it: an interval with square brackets, like $[0, 3]$ ⁠, and "maximum value" or "absolute maximum".

Find the critical points inside the interval: solve $f'(x) = 0$ ⁠, and add any point where $f$ is not differentiable.
Make a table of $f$ at those points and at both end points.
The largest value is the absolute maximum, the smallest is the absolute minimum.

Exercise 6.3 Q7. Find the maximum and minimum values of the function below on $[0, 3]$ ⁠.

\begin{aligned} f(x) &= 3x^4 - 8x^3 + 12x^2 \\ &\quad - 48x + 25 \\ f'(x) &= 12x^3 - 24x^2 \\ &\quad + 24x - 48 \\ &= 12(x - 2)(x^2 + 2) \end{aligned}

As $x^2 + 2 > 0$ ⁠, $f'(x) = 0$ only at $x = 2$ ⁠, which is in $[0, 3]$ ⁠.

\begin{aligned} f(0) &= 25 \\ f(2) &= -39 \\ f(3) &= 16 \end{aligned}

∴ maximum value 25, at $x = 0$ ⁠; minimum value $-39$ ⁠, at $x = 2$

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In past papers (5): click to see

Board 2026 Q7 · Board 2025 Q8 · Board 2025 Q33 · Sample 2025-26 Q37 III B · Sample 2022-23 Q36(iii)(b)

10. Word problems: the biggest box, the least cost

Spot it: a real object and the words "maximum", "minimum", "largest" or "least".

Name one variable, write its allowed range, and write the quantity to make biggest or smallest in that one variable.
Differentiate and solve $= 0$ ⁠. Keep only the solutions inside the range.
Use the second derivative (or the sign of the first) to show it is a maximum or a minimum. If it is the only critical point in the range, it gives the greatest (or least) value.
Answer the question asked, with units.

Exercise 6.3 Q17. An open box is made from a square sheet of tin of side 18 cm by cutting equal squares from the corners and folding up the flaps. What side of square should be cut off to make the volume largest?

Let the side cut off be $x$ cm, $0 < x < 9$ ⁠.

\begin{aligned} V &= x(18 - 2x)^2 \\ &= 4x^3 - 72x^2 + 324x \\ V' &= 12x^2 - 144x + 324 \\ &= 12(x - 3)(x - 9) \end{aligned}

$V' = 0$ gives $x = 3$ or $x = 9$ ⁠. As $x < 9$ ⁠, $x = 3$ ⁠.

\begin{aligned} V'' &= 24x - 144 \\ V''(3) &= -72 < 0 \end{aligned}

So $V$ is maximum at $x = 3$ ⁠.

∴ cut off squares of side 3 cm

See a full exam answer · Go deeper: see it visually · Basics: building the function

In past papers (5): click to see

Board 2026 Q36(iii)(a) · Board 2025 Q36 · Sample 2024-25 Q36(iii)(a) · Sample 2022-23 Q37 · Sample 2020-21 Q17

Shortcuts

Shortcuts and quick tips

Small habits that save time and marks.

Differentiate first, then put in the numbers for the moment. If you put $r = 10$ in first, the area becomes a constant and its rate wrongly becomes 0.
A quantity that is shrinking has a negative rate. In Exercise 6.1 Q7 the length falls at 5 cm/min, so $\frac{dx}{dt} = -5$ ⁠.
A negative answer means "decreasing". Say it in words: "the perimeter is decreasing at 2 cm/min".
Always write units: area per second is cm²/s, and volume per cm of radius is cm³/cm.
Factorise $f'(x)$ fully. Then one test number in each interval gives its sign.
A squared factor such as $(x - 1)^2$ is never negative, so it never changes the sign of $f'(x)$ ⁠. If $f'(x) \ge 0$ and it is 0 only at single points, $f$ is still strictly increasing: in Exercise 6.2 Q18, $f'(1) = 0$ ⁠, yet $f$ is strictly increasing on $\R$ ⁠.
For "increasing" (not "strictly"), you may also include an end point where $f' = 0$ ⁠: the function of type 5 is increasing on $(-\infty, -2]$ as well. The book's answers use open intervals. Both are correct.
On a closed interval, the answer can sit at an end point. In Exercise 6.3 Q12 the largest value of $x + \sin 2x$ on $[0, 2\pi]$ is at $x = 2\pi$ ⁠.
In word problems, throw away answers outside the allowed range, like $x = 9$ for the box in Exercise 6.3 Q17 (it leaves no base).
In an MCQ that asks where $f$ is increasing, factorise $f'(x)$ and check its sign in each option. Only the right option earns the mark.

Watch out

Mistakes that cost marks

Each one is common, and each has an easy fix.

Putting in the numbers before differentiating → differentiate first, then substitute.
Leaving out the rate of the radius → by the chain rule, $\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$ ⁠, not just $2\pi r$ ⁠.
A positive rate for something that shrinks → use a minus sign, as in $\frac{dx}{dt} = -5$ ⁠.
Mixing up the point and the value → in Exercise 6.3 Q3 (v), the point is $x = 1$ and the value is $f(1) = 19$ ⁠.
Saying "neither" as soon as $f''(c) = 0$ → the test only fails; use the first derivative test.
Forgetting the end points on a closed interval → put both ends in the table.
Writing "increasing" when the question says "strictly increasing" → show $f'(x) > 0$ inside the interval (except perhaps at single points) and use the word "strictly".
No range for the variable in a word problem → write it, like $0 < x < 9$ ⁠, and check your answer lies inside it.

Marks

How the examiner gives marks

From CBSE's sample-paper marking schemes. The mark splits are examples from those papers, not fixed rules. Writing these lines is what earns marks.

Rates: in a CBSE sample paper (2025-26), each relation and each derivative line earned ½ or 1 mark, and the answer ended with a sentence in words. Add units too.
Increasing and decreasing: in a 2-mark sample-paper question (2023-24), one of its two marking methods gave ½ for $f'(x)$ ⁠, 1 for the sign of $f'(x)$ and ½ for the interval in the answer.
Maxima and minima: the critical point from $f'(x) = 0$ ⁠, then $f''$ and its sign at that point, then the word "maximum" or "minimum". In a 2025-26 sample-paper case study each of these earned 1 mark.
Conclusions in words carry marks, for example " $f''(3) > 0$ ⁠, so $x = 3$ is a point of local minima".
A slip is penalised only once, so keep going. A correct different method, such as the first derivative test instead of the second, gets full marks.
An MCQ earns 1 mark for the right option only; working earns nothing there.

Practice

Practise these first

Easiest first. Together they cover every type above.

Exercise 6.1 Q1 rate of change of a circle's area with its radius
Exercise 6.1 Q15 marginal cost
Exercise 6.2 Q2 show that $e^{2x}$ is increasing
Exercise 6.2 Q4 where a function increases and decreases
Exercise 6.3 Q2 maximum and minimum values without calculus
Exercise 6.1 Q7 a rate in time, with a negative rate
Exercise 6.3 Q6 maximum profit by the second derivative test
Example 19 the first derivative test at a corner, where $f$ is not differentiable
Exercise 6.3 Q10 absolute maximum on two closed intervals
Exercise 6.3 Q13 a word problem: two numbers with the largest product
Exercise 6.1 Q10 a sliding ladder: rates in time
Exercise 6.3 Q18 a word problem: the biggest open box

You can do this

Ten recipes, one at a time. Work through the practice list, write units and words in every answer, and you are ready for the exam.

Ready to go deeper? Open the interactive lesson

Past papers

37 CBSE board and sample-paper questions on Application of Derivatives, each with a full-marks exam answer. Practise them →