MathsCBSE Class 12 · Chapter 6

Application of Derivatives Class 12: Notes, Methods and Solved Questions

Updated 14 October 2026 · Fact-checked

Application of Derivatives is Chapter 6 of CBSE Class 12 Maths. It uses the derivative to find rates of change, to decide where a function is increasing or decreasing, and to locate local and absolute maxima and minima with the first and second derivative tests, ending in real-life optimisation problems on area, volume, cost and profit.

What Application of Derivatives is about

This chapter turns the differentiation skills of Chapter 5 into tools. The derivative $\frac{d y}{d x}$ measures how fast y changes when x changes, so it answers questions such as "how fast is the area of a ripple growing?" or "what does the next unit cost to make?" (marginal cost). When two quantities both change with time, the chain rule links their rates: $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ .

The sign of $f^{'} (x)$ tells you the shape of a graph: $f^{'} > 0$ on an interval means f is increasing there, $f^{'} < 0$ means decreasing. Points where $f^{'} (c) = 0$ (or where f is not differentiable) are critical points, the only candidates for local maxima and minima. You will classify them with the first derivative test (watch the sign of $f^{'}$ change) and the second derivative test ( $f^{''} (c) < 0$ for a maximum, $f^{''} (c) > 0$ for a minimum), find absolute maximum and minimum values on a closed interval, and solve optimisation problems: the largest box, the cheapest tank, the highest revenue.

Tangents and normals and approximations using differentials are not in the 2026-27 syllabus. Calculus (Unit III) carries 35 of the 80 marks. In our question bank of eight board papers (2023 to 2026) and four sample papers, this chapter has 45 questions and appears in all eight board papers, including case studies on rates and on maximising area, volume or revenue.

Weightage: This chapter is part of Unit III, Calculus, which carries 35 of the 80 theory marks together with Continuity and Differentiability, Integrals, Application of Integrals and Differential Equations.Source: CBSE Mathematics syllabus 2026-27, Unit III Calculus

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7.8 marks per paper. See the questions

Key concepts

Rate of change of a quantity

NCERT section 6.2

If y = f(x), then $\frac{d y}{d x}$ is the rate of change of y with respect to x, and $(\frac{d y}{d x})_{x = x_{0}}$ is that rate at the particular value $x_{0}$ . A positive rate means y increases as x increases; a negative rate means y decreases.

When x and y both depend on time t, the chain rule connects their rates:

$\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ , and $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ (when $\frac{d x}{d t} \neq = 0$ ).

Always write the units of a rate, such as cm²/s or ₹ per unit.

Worked example

The radius of a spherical balloon is increasing at 0.5 cm/s. How fast is its volume increasing when the radius is 6 cm? [2 marks]

Try it first, then show the solutionSolution

$V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
At r = 6: $\frac{d V}{d t} = 4 π (36) (0.5) = 72 π$

Answer: $72 π$ cm³/s

Related rates: two quantities linked by an equation

NCERT section 6.2

In a related-rates problem, several quantities change with time and are tied together by a geometric or physical relation (Pythagoras, similar triangles, a volume formula). Method:

1. Draw a figure and name the changing quantities with letters (not numbers).
2. Write the relation between them.
3. Differentiate both sides with respect to t.
4. Only now substitute the values at the given instant, and solve for the unknown rate.

Substituting numbers before differentiating freezes a changing quantity and loses its rate, which is the classic error.

Worked example

A 10 m ladder leans against a vertical wall. Its foot slides away from the wall at 0.6 m/s. How fast is the top sliding down when the foot is 6 m from the wall? [3 marks]

Try it first, then show the solutionSolution

Let x = distance of the foot from the wall and y = height of the top. Then $x^{2} + y^{2} = 100$ .
Differentiate: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0 \Rightarrow \frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t}$
When x = 6, y = 8: $\frac{d y}{d t} = - \frac{6}{8} (0.6) = - 0.45$

Answer: The top slides down at 0.45 m/s (the negative sign shows the height is decreasing).

Marginal cost and marginal revenue

NCERT section 6.2

In economics the derivative has a special name. If C(x) is the total cost of producing x units, the marginal cost is $M C = \frac{d C}{d x}$ , the instantaneous rate of change of cost with output (roughly the cost of making one more unit). If R(x) is the total revenue from selling x units, the marginal revenue is $M R = \frac{d R}{d x}$ .

Profit is $P (x) = R (x) - C (x)$ , so the marginal profit is $M R - M C$ .

Worked example

The total cost (in ₹) of producing x units is $C (x) = 0.02 x^{3} - 0.6 x^{2} + 25 x + 800$ . Find the marginal cost when 20 units are produced. [2 marks]

Try it first, then show the solutionSolution

$M C = \frac{d C}{d x} = 0.06 x^{2} - 1.2 x + 25$
At x = 20: $0.06 (400) - 1.2 (20) + 25 = 24 - 24 + 25 = 25$

Answer: Marginal cost = ₹25

Increasing and decreasing functions

NCERT section 6.3

On an interval I, f is increasing if $x_{1} < x_{2} \Rightarrow f (x_{1}) < f (x_{2})$ (the graph rises from left to right) and decreasing if $x_{1} < x_{2} \Rightarrow f (x_{1}) > f (x_{2})$ . NCERT also uses "strictly increasing" for the same strict inequality.

Derivative test (for f continuous on [a, b] and differentiable on (a, b)):

$f^{'} (x) > 0$ for every x in (a, b) ⇒ f is increasing on [a, b]
$f^{'} (x) < 0$ for every x in (a, b) ⇒ f is decreasing on [a, b]
$f^{'} (x) = 0$ for every x in (a, b) ⇒ f is constant on [a, b]

A function whose derivative is positive except at isolated points where it is 0 (such as $x^{3}$ at 0) is still increasing.

Worked example

Show that $f (x) = x^{3} + 2 x - 5$ is increasing on R. [2 marks]

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} + 2$
$3 x^{2} \geq 0$ for every real x, so $f^{'} (x) \geq 2 > 0$ on R. Hence f is increasing on R.

Answer: Shown: $f^{'} (x) = 3 x^{2} + 2 > 0$ for all x.

Finding intervals of increase and decrease

NCERT section 6.3

To find where f increases or decreases:

1. Find $f^{'} (x)$ and factorise it.
2. Solve $f^{'} (x) = 0$ (and note points where $f^{'}$ does not exist). These points split the domain into intervals.
3. Test the sign of $f^{'}$ in each interval (a sign table or a test value in each).
4. $f^{'} > 0$ : increasing; $f^{'} < 0$ : decreasing.

You may include an endpoint where f is continuous, so both $(- \infty, 1)$ and $(- \infty, 1]$ are accepted.

Worked example

Find the intervals in which $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 5$ is increasing or decreasing. [3 marks]

Try it first, then show the solutionSolution

$f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$ ; $f^{'} (x) = 0$ at x = 1, 2.
On $(- \infty, 1)$ : both factors negative, $f^{'} > 0$ . On (1, 2): $f^{'} < 0$ . On $(2, \infty)$ : $f^{'} > 0$ .
Increasing on (−∞, 1] and [2, ∞); decreasing on [1, 2].

Answer: Increasing on (−∞, 1] ∪ [2, ∞); decreasing on [1, 2].

Critical points, local maxima and local minima

NCERT section 6.4

f has a local maximum at an interior point c if f(c) ≥ f(x) for all x close to c, and a local minimum if f(c) ≤ f(x) for all x close to c. Together these are local extrema (turning points).

Theorem: if f has a local extremum at c and f is differentiable at c, then $f^{'} (c) = 0$ . The converse is false: $f (x) = x^{3}$ has $f^{'} (0) = 0$ but no extremum at 0.

A critical point of f is a point c in its domain where $f^{'} (c) = 0$ or f is not differentiable. Every local extremum occurs at a critical point, so critical points are the only candidates you need to test.

Worked example

Find the critical point and the local minimum value of $f (x) = ∣ x - 3∣ + 1$ . [1 mark]

Try it first, then show the solutionSolution

f is not differentiable at x = 3 (a corner), so x = 3 is a critical point.
$∣ x - 3∣ \geq 0$ , so $f (x) \geq 1 = f (3)$ for every x.

Answer: Critical point x = 3; local (and absolute) minimum value 1.

First derivative test

NCERT section 6.4

Let f be continuous at a critical point c. As x increases through c:

if $f^{'}$ changes from positive to negative, c is a point of local maximum;
if $f^{'}$ changes from negative to positive, c is a point of local minimum;
if $f^{'}$ does not change sign, c is neither (a point of inflection, as for $x^{3}$ at 0).

Geometrically: the graph climbs to a peak and then falls (maximum), or falls into a valley and then climbs (minimum). This test works even where f is not differentiable at c.

Worked example

Find the points of local maxima and minima of $f (x) = x^{3} - 6 x^{2} + 9 x + 4$ and the corresponding values. [3 marks]

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ ; critical points x = 1, 3.
Near x = 1: $f^{'}$ goes from + to − ⇒ local maximum, f(1) = 1 − 6 + 9 + 4 = 8.
Near x = 3: $f^{'}$ goes from − to + ⇒ local minimum, f(3) = 27 − 54 + 27 + 4 = 4.

Answer: Local maximum 8 at x = 1; local minimum 4 at x = 3.

Second derivative test

NCERT section 6.4

Let f be twice differentiable at c with $f^{'} (c) = 0$ . Then:

$f^{''} (c) < 0$ ⇒ c is a point of local maximum (the curve bends down like ∩);
$f^{''} (c) > 0$ ⇒ c is a point of local minimum (the curve bends up like ∪);
$f^{''} (c) = 0$ ⇒ the test fails; go back to the first derivative test.

The syllabus asks you to use this test as a tool; it is quicker than a sign table when $f^{''}$ is easy to compute. Example of failure: for $x^{4}$ and $x^{3}$ , $f^{''} (0) = 0$ , but $x^{4}$ has a minimum at 0 and $x^{3}$ has none.

Worked example

Use the second derivative test to find the local maximum and minimum values of $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 7$ . [3 marks]

Try it first, then show the solutionSolution

$f^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x - 2) (x + 1)$ ; critical points x = −1, 2.
$f^{''} (x) = 12 x - 6$ . $f^{''} (- 1) = - 18 < 0$ ⇒ local maximum; $f^{''} (2) = 18 > 0$ ⇒ local minimum.
f(−1) = −2 − 3 + 12 + 7 = 14; f(2) = 16 − 12 − 24 + 7 = −13.

Answer: Local maximum value 14 at x = −1; local minimum value −13 at x = 2.

Absolute maximum and minimum on a closed interval

NCERT section 6.4.1

A function continuous on a closed interval [a, b] always has an absolute maximum and an absolute minimum there. They occur either at a critical point inside the interval or at an end point.

Working rule:

1. Find all critical points of f in (a, b).
2. Evaluate f at these points and at the end points a and b.
3. The largest of these values is the absolute maximum; the smallest is the absolute minimum.

No derivative test is needed here: just compare values. On an open interval or on R, an absolute maximum or minimum may not exist.

Worked example

Find the absolute maximum and minimum values of $f (x) = x^{3} - 3 x^{2} - 9 x + 2$ on [−2, 4]. [3 marks]

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1)$ ; critical points x = −1 and x = 3, both inside [−2, 4].
f(−2) = −8 − 12 + 18 + 2 = 0; f(−1) = −1 − 3 + 9 + 2 = 7; f(3) = 27 − 27 − 27 + 2 = −25; f(4) = 64 − 48 − 36 + 2 = −18.
Largest value 7, smallest value −25.

Answer: Absolute maximum 7 at x = −1; absolute minimum −25 at x = 3.

Optimisation (word problems on maxima and minima)

NCERT section 6.4

Most 5-mark questions in this chapter are optimisation problems. A reliable method:

1. Name the quantity to be maximised or minimised (area, volume, cost, distance).
2. Write it in terms of the variables, then use the given condition (fixed perimeter, fixed volume) to reduce it to one variable.
3. Note the allowed range of that variable (lengths are positive).
4. Differentiate, set the derivative equal to 0 and solve.
5. Justify maximum or minimum with the second derivative test (or the first derivative test).
6. Answer the question actually asked: the dimensions, the value, or both.

A neat trick: to maximise $g (x)$ or a distance, maximise its square instead; the same x works.

Worked example

An open box is made from a 24 cm × 24 cm square sheet by cutting equal squares of side x from each corner and folding up the sides. Find x for maximum volume and the maximum volume. [5 marks]

Try it first, then show the solutionSolution

$V = x (24 - 2 x)^{2} = 4 x^{3} - 96 x^{2} + 576 x$ , with 0 < x < 12.
$\frac{d V}{d x} = 12 x^{2} - 192 x + 576 = 12 (x - 4) (x - 12)$ ; in the allowed range, x = 4.
$\frac{d ^{2} V}{d x ^{2}} = 24 x - 192 = - 96 < 0$ at x = 4, so V is maximum.
$V (4) = 4 (16)^{2} = 1024$

Answer: Cut squares of side 4 cm; maximum volume 1024 cm³.

Maths in real life: Application of Derivatives

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Rate of change (spreading oil)

Suppose oil leaking from a tanker spreads on calm water as a circle whose radius grows at 2 m/min, and the coast guard wants to know how fast the polluted area is growing.

Suppose the radius of the circular oil patch increases at 2 m/min. How fast is the area increasing when the radius is 50 m?

Try it first, then show the solutionSolution

$A = π r^{2} \Rightarrow \frac{d A}{d t} = 2 π r \frac{d r}{d t}$
At r = 50: $\frac{d A}{d t} = 2 π (50) (2) = 200 π \approx 628$

Answer: About 200π ≈ 628 m² per minute, and the rate keeps growing as the patch widens.

Memory hook: Area rate = circumference × radius rate: the rim sweeps out the new area.

Optimisation (fencing a field)

Imagine a farmer with 200 m of fencing who wants a rectangular plot beside a straight river, fencing only the three sides away from the river.

Suppose the side parallel to the river is y m and the other two sides are x m each, with 2x + y = 200. Find x and y for the largest area.

Try it first, then show the solutionSolution

$A = x y = x (200 - 2 x) = 200 x - 2 x^{2}$ , 0 < x < 100
$A^{'} = 200 - 4 x = 0 \Rightarrow x = 50$ ; $A^{''} = - 4 < 0$ , so maximum
y = 200 − 100 = 100; A = 50 × 100 = 5000

Answer: x = 50 m, y = 100 m, giving the largest area 5000 m².

Memory hook: The river saves a side, so the long side is twice the short side.

Marginal revenue

Suppose a small bakery finds that its daily revenue in rupees from selling x cakes is R(x) = 400x − 2x², and the owner wants to know whether baking one more cake still adds revenue.

Find the marginal revenue at x = 80 and at x = 110, and the number of cakes that gives maximum revenue.

Try it first, then show the solutionSolution

$M R = R^{'} (x) = 400 - 4 x$
MR(80) = 80 > 0 (more cakes still add revenue); MR(110) = −40 < 0 (revenue now falls)
MR = 0 at x = 100; $R^{''} = - 4 < 0$ , so revenue is maximum at 100 cakes

Answer: MR is ₹80 at 80 cakes and −₹40 at 110 cakes; revenue is greatest at 100 cakes (₹20,000).

Memory hook: Keep producing while the marginal revenue is positive; stop where it hits zero.

Topics in this chapter

Application of Derivatives formula sheet

Rate of change at a point: $(\frac{d y}{d x})_{x = x_{0}} = f^{'} (x_{0})$
Related rates (chain rule): $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$
Marginal cost and revenue: $M C = \frac{d C}{d x}, M R = \frac{d R}{d x}, P (x) = R (x) - C (x)$
Increasing or decreasing on (a, b): $f^{'} (x) > 0 \Rightarrow increasing, f^{'} (x) < 0 \Rightarrow decreasing$
Critical point: $f^{'} (c) = 0 or f^{'} (c) does not exist$
First derivative test: $f^{'} : + \to - \Rightarrow local max; f^{'} : - \to + \Rightarrow local min$
Second derivative test: $f^{'} (c) = 0 : f^{''} (c) < 0 \Rightarrow max, f^{''} (c) > 0 \Rightarrow min, f^{''} (c) = 0 \Rightarrow test fails$
Absolute extrema on [a, b]: $max / min of {f (a), f (b), f (c_{1}), \dots, f (c_{k})}$
Sphere: $V = \frac{4}{3} π r^{3}, S = 4 π r^{2}$
Cone and cylinder: $V_{cone} = \frac{1}{3} π r^{2} h, V_{cyl} = π r^{2} h, CSA_{cyl} = 2 π r h$
Circle: $A = π r^{2}, C = 2 π r$

How the board asks Application of Derivatives

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Rate of change and related rates	Balloons, ripples, ladders, shadows, cones filling or draining; 2-3 marks or a case study.	Write the relation in letters, differentiate with respect to t, then substitute the instant's values; give units and the meaning of the sign.
Marginal cost and revenue	Find MC or MR at a given output from a cubic or quadratic cost/revenue function; 1-2 marks.	Differentiate and substitute; write the answer in rupees.
Increasing and decreasing functions	Find intervals; prove a function is increasing or decreasing on an interval; MCQs on e^x, log x, trigonometric functions; find a parameter so f is increasing on R.	Factorise $f^{'}$ , make a sign table; for "increasing on R" use $f^{'} \geq 0$ for all x (often a discriminant condition).
Local maxima and minima	Find local extrema and values; assertion-reason on the first and second derivative tests; MCQs on critical points.	Critical points from $f^{'} = 0$ ; classify with the sign change of $f^{'}$ or the sign of $f^{''}$ .
Absolute extrema and optimisation	Absolute max/min on [a, b]; 5-mark word problems on boxes, cylinders, fencing, cost and profit; case studies on revenue or area.	Closed interval: compare values at critical points and end points. Word problem: reduce to one variable, differentiate, justify, answer the question asked.

Tricks to remember Application of Derivatives

Short memory aids. Say each one aloud once, then use it on a question the same day.

Letters first, numbers last
In related rates, differentiate the relation while everything is still a letter; plug in the instant's values only at the end.
Use it when: Every related-rates question.
Plus-to-minus is a peak
$f^{'}$ : + then − is a hill top (max); − then + is a valley (min). Same sign on both sides: just passing through.
Use it when: First derivative test.
Frown is max, smile is min
$f^{''} < 0$ means the curve frowns (∩): maximum. $f^{''} > 0$ smiles (∪): minimum.
Use it when: Second derivative test.
Ends and middles
Absolute extrema on [a, b]: check the two ends and every critical point in the middle, then pick the biggest and smallest.
Use it when: Closed-interval questions.
Square the root
To maximise a distance or a square root, maximise its square; the answer for x does not change.
Use it when: Nearest-point and area-with-root problems.

Application of Derivatives: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Interval of decrease · 1 mark · MCQBoard level

The function $f (x) = x^{3} - 3 x$ is decreasing on:
(a) $(- \infty, - 1)$
(b) $(- 1, 1)$
(c) $(1, \infty)$
(d) R

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} - 3 = 3 (x - 1) (x + 1) < 0$ when $- 1 < x < 1$ .

Answer: (b) (−1, 1)

Example 2: A zero derivative is not enough · 1 mark · Assertion-reasonBoard level

Assertion (A): $f (x) = x^{3}$ has neither a maximum nor a minimum at x = 0.
Reason (R): $f^{'} (0) = 0$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.

Try it first, then show the solutionSolution

A is true: $f^{'} (x) = 3 x^{2}$ does not change sign at 0. R is true, but $f^{'} (0) = 0$ happens at maxima and minima too, so it cannot explain why there is no extremum.

Answer: (b)

Example 3: A leaking balloon · 2 marks · Very short answerHarder board level

Air leaks from a spherical balloon so that its volume decreases at 50 cm³/min. Find the rate at which its surface area decreases when the radius is 10 cm. [2 marks]

Try it first, then show the solutionSolution

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow - 50 = 400 π \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = - \frac{1}{8 π}$
$S = 4 π r^{2} \Rightarrow \frac{d S}{d t} = 8 π r \frac{d r}{d t} = 80 π (- \frac{1}{8 π}) = - 10$

Answer: The surface area decreases at 10 cm²/min.

Example 4: Monotonicity of a log and inverse-tan mix · 3 marks · Short answerHarder board level

Find the intervals in which $f (x) = lo g (1 + x^{2}) - 2 tan^{- 1} x$ is increasing or decreasing, and hence find the least value of f. [3 marks]

Try it first, then show the solutionSolution

$f^{'} (x) = \frac{2 x}{1 + x ^{2}} - \frac{2}{1 + x ^{2}} = \frac{2 ( x - 1 )}{1 + x ^{2}}$
The denominator is always positive, so the sign of $f^{'}$ is the sign of x − 1: $f^{'} < 0$ for x < 1 and $f^{'} > 0$ for x > 1.
So f is decreasing on (−∞, 1] and increasing on [1, ∞); it therefore has its least value at x = 1: $f (1) = lo g 2 - 2 \cdot \frac{π}{4} = lo g 2 - \frac{π}{2}$ .

Answer: Decreasing on (−∞, 1], increasing on [1, ∞); least value $lo g 2 - \frac{π}{2}$ at x = 1.

Example 5: Local extrema of a product with an exponential · 3 marks · Short answerHarder board level

Find the local maximum and local minimum values of $f (x) = x^{2} e^{- x}$ . [3 marks]

Try it first, then show the solutionSolution

$f^{'} (x) = 2 x e^{- x} - x^{2} e^{- x} = x e^{- x} (2 - x)$ ; critical points x = 0, 2 (since $e^{- x} > 0$ ).
x < 0: $f^{'} < 0$ ; 0 < x < 2: $f^{'} > 0$ ; x > 2: $f^{'} < 0$ .
Local minimum at x = 0: f(0) = 0. Local maximum at x = 2: $f (2) = \frac{4}{e ^{2}}$ .

Answer: Local minimum 0 at x = 0; local maximum $\frac{4}{e ^{2}}$ at x = 2.

Example 6: Cutting a wire for least area · 5 marks · Long answerHarder board level

A wire 36 m long is cut into two pieces. One piece is bent into a square and the other into a circle. Find the lengths of the two pieces for which the total enclosed area is least. [5 marks]

Try it first, then show the solutionSolution

Let the square use x m and the circle (36 − x) m. Square side $\frac{x}{4}$ ; circle radius $\frac{36 - x}{2 π}$ .
$A = \frac{x ^{2}}{16} + π (\frac{36 - x}{2 π})^{2} = \frac{x ^{2}}{16} + \frac{( 36 - x ) ^{2}}{4 π}$
$\frac{d A}{d x} = \frac{x}{8} - \frac{36 - x}{2 π} = 0 \Rightarrow π x = 4 (36 - x) \Rightarrow x = \frac{144}{π + 4}$
$\frac{d ^{2} A}{d x ^{2}} = \frac{1}{8} + \frac{1}{2 π} > 0$ , so the area is least.
Circle piece: $36 - \frac{144}{π + 4} = \frac{36 π}{π + 4}$

Answer: Square piece $\frac{144}{π + 4}$ m (about 20.2 m), circle piece $\frac{36 π}{π + 4}$ m (about 15.8 m).

Example 7: Parameter for an increasing function · 3 marks · Short answerHOTS / competency

Find all values of a for which $f (x) = x^{3} + a x^{2} + 3 x + 5$ is increasing on R. [3 marks]

Try it first, then show the solutionSolution

$f^{'} (x) = 3 x^{2} + 2 a x + 3$ . We need $f^{'} (x) \geq 0$ for all x, with equality at no more than isolated points.
A quadratic with positive leading coefficient is never negative when its discriminant is ≤ 0: $4 a^{2} - 36 \leq 0 \Rightarrow - 3 \leq a \leq 3$ .
At a = ±3, $f^{'} (x) = 3 (x \pm 1)^{2}$ is zero at only one point, so f is still increasing.

Answer: $- 3 \leq a \leq 3$

Example 8: Nearest point on a parabola · 5 marks · Long answerHOTS / competency

Find the points on the curve $y = x^{2}$ nearest to the point (0, 5), and the least distance. [5 marks]

Try it first, then show the solutionSolution

For a point $(x, x^{2})$ , the square of the distance is $D = x^{2} + (x^{2} - 5)^{2}$ (minimise D instead of the distance).
$D^{'} = 2 x + 4 x (x^{2} - 5) = 2 x (2 x^{2} - 9) = 0 \Rightarrow x = 0$ or $x^{2} = \frac{9}{2}$ .
$D^{''} = 12 x^{2} - 18$ : at x = 0, D″ = −18 < 0 (a local maximum of D); at $x^{2} = \frac{9}{2}$ , D″ = 36 > 0 (minimum).
$D = \frac{9}{2} + (- \frac{1}{2})^{2} = \frac{19}{4}$ , so the distance is $\frac{19}{2}$ .
Points: $(\pm \frac{3}{2}, \frac{9}{2})$ .

Answer: $(\pm \frac{3}{2}, \frac{9}{2})$ ; least distance $\frac{19}{2}$ units.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

A conical coffee filter

Suppose a conical filter, vertex down, has radius 6 cm and height 12 cm. Coffee drips out of it at a steady 2 cm³/s. Let h be the depth of liquid and r the radius of its surface at time t.

(i) · 1 mark
Express r in terms of h.
Answer
Answer: $r = \frac{h}{2}$
Similar triangles: $\frac{r}{h} = \frac{6}{12}$ .
(ii) · 1 mark
Express the volume of liquid V in terms of h.
Answer
Answer: $V = \frac{π h ^{3}}{12}$
$V = \frac{1}{3} π (\frac{h}{2})^{2} h$ .
(iii) · 2 marks
How fast is the level falling when h = 4 cm?
Answer
Answer: $\frac{1}{2 π}$ cm/s (about 0.16 cm/s)
$\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}$ ; $- 2 = 4 π \frac{d h}{d t} \Rightarrow \frac{d h}{d t} = - \frac{1}{2 π}$ .

Pricing a gym membership

Suppose a gym has 400 members paying ₹1000 a month. A survey suggests that for every ₹10 increase in the fee, 2 members will leave. Let x be the number of ₹10 increases.

(i) · 1 mark
Write the monthly revenue R(x).
Answer
Answer: $R (x) = (1000 + 10 x) (400 - 2 x)$
Fee times number of members.
(ii) · 1 mark
Find x for which R is maximum.
Answer
Answer: x = 50
$R^{'} (x) = 10 (400 - 2 x) - 2 (1000 + 10 x) = 2000 - 40 x = 0$ .
(iii) · 2 marks
Find the best fee, the number of members then, and the maximum revenue; justify that it is a maximum.
Answer
Answer: Fee ₹1500, 300 members, revenue ₹4,50,000
$R^{''} (x) = - 40 < 0$ , so maximum. Fee = 1000 + 500, members = 400 − 100, R = 1500 × 300 = 4,50,000 (up from ₹4,00,000).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Show that $f (x) = x^{3} - 3 x^{2} + 3 x + 5$ is increasing on R. [2 marks] (2 marks)

Step	Marks
$f^{'} (x) = 3 x^{2} - 6 x + 3 = 3 (x - 1)^{2}$	1
$f^{'} (x) \geq 0$ for all x, and $f^{'} (x) = 0$ only at the single point x = 1, so f is increasing on R.	1

A man 1.8 m tall walks away from a lamp post 4.5 m high at 1.2 m/s. How fast is the length of his shadow increasing? [3 marks] (3 marks)

Step	Marks
Let x = distance of the man from the post and s = shadow length. Similar triangles: $\frac{s}{1.8} = \frac{x + s}{4.5}$	1
$4.5 s = 1.8 x + 1.8 s \Rightarrow s = \frac{2}{3} x$	1
$\frac{d s}{d t} = \frac{2}{3} \cdot \frac{d x}{d t} = \frac{2}{3} (1.2) = 0.8$ m/s	1

A closed cylindrical can must hold 1000 cm³. Show that its total surface area is least when its height equals its diameter. [5 marks] (5 marks)

Step	Marks
$π r^{2} h = 1000 \Rightarrow h = \frac{1000}{π r ^{2}}$	1
$S = 2 π r^{2} + 2 π r h = 2 π r^{2} + \frac{2000}{r}$	1
$\frac{d S}{d r} = 4 π r - \frac{2000}{r ^{2}} = 0 \Rightarrow r^{3} = \frac{500}{π}$	1
$\frac{d ^{2} S}{d r ^{2}} = 4 π + \frac{4000}{r ^{3}} > 0$ , so S is least there	1
$h = \frac{1000}{π r ^{2}} = \frac{1000 r}{π r ^{3}} = \frac{1000 r}{500} = 2 r$ : height = diameter	1

Where students lose marks in Application of Derivatives

The slips examiners see most often, and what to write instead.

Mistake: Substituting values before differentiating
Putting x = 6 into the relation first turns a changing length into a constant, so its rate becomes 0.
Wrong working
$6^{2} + y^{2} = 100 \Rightarrow 0 + 2 y \frac{d y}{d t} = 0$
Right working
$2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ , then put x = 6, y = 8
Do this: Differentiate the general relation, then substitute.
Usually costs: Most of the marks for the question.
Mistake: Ignoring the sign or the units of a rate
Writing "the height changes at 0.45" when the answer is −0.45 m/s.
Do this: State "decreasing at 0.45 m/s" and always give units such as cm³/s.
Usually costs: ½ to 1 mark.
Mistake: Not justifying maximum or minimum
Solving $f^{'} (x) = 0$ and declaring a maximum without a test.
Do this: Show $f^{''} < 0$ (or the sign change of $f^{'}$ ) at the point; marking schemes give a separate mark for this.
Usually costs: 1 mark in a 5-mark question.
Mistake: Forgetting the end points for absolute extrema
Comparing only the critical values.
Do this: Always evaluate f(a) and f(b) as well; in many questions the absolute maximum is at an end point.
Usually costs: 1-2 marks.
Mistake: Treating f″(c) = 0 as "no extremum"
The second derivative test is inconclusive there, not negative.
Wrong working
$f (x) = x^{4}$ : $f^{''} (0) = 0$ , so no extremum at 0.
Right working
$f^{'}$ changes from − to + at 0, so x = 0 is a local minimum.
Do this: When f″(c) = 0, switch to the first derivative test.
Usually costs: 1-2 marks.

NCERT exercise guide: Application of Derivatives

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 6.1 (rate of change of quantities)

What it tests: Rates of change of area, volume and circumference, related rates (ladders, balloons, sand cones), marginal cost and marginal revenue.

How to approach it: Write the formula, differentiate with respect to t (chain rule), substitute the given instant; give units.

Same kind as Exercise 6.1

Air is pumped into a spherical balloon at 400 cm³/s. How fast is the radius increasing when the radius is 10 cm?

Show the solutionSolution

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
$400 = 4 π (100) \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{1}{π}$

Answer: $\frac{1}{π}$ cm/s

Exercise 6.2 (increasing and decreasing functions)

What it tests: Proving functions increasing or decreasing on R or on an interval, finding intervals of monotonicity, trigonometric and logarithmic functions, and a parameter for which f is increasing.

How to approach it: Find $f^{'}$ , factorise, sign table; for trigonometric functions check the sign of sin or cos on the interval.

Same kind as Exercise 6.2

Find the intervals in which $f (x) = x^{4} - 2 x^{2}$ is increasing or decreasing.

Show the solutionSolution

$f^{'} (x) = 4 x^{3} - 4 x = 4 x (x - 1) (x + 1)$ ; zero at x = −1, 0, 1.
Signs: $(- \infty, - 1)$ : −; (−1, 0): +; (0, 1): −; $(1, \infty)$ : +.

Answer: Increasing on [−1, 0] and [1, ∞); decreasing on (−∞, −1] and [0, 1].

Exercise 6.3 (maxima and minima)

What it tests: Maximum and minimum values by inspection, local extrema by the first and second derivative tests, absolute extrema on closed intervals and optimisation problems.

How to approach it: Critical points, then a test; for closed intervals compare values; for word problems reduce to one variable and justify.

Same kind as Exercise 6.3

Find two positive numbers x and y with x + y = 20 such that $x^{2} y$ is maximum.

Show the solutionSolution

$P = x^{2} (20 - x) = 20 x^{2} - x^{3}$ , 0 < x < 20
$P^{'} = 40 x - 3 x^{2} = x (40 - 3 x) = 0 \Rightarrow x = \frac{40}{3}$
$P^{''} = 40 - 6 x = - 40 < 0$ at $x = \frac{40}{3}$ : maximum; $y = \frac{20}{3}$

Answer: $x = \frac{40}{3}$ , $y = \frac{20}{3}$

Miscellaneous Exercise on Chapter 6

What it tests: Mixed questions: related rates with triangles, monotonicity of harder functions, inscribed figures of maximum area or volume, least-cost tanks and absolute extrema of trigonometric functions.

How to approach it: Identify which tool the question needs; draw a figure for geometry questions and express everything in one variable.

Same kind as the Miscellaneous Exercise

Find the absolute maximum and minimum values of $f (x) = x + 2 cos x$ on $[0, π]$ .

Show the solutionSolution

$f^{'} (x) = 1 - 2 sin x = 0 \Rightarrow sin x = \frac{1}{2} \Rightarrow x = \frac{π}{6}, \frac{5 π}{6}$
$f (0) = 2$ , $f (\frac{π}{6}) = \frac{π}{6} + 3 \approx 2.26$ , $f (\frac{5 π}{6}) = \frac{5 π}{6} - 3 \approx 0.89$ , $f (π) = π - 2 \approx 1.14$

Answer: Absolute maximum $\frac{π}{6} + 3$ at $x = \frac{π}{6}$ ; absolute minimum $\frac{5 π}{6} - 3$ at $x = \frac{5 π}{6}$ .

Application of Derivatives: previous year questions

Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 7.8 marks per paper; 27 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Application of Derivatives

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 65/1/1, Q7 · 1 mark · MCQ

The least value of $f (x) = x^{3} - 12 x$ , $x \in [0, 3]$ is

(A)–16
(B)–9
(C)0
(D)16

Try it first, then show our solutionSolution

$f^{'} (x) = 3 x^{2} - 12 = 0 \Rightarrow x = 2$ in $[0, 3]$ .
$f (0) = 0$ , $f (2) = 8 - 24 = - 16$ , $f (3) = 27 - 36 = - 9$ . Least value $- 16$ . Option (A).
1 mark

Answer: $- 16$ , option (A)

Where the marks are

Finds critical point $x = 2$ and compares values at endpoints (1 mark)

See it in the full 2026 board paper, set 65/1/1

2026, set 65/1/1, Q22 · 2 marks · Very short answerCompetency-based

A room freshner bottle in the shape of an inverted cone sprays the perfume at regular intervals such that volume of the perfume in the bottle decreases at the steady rate of 1 mm $^{3}$ /min. Find the rate at which level of perfume is dropping at an instant when level of perfume in the bottle is 10 mm, if the semi-vertical angle of conical bottle is $\frac{π}{6}$ .

Figure in the paper: Left: a room freshener bottle shaped as an inverted cone spraying perfume. Right: an inverted cone with shaded perfume; the semi-vertical angle is marked 30° and the perfume level height is marked 10 mm. See it in the official paper.

Try it first, then show our solutionSolution

Let $V$ , $r$ , $h$ be the volume, radius and height of perfume at time $t$ . Given $\frac{d V}{d t} = - 1$ mm³/min.
From the semi-vertical angle: $\frac{r}{h} = tan \frac{π}{6} = \frac{1}{3} \Rightarrow r = \frac{h}{3}$
½ mark
$V = \frac{1}{3} π r^{2} h = \frac{1}{3} π \cdot \frac{h ^{2}}{3} \cdot h = \frac{π h ^{3}}{9}$
½ mark
Differentiating w.r.t. $t$ : $\frac{d V}{d t} = \frac{π}{9} \cdot 3 h^{2} \frac{d h}{d t} = \frac{π h ^{2}}{3} \frac{d h}{d t}$
½ mark
So $\frac{d h}{d t} = - \frac{3}{π h ^{2}}$ . At $h = 10$ mm: $\frac{d h}{d t} = - \frac{3}{100 π}$ .
The level of perfume is dropping at $\frac{3}{100 π}$ mm/min.
½ mark

Answer: Level drops at $\frac{3}{100 π}$ mm/min

Where the marks are

Uses $tan \frac{π}{6}$ to get $r = \frac{h}{3}$ (½ mark)
Writes $V = \frac{π h ^{3}}{9}$ (½ mark)
Differentiates: $\frac{d V}{d t} = \frac{π h ^{2}}{3} \frac{d h}{d t}$ (½ mark)
Gets rate $\frac{3}{100 π}$ mm/min at $h = 10$ (½ mark)

See it in the full 2026 board paper, set 65/1/1

2026, set 65/1/1, Q36 · 4 marks · Case-basedCompetency-based

An online delivery company in a city has 5000 subscribers and collects annual subscription fees of ₹ 300 per subscriber for unlimited free deliveries.

The company wishes to increase the annual subscription fee. It is predicted that, for every increase of ₹ 1, ten subscribers will discontinue. Assume that the company increased the annual fee by ₹ $x$ .

Based on the given information, answer the following questions :

(i)
How many subscribers will discontinue after an increase of ₹ $x$ in annual fee ?
1 mark
(ii)
If R( $x$ ) denotes the total revenue collected after the increase of ₹ $x$ in subscription fee, express R( $x$ ) as a function of $x$ .
1 mark
(iii)(a)
Find the value of $x$ for which R( $x$ ) is maximum.
2 marks
OR
(iii)(b)
Find the sub-intervals of (0, 5000) in which R( $x$ ) is increasing and decreasing.
2 marks

Figure in the paper: Illustration of an online delivery app and a delivery person, alongside a graph with 'Subscription Fee' on x-axis and 'No. of Subscribers' on y-axis showing a straight line decreasing. See it in the official paper.

Try it first, then show our solutionSolution

(i) Each ₹1 increase loses 10 subscribers, so an increase of ₹ $x$ makes $10 x$ subscribers discontinue.
1 mark
(ii) Subscribers left $= 5000 - 10 x$ and the new fee $= ₹ (300 + x)$ .
So $R (x) = (5000 - 10 x) (300 + x) = 1500000 + 2000 x - 10 x^{2}$ .
1 mark
(iii)(a) Differentiate: $R^{'} (x) = 2000 - 20 x$ .
½ mark
(iii)(a) Set $R^{'} (x) = 0$ : $2000 - 20 x = 0 \Rightarrow x = 100$ .
½ mark
(iii)(a) Second derivative: $R^{''} (x) = - 20 < 0$ at $x = 100$ .
½ mark
(iii)(a) So $R (x)$ is maximum at $x = 100$ , i.e. when the fee is raised by ₹100. The maximum revenue is $R (100) = 4000 \times 400 = ₹16, 00, 000$ .
½ mark
(iii)(b) $R^{'} (x) = 2000 - 20 x$ .
½ mark
(iii)(b) $R^{'} (x) = 0 \Rightarrow x = 100$ , which splits $(0, 5000)$ into $(0, 100)$ and $(100, 5000)$ .
½ mark
(iii)(b) For $0 < x < 100$ , $R^{'} (x) > 0$ , so $R (x)$ is increasing on $(0, 100)$ (or $(0, 100]$ ).
½ mark
(iii)(b) For $100 < x < 5000$ , $R^{'} (x) < 0$ , so $R (x)$ is decreasing on $(100, 5000)$ (or $[100, 5000)$ ).
½ mark

Answer: (i) $10 x$ ; (ii) $R (x) = (5000 - 10 x) (300 + x)$ ; (iii)(a) $x = 100$ ; (iii)(b) increasing on $(0, 100)$ , decreasing on $(100, 5000)$

Where the marks are

(i) States $10 x$ subscribers discontinue (1 mark)
(ii) Writes $R (x) = (5000 - 10 x) (300 + x)$ (1 mark)
(iii)(a) Finds $R^{'} (x) = 2000 - 20 x$ (½ mark)
(iii)(a) Solves $R^{'} (x) = 0$ to get $x = 100$ (½ mark)
(iii)(a) Shows $R^{''} (x) = - 20 < 0$ (½ mark)
(iii)(a) Concludes $R$ is maximum at $x = 100$ (½ mark)
(iii)(b) Finds $R^{'} (x)$ (½ mark)
(iii)(b) Gets the critical point $x = 100$ (½ mark)
(iii)(b) States increasing on $(0, 100)$ with reason $R^{'} > 0$ (½ mark)
(iii)(b) States decreasing on $(100, 5000)$ with reason $R^{'} < 0$ (½ mark)

See it in the full 2026 board paper, set 65/1/1

Every Application of Derivatives question, by year

For 2027

Sample paper 2026-27: Q9 (1 mark, solved), Q10 (1 mark, solved), Q27 (3 marks, solved), Q37 (4 marks, solved)

2026

Set 65/1/1: Q7 (1 mark, solved), Q22 (2 marks, solved), Q36 (4 marks, solved)

Set 65/5/1: Q9 (1 mark, solved), Q23 (2 marks, solved), Q38 (4 marks, solved)

Sample paper 2025-26: Q9 (1 mark, solved), Q27 (3 marks, solved), Q37 (4 marks, solved)

2025

Set 65/1/1: Q8 (1 mark, solved), Q24 (2 marks, solved), Q26 (3 marks, solved), Q33 (5 marks, solved), Q36 (4 marks, solved)

Set 65/7/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q29 (3 marks, solved)

Sample paper 2024-25: Q3 (1 mark, solved), Q22 (2 marks, solved), Q26 (3 marks, solved), Q27 (3 marks, solved), Q36 (4 marks, solved)

2024

Set 65/1/1: Q4 (1 mark, solved), Q22 (2 marks, solved), Q36 (4 marks, solved)

Set 65/5/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q23 (2 marks, solved), Q37 (4 marks, solved)

Sample paper 2023-24: Q19 (1 mark, solved), Q22 (2 marks, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q25 (2 marks, solved), Q38 (4 marks, solved)

2023

Set 65/1/1: Q16 (1 mark, solved), Q25 (2 marks, solved), Q37 (4 marks, solved)

Set 65/5/1: Q9 (1 mark, solved), Q32(a) (5 marks, solved), Q32(b) (5 marks, solved)

Practice questions on Application of Derivatives

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The rate of change of the volume of a sphere with respect to its radius, when r = 3 cm, is:
(a) $36 π$ cm³/cm
(b) $12 π$ cm³/cm
(c) $27 π$ cm³/cm
(d) $9 π$ cm³/cm
Hint
$\frac{d V}{d r} = 4 π r^{2}$ .
Answer
(a) $36 π$
Q2 · 1 mark · MCQBoard level
The function $f (x) = e^{- x}$ on R is:
(a) increasing
(b) decreasing
(c) neither increasing nor decreasing
(d) constant
Hint
Sign of the derivative.
Answer
(b) decreasing, since $f^{'} (x) = - e^{- x} < 0$
Q3 · 1 mark · MCQBoard level
The maximum value of $f (x) = - x^{2} + 4 x + 1$ is:
(a) 1
(b) 3
(c) 5
(d) 4
Hint
Solve $f^{'} (x) = 0$ .
Answer
(c) 5, at x = 2
Q4 · 1 mark · Assertion-reasonBoard level
Assertion (A): $f (x) = x + \frac{1}{x}$ , x > 0, has a local minimum at x = 1.
Reason (R): $f^{'} (1) = 0$ and $f^{''} (1) > 0$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$f^{'} (x) = 1 - \frac{1}{x ^{2}}$ , $f^{''} (x) = \frac{2}{x ^{3}}$ .
Answer
(a). f′(1) = 0 and f″(1) = 2 > 0, which is exactly the second derivative test for a minimum (value 2).
Q5 · 2 marks · Very short answerHarder board level
The total cost (in ₹) of producing x items is $C (x) = 0.004 x^{3} - 0.06 x^{2} + 20 x + 1500$ . Find the marginal cost when 50 items are produced.
Hint
Differentiate, then put x = 50.
Answer
$M C = 0.012 x^{2} - 0.12 x + 20 = 30 - 6 + 20 = 44$ , so ₹44.
Q6 · 2 marks · Very short answerHarder board level
Find the intervals in which $f (x) = (x - 1)^{2} (x + 2)$ is increasing or decreasing.
Hint
Product rule, then factorise.
Answer
$f^{'} (x) = 2 (x - 1) (x + 2) + (x - 1)^{2} = 3 (x - 1) (x + 1)$ . Increasing on (−∞, −1] and [1, ∞); decreasing on [−1, 1].
Q7 · 3 marks · Short answerHarder board level
A balloon rises vertically at 3 m/s from a point on the ground 40 m from an observer. How fast is the distance between the balloon and the observer increasing when the balloon is 30 m high?
Hint
$s^{2} = 4 0^{2} + h^{2}$ .
Answer
$2 s \frac{d s}{d t} = 2 h \frac{d h}{d t}$ . At h = 30, s = 50: $\frac{d s}{d t} = \frac{30}{50} (3) = 1.8$ m/s.
Q8 · 3 marks · Short answerHarder board level
Show that $f (x) = tan^{- 1} (sin x + cos x)$ is decreasing on $(\frac{π}{4}, \frac{π}{2})$ .
Hint
Compare sin x and cos x for $x > \frac{π}{4}$ .
Answer
$f^{'} (x) = \frac{c o s x - s i n x}{1 + ( s i n x + c o s x ) ^{2}}$ . The denominator is positive, and on $(\frac{π}{4}, \frac{π}{2})$ , $sin x > cos x$ , so $f^{'} (x) < 0$ : f is decreasing.
Q9 · 3 marks · Short answerHarder board level
Find the absolute maximum and minimum values of $f (x) = 2 x^{3} - 9 x^{2} + 12 x - 3$ on [0, 3].
Hint
Critical points 1 and 2; check the ends too.
Answer
f(0) = −3, f(1) = 2, f(2) = 1, f(3) = 6. Absolute maximum 6 at x = 3; absolute minimum −3 at x = 0.
Q10 · 5 marks · Long answerHarder board level
An open tank with a square base and vertical sides must hold 32 m³. Find its dimensions so that the area of metal sheet used is least.
Hint
$h = \frac{32}{x ^{2}}$ ; area = base + four sides.
Answer
$S = x^{2} + 4 x h = x^{2} + \frac{128}{x}$ . $S^{'} = 2 x - \frac{128}{x ^{2}} = 0 \Rightarrow x = 4$ ; $S^{''} = 2 + \frac{256}{x ^{3}} > 0$ (minimum). Base 4 m × 4 m, depth 2 m, least area 48 m².
Q11 · 3 marks · Short answerHOTS / competency
Find all values of k for which $f (x) = k x^{3} - 9 k x^{2} + 9 x + 3$ is increasing on R.
Hint
Need $3 k x^{2} - 18 k x + 9 \geq 0$ for every x; treat k = 0 separately.
Answer
k = 0 gives $f^{'} = 9 > 0$ . For k ≠ 0 we need k > 0 and discriminant $324 k^{2} - 108 k \leq 0$ , i.e. $0 < k \leq \frac{1}{3}$ (at $k = \frac{1}{3}$ , $f^{'} = (x - 3)^{2}$ vanishes at one point only). So $0 \leq k \leq \frac{1}{3}$ .
Q12 · 2 marks · Very short answerHOTS / competency
Using a suitable increasing function, prove that $e^{x} > 1 + x$ for all x > 0.
Hint
Let $g (x) = e^{x} - 1 - x$ and compare with g(0).
Answer
$g^{'} (x) = e^{x} - 1 > 0$ for x > 0, so g is increasing on $[0, \infty)$ . Hence $g (x) > g (0) = 0$ for x > 0, i.e. $e^{x} > 1 + x$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does a negative value of dV/dt mean?Show answer
V is decreasing with time.
2.What is marginal cost?Show answer
The rate of change of total cost with output, dC/dx.
3.State the derivative test for an increasing function.Show answer
If f′(x) > 0 on (a, b) and f is continuous on [a, b], then f is increasing on [a, b].
4.Define a critical point.Show answer
A point in the domain where f′(c) = 0 or f is not differentiable.
5.If f′(c) = 0 and f″(c) > 0, what is c?Show answer
A point of local minimum.
6.When does the second derivative test fail?Show answer
When f″(c) = 0; use the first derivative test instead.
7.Where can the absolute maximum on [a, b] occur?Show answer
At a critical point inside the interval or at an end point.
8.Is e^x increasing or decreasing on R?Show answer
Increasing, since its derivative e^x is always positive.

Best order to study it

Revise the chain rule and implicit differentiation from Chapter 5, and the mensuration formulas for spheres, cones and cylinders.
Learn rate of change, marginal cost and marginal revenue; then practise related-rates problems with a figure every time.
Learn the definitions of increasing and decreasing functions and the derivative test; practise sign tables.
Study critical points, the first derivative test and the second derivative test.
Practise absolute maximum and minimum on closed intervals.
Finish with optimisation word problems and case studies, writing the justification step every time.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find the rate of change of area or volume at a given radius or edge?
Can you solve a ladder, shadow or cone related-rates problem with correct signs and units?
Can you find marginal cost and marginal revenue at a given output?
Can you prove that a function is increasing or decreasing on an interval?
Can you find the intervals of increase and decrease using a sign table?
Can you find a parameter k so that f is increasing on R?
Can you find critical points, including points where f is not differentiable?
Can you apply the first and second derivative tests, and know when the second fails?
Can you find absolute extrema on [a, b] by comparing values?
Can you set up and solve an optimisation problem with a full justification?

Application of Derivatives: frequently asked questions

Are tangents and normals in the 2026-27 Class 12 Maths syllabus?

No. Tangents and normals and approximations using differentials were removed in the NCERT rationalisation, and the 2026-27 syllabus lists only rate of change, increasing and decreasing functions, and maxima and minima.

Should I use the first or the second derivative test?

Use the second derivative test when the second derivative is easy to find, since it is quicker. If it gives zero, or the function is not differentiable at the point, use the first derivative test.

Do I write intervals with round or square brackets?

Both are accepted when the function is continuous at the end point. NCERT often uses closed intervals for increasing and decreasing, while open intervals are also correct.

How important is this chapter?

It belongs to Calculus, which carries 35 of the 80 marks. In our question bank it has 45 questions and appears in all eight board papers from 2023 to 2026, with case studies on rates and optimisation in several of them.

What is the difference between local and absolute maximum?

A local maximum is the highest value only near a point. The absolute maximum is the highest value over the whole interval, and it can be at an end point.

Is a critical point always a maximum or a minimum?

No. At a critical point the derivative is zero or does not exist, but the function may just flatten and carry on, as x cubed does at zero. Such a point is a point of inflection.

Based on: CBSE Mathematics (041) syllabus 2026-27; NCERT Mathematics Part I, Textbook for Class XII, Chapter 6 (Reprint 2026-27)