MathsISC Class 12 · Chapter 6

Applications of Derivatives ISC Class 12: Notes, Methods and Solved Problems

Updated 14 October 2026 · Fact-checked

Applications of Derivatives is Chapter 6 of ISC Class 12 Mathematics. It covers rates of change, increasing and decreasing functions, equations of tangents and normals, the angle between two curves, and maxima and minima using the first and second derivative tests, including absolute extrema and real-life optimisation problems.

What Applications of Derivatives is about

The derivative measures how fast one quantity changes with another, and this chapter puts that to work. It is one of the most heavily tested parts of the 35 Calculus marks, and long application problems regularly appear in Section D.

Rate of change: if y depends on x and x changes with time, $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ . Typical problems involve growing circles, inflating spheres, filling tanks and sliding ladders.

Increasing and decreasing functions: f is increasing where f′(x) ≥ 0 and strictly increasing where f′(x) > 0 (similarly decreasing). Find the critical points (f′ = 0), split the number line, and test the sign of f′ in each interval.

Tangents and normals: the slope of the tangent at $(x_{1}, y_{1})$ is $m = (\frac{d y}{d x})_{(x_{1}, y_{1})}$ ; the tangent is $y - y_{1} = m (x - x_{1})$ and the normal $y - y_{1} = - \frac{1}{m} (x - x_{1})$ . The angle between two curves at a common point is the angle between their tangents, $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ ; they cut orthogonally if $m_{1} m_{2} = - 1$ .

Maxima and minima: critical points, stationary (turning) points and extreme points. The first derivative test looks at the change in sign of f′; the second derivative test uses f″(c) < 0 (local maximum) or f″(c) > 0 (local minimum). Absolute maxima and minima on a closed interval come from comparing values at critical points and end points. Finally, real-life optimisation: write the quantity in one variable, differentiate, find the critical point and confirm it with the second derivative test.

The 2027 syllabus no longer includes the commerce applications (cost, revenue, profit) that appeared in earlier Section C papers.

Weightage: Applications of Derivatives is part of the unit Calculus, which carries 35 of the 80 marks in the ISC Mathematics, shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Rate of change: $\frac{d y}{d x}$ is the rate of change of y with respect to x. When both depend on time, use the chain rule: $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$ . Method: (1) write the formula linking the quantities (area, volume, Pythagoras); (2) differentiate with respect to t; (3) substitute the values at the given instant only after differentiating. Signs matter: a decreasing quantity has a negative rate.; Worked example
A spherical balloon is inflated so that its volume increases at 100 cm³ s⁻¹. How fast is its radius increasing when the radius is 5 cm?
Try it first, then show the solutionSolution
$V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
$100 = 4 π (25) \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{1}{π} c m s^{- 1}$
Answer: $\frac{1}{π} \approx 0.32$ cm s⁻¹
Increasing and decreasing functions: On an interval, f is increasing if $x_{1} < x_{2} \Rightarrow f (x_{1}) \leq f (x_{2})$ and strictly increasing if $f (x_{1}) < f (x_{2})$ ; similarly for decreasing. Test: f′(x) > 0 on an interval ⇒ strictly increasing; f′(x) < 0 ⇒ strictly decreasing; f′(x) ≥ 0 ⇒ increasing. Method: solve f′(x) = 0, mark the roots on the number line, find the sign of f′ in each part. Use open intervals for strictly increasing/decreasing and be careful with brackets.; Worked example
Find the intervals in which $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 5$ is strictly increasing or strictly decreasing.
Try it first, then show the solutionSolution
$f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$
f′ > 0 for x < 1 and x > 2; f′ < 0 for 1 < x < 2.
Answer: Strictly increasing on (−∞, 1) ∪ (2, ∞); strictly decreasing on (1, 2)
Tangents and normals: The slope of the tangent to y = f(x) at $(x_{1}, y_{1})$ is $m = f^{'} (x_{1})$ . Tangent: $y - y_{1} = m (x - x_{1})$ . Normal (perpendicular to the tangent): $y - y_{1} = - \frac{1}{m} (x - x_{1})$ . Special cases: if m = 0 the tangent is horizontal (y = $y_{1}$ ) and the normal vertical; if the slope is infinite, the tangent is vertical. A tangent parallel to a given line has the same slope; perpendicular, slope product −1. For a parametric curve, use $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ .; Worked example
Find the equations of the tangent and normal to $y = x^{2} - 2 x + 3$ at x = 2.
Try it first, then show the solutionSolution
y(2) = 3; $\frac{d y}{d x} = 2 x - 2$ , so m = 2.
Tangent: y − 3 = 2(x − 2), i.e. y = 2x − 1.
Normal: $y - 3 = - \frac{1}{2} (x - 2)$ , i.e. x + 2y − 8 = 0.
Answer: Tangent y = 2x − 1; normal x + 2y = 8
Angle between two curves: The angle between two intersecting curves at a common point is the angle θ between their tangents there: $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ . The curves cut orthogonally if $m_{1} m_{2} = - 1$ and touch if $m_{1} = m_{2}$ . Method: find the point(s) of intersection first, then each slope by differentiating (implicitly if necessary).; Worked example
Find the angle between the curves $y^{2} = x$ and xy = 1 at their point of intersection.
Try it first, then show the solutionSolution
Intersection: $x = y^{2}$ and $x y = 1$ give $y^{3} = 1$ , so (1, 1).
$y^{2} = x$ : $2 y y^{'} = 1$ , $m_{1} = \frac{1}{2}$ . xy = 1: $y^{'} = - \frac{y}{x}$ , $m_{2} = - 1$ .
$tan θ = \frac{\frac{1}{2} + 1}{1 - \frac{1}{2}} = 3$
Answer: $θ = tan^{- 1} 3$ (about 71.6°)
Local maxima and minima: first and second derivative tests: A critical point is where f′(c) = 0 or f′ does not exist; points with f′(c) = 0 are stationary (turning) points. First derivative test: if f′ changes from + to − at c, f has a local maximum; from − to +, a local minimum; no change, neither (a point of inflection, as for x³ at 0). Second derivative test: if f′(c) = 0 and f″(c) < 0, local maximum; f″(c) > 0, local minimum; f″(c) = 0, the test fails and you return to the first derivative test. A local maximum can be smaller than a local minimum elsewhere.; Worked example
Find the local maximum and minimum values of $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 5$ .
Try it first, then show the solutionSolution
f′(x) = 6(x − 1)(x − 2) = 0 at x = 1, 2; f″(x) = 12x − 18.
f″(1) = −6 < 0: local maximum f(1) = 2 − 9 + 12 + 5 = 10.
f″(2) = 6 > 0: local minimum f(2) = 16 − 36 + 24 + 5 = 9.
Answer: Local maximum 10 at x = 1; local minimum 9 at x = 2
Absolute maxima and minima on a closed interval: A continuous function on [a, b] attains an absolute maximum and an absolute minimum. They occur either at critical points inside the interval or at the end points. Method: find the critical points in (a, b), evaluate f at them and at a and b, and pick the largest and smallest values.; Worked example
Find the absolute maximum and minimum of $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 5$ on [0, 3].
Try it first, then show the solutionSolution
Critical points in (0, 3): x = 1, 2.
f(0) = 5, f(1) = 10, f(2) = 9, f(3) = 54 − 81 + 36 + 5 = 14.
Answer: Absolute maximum 14 at x = 3; absolute minimum 5 at x = 0
Optimisation problems: Steps: (1) draw a diagram and name the variables; (2) write the quantity to be maximised or minimised; (3) use the given condition (fixed perimeter, volume, area) to express it in one variable; (4) differentiate, set the derivative to zero and solve; (5) confirm maximum or minimum with the second derivative test; (6) answer the question asked (dimensions, value) with units. Standard results worth knowing: a closed cylinder of fixed volume has least surface area when h = 2r; among rectangles of fixed perimeter, the square has the greatest area.; Worked example
A closed cylindrical can must hold 128π cm³. Find the radius and height that use the least material.
Try it first, then show the solutionSolution
$π r^{2} h = 128 π \Rightarrow h = \frac{128}{r ^{2}}$ ; surface $S = 2 π r^{2} + 2 π r h = 2 π r^{2} + \frac{256 π}{r}$ .
$S^{'} = 4 π r - \frac{256 π}{r ^{2}} = 0 \Rightarrow r^{3} = 64 \Rightarrow r = 4$ ; h = 128/16 = 8.
$S^{''} = 4 π + \frac{512 π}{r ^{3}} > 0$ : minimum.
Answer: r = 4 cm, h = 8 cm (height equals diameter)

Maths in real life: Applications 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.

Related rates

Suppose a 5 m ladder leans against a wall and its foot slides away from the wall at 0.3 m s⁻¹.

How fast is the top sliding down when the foot is 3 m from the wall?

Try it first, then show the solutionSolution

$x^{2} + y^{2} = 25 \Rightarrow 2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$
At x = 3, y = 4: $\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} = - \frac{3}{4} (0.3) = - 0.225$

Answer: The top slides down at 0.225 m s⁻¹

Memory hook: Write the relation, differentiate in time, then plug in.

Rate of change of volume

Imagine water being pumped into a cylindrical tank of radius 2 m at 6.28 m³ per minute.

How fast is the water level rising? (Take π = 3.14.)

Try it first, then show the solutionSolution

$V = π r^{2} h = 4 π h \Rightarrow \frac{d V}{d t} = 4 π \frac{d h}{d t}$
$\frac{d h}{d t} = \frac{6.28}{4 \times 3.14} = 0.5$

Answer: 0.5 m per minute

Memory hook: Volume rate ÷ cross-section area = level rate.

Optimisation

Suppose a farmer has 100 m of fencing to enclose a rectangular plot beside a straight river; no fence is needed along the river.

Find the dimensions giving the largest area.

Try it first, then show the solutionSolution

Width w (two sides) and length l (one side): 2w + l = 100, so A = w(100 − 2w).
A′ = 100 − 4w = 0 ⇒ w = 25, l = 50; A″ = −4 < 0: maximum. A = 1250 m².

Answer: 25 m × 50 m (long side along the river), area 1250 m²

Memory hook: One variable, derivative zero, check the sign of the second derivative.

Topics in this chapter

Applications of Derivatives formula sheet

Related rates: $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$; Substitute values after differentiating.
Increasing/decreasing test: $f^{'} (x) > 0 \Rightarrow strictly increasing; f^{'} (x) < 0 \Rightarrow strictly decreasing$; f′ ≥ 0: increasing.
Tangent: $y - y_{1} = m (x - x_{1}), m = (\frac{d y}{d x})_{(x_{1}, y_{1})}$
Normal: $y - y_{1} = - \frac{1}{m} (x - x_{1})$; If m = 0 the normal is x = $x_{1}$ .
Angle between curves: $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$; Orthogonal: $m_{1} m_{2} = - 1$ .
Second derivative test: $f^{'} (c) = 0 : f^{''} (c) < 0 \Rightarrow local max; f^{''} (c) > 0 \Rightarrow local min$; f″(c) = 0: use the first derivative test.
Useful mensuration: $V_{s p h er e} = \frac{4}{3} π r^{3}, S_{s p h er e} = 4 π r^{2}, V_{cy l} = π r^{2} h, S_{cy l} = 2 π r^{2} + 2 π r h, V_{co n e} = \frac{1}{3} π r^{2} h$; Equilateral triangle area $\frac{3}{4} a^{2}$ .

How the board asks Applications of Derivatives

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

Question type	How it is asked	Method
Rates of change (1-2 marks)	Circle, sphere, cylinder or ladder with a given rate; find another rate at an instant.	Formula, differentiate with respect to t, then substitute.
Increasing/decreasing	Find intervals; Statement I/II on 'increasing' versus 'strictly increasing'; prove a function is increasing.	Sign table of f′; use correct brackets.
Tangents, normals, angles	Equations at a point; point where the tangent is parallel to a line or the x-axis; angle between curves; orthogonal curves.	Slope from dy/dx; parallel to x-axis means dy/dx = 0.
Local extrema (2-3 marks)	Find local max/min values; show a function has a minimum at a given point.	f′ = 0, then the second derivative test; state the value.
Absolute extrema	Greatest and least values on a closed interval.	Compare values at critical points and end points.
Optimisation (5 marks)	Windows, boxes, cans, fences, inscribed shapes with real-life contexts.	One-variable function, derivative zero, second derivative test, answer with units.

Tricks to remember Applications of Derivatives

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

Plus to minus is a peak
f′ goes + → −: maximum (hill top); − → +: minimum (valley).
Use it when: First derivative test.
Smile and frown
f″ > 0: curve smiles (minimum); f″ < 0: curve frowns (maximum).
Use it when: Second derivative test.
Normal is negative reciprocal
Tangent slope m, normal slope −1/m.
Use it when: Writing normals.
Differentiate before you substitute
In rates problems, put in the instant's values only after differentiating.
Use it when: Avoiding the 'constant' mistake.
Ends count too
Absolute extremes on [a, b]: check the end points as well as the critical points.
Use it when: Closed-interval questions.

Applications 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: Where is x³ − 3x increasing? · 1 mark · MCQBoard level

The function $f (x) = x^{3} - 3 x$ is strictly increasing on:
(a) (−1, 1)
(b) (−∞, −1) ∪ (1, ∞)
(c) (−1, ∞)
(d) (−∞, 1)

Try it first, then show the solutionSolution

f′(x) = 3(x − 1)(x + 1) > 0 when x < −1 or x > 1.

Answer: (b)

Example 2: Stationary but not extreme · 1 mark · Assertion-reasonBoard level

Assertion: f(x) = x³ has no local maximum or minimum at x = 0, although f′(0) = 0.
Reason: f′(x) = 3x² does not change sign as x passes through 0.
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

f′ ≥ 0 on both sides of 0, so by the first derivative test there is no extremum (a point of inflection). The reason explains it.

Answer: (a)

Example 3: Horizontal tangent · 2 marks · Short answerHarder board level

Find the point on the curve $y = x^{2} - 4 x + 5$ at which the tangent is parallel to the x-axis. [2]

Try it first, then show the solutionSolution

Parallel to the x-axis means slope 0: $\frac{d y}{d x} = 2 x - 4 = 0 \Rightarrow x = 2$ .
y = 4 − 8 + 5 = 1.

Answer: (2, 1)

Example 4: Tangent and normal · 3 marks · Short answerHarder board level

Find the equations of the tangent and the normal to $y = x^{2} - 2 x + 3$ at the point where x = 2. [3]

Try it first, then show the solutionSolution

Point (2, 3); slope $\frac{d y}{d x} = 2 x - 2 = 2$ .
Tangent: y − 3 = 2(x − 2) ⇒ 2x − y − 1 = 0.
Normal: y − 3 = −½(x − 2) ⇒ x + 2y − 8 = 0.

Answer: 2x − y − 1 = 0; x + 2y − 8 = 0

Example 5: Angle between two curves · 3 marks · Short answerHarder board level

Find the angle of intersection of the curves $y^{2} = x$ and xy = 1. [3]

Try it first, then show the solutionSolution

Solving: $y^{3} = 1$ , so they meet at (1, 1).
Slopes: $m_{1} = \frac{1}{2 y} = \frac{1}{2}$ , $m_{2} = - \frac{y}{x} = - 1$ .
$tan θ = \frac{\frac{1}{2} - ( - 1 )}{1 + \frac{1}{2} ( - 1 )} = \frac{3/2}{1/2} = 3$ .

Answer: $tan^{- 1} 3$

Example 6: A Norman window · 5 marks · Long answerHarder board level

A window is in the shape of a rectangle surmounted by a semicircle, the semicircle's diameter being the top side of the rectangle. The total perimeter is 10 m.
(a) Taking the radius as r, express the area A in terms of r. [2]
(b) Find r for maximum light (maximum area), prove it is a maximum, and find the maximum area. [3]

Try it first, then show the solutionSolution

(a) Rectangle 2r wide and h high. Perimeter: 2r + 2h + πr = 10 ⇒ $h = \frac{10 - ( 2 + π ) r}{2}$ .
$A = 2 r h + \frac{1}{2} π r^{2} = r [10 - (2 + π) r] + \frac{π r ^{2}}{2} = 10 r - 2 r^{2} - \frac{π r ^{2}}{2}$ .
(b) $\frac{d A}{d r} = 10 - 4 r - π r = 0 \Rightarrow r = \frac{10}{4 + π} \approx 1.40 m$ .
$\frac{d ^{2} A}{d r ^{2}} = - (4 + π) < 0$ : maximum. Then h = r ≈ 1.40 m (width 2r ≈ 2.80 m).
$A_{ma x} = \frac{50}{4 + π} \approx 7.0 m^{2}$ .

Answer: r = 10/(4 + π) ≈ 1.40 m (rectangle 2.80 m wide, 1.40 m high); maximum area ≈ 7.0 m²

Example 7: Absolute extrema on a closed interval · 3 marks · Short answerHOTS / competency

Find the absolute maximum and absolute minimum values of $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 5$ on [0, 3]. [3]

Try it first, then show the solutionSolution

f′(x) = 6(x − 1)(x − 2): critical points 1 and 2 lie inside.
f(0) = 5, f(1) = 10, f(2) = 9, f(3) = 14.
The largest is 14 (at an end point), the smallest 5 (at the other end).

Answer: Absolute maximum 14 at x = 3; absolute minimum 5 at x = 0

Example 8: Inflating a balloon · 2 marks · Short answerHOTS / competency

Air is pumped into a spherical balloon at 100 cm³ s⁻¹. Find the rate at which its radius and its surface area increase when r = 5 cm. [2]

Try it first, then show the solutionSolution

$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{100}{100 π} = \frac{1}{π}$ cm s⁻¹.
$S = 4 π r^{2} \Rightarrow \frac{d S}{d t} = 8 π r \frac{d r}{d t} = 8 π (5) \frac{1}{π} = 40$ cm² s⁻¹.

Answer: $\frac{1}{π}$ cm s⁻¹; 40 cm² s⁻¹

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Making the biggest open box

Suppose a craft club cuts equal squares of side x cm from the corners of a 24 cm × 24 cm card and folds up the sides to make an open box.

(i) · 1 mark
Express the volume V in terms of x and state the possible values of x.
Answer
Answer: $V = x (24 - 2 x)^{2}$ , 0 < x < 12
Base is (24 − 2x) square, height x.
(ii) · 1 mark
Find $\frac{d V}{d x}$ and the critical points.
Answer
Answer: $\frac{d V}{d x} = (24 - 2 x) (24 - 6 x)$ ; x = 4 (x = 12 is excluded).
Product rule: $(24 - 2 x)^{2} - 4 x (24 - 2 x) = (24 - 2 x) (24 - 6 x)$ .
(iii) · 1 mark
Show that x = 4 gives a maximum.
Answer
Answer: V″(4) < 0
$V = 4 x^{3} - 96 x^{2} + 576 x$ , $V^{''} = 24 x - 192$ ; V″(4) = −96 < 0.
(iv) · 2 marks
Find the maximum volume and the box's dimensions.
Answer
Answer: 1024 cm³; base 16 cm × 16 cm, height 4 cm
V = 4 × 16² = 1024.

Watching a ladder slip

Suppose a 5 m ladder rests against a vertical wall on level ground. Its foot is pulled away from the wall at a steady 0.3 m s⁻¹.

(i) · 1 mark
Write the relation between the distance x of the foot from the wall and the height y of the top.
Answer
Answer: $x^{2} + y^{2} = 25$
Pythagoras.
(ii) · 2 marks
Find the rate at which the top slides down when x = 3 m.
Answer
Answer: 0.225 m s⁻¹
$\frac{d y}{d t} = - \frac{x}{y} \frac{d x}{d t} = - \frac{3}{4} (0.3)$
(iii) · 1 mark
Find this rate when x = 4 m.
Answer
Answer: 0.4 m s⁻¹
y = 3: $\frac{4}{3} (0.3) = 0.4$ .
(iv) · 1 mark
What happens to the speed of the top as the ladder nears the ground?
Answer
Answer: It grows without limit in this model.
As y → 0, $\frac{x}{y} \to \infty$ ; in reality the top leaves the wall before that.

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.

Find the point on $y = x^{2} - 4 x + 5$ where the tangent is parallel to the x-axis. (2 marks)

Step	Marks
dy/dx = 2x − 4 equated to 0	1
Point (2, 1)	1

Find the local maximum and minimum values of $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 5$ . (3 marks)

Step	Marks
f′(x) = 6(x − 1)(x − 2) = 0 gives x = 1, 2	1
f″(x) = 12x − 18; f″(1) < 0, f″(2) > 0	1
Local max f(1) = 10; local min f(2) = 9	1

A closed cylinder has volume 128π cm³. Find the dimensions for least total surface area. (5 marks)

Step	Marks
$h = 128/ r^{2}$ from the volume	1
$S = 2 π r^{2} + \frac{256 π}{r}$ in one variable	1
S′ = 0 gives r = 4	1
S″ > 0: minimum	1
r = 4 cm, h = 8 cm	1

Where students lose marks in Applications of Derivatives

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

Mistake: Not expressing the quantity in one variable
CISCE examiners' comments, 2024: in a window-design problem candidates could not express the area in one variable, used a wrong formula for the area of an equilateral triangle, failed to simplify and did not complete the second derivative test.
Wrong working
A = xy + (√3/4)x² differentiated with respect to x, keeping y.
Right working
Use the perimeter to write y in terms of x first, then differentiate A(x).
Do this: Write the constraint, substitute, then differentiate.
Usually costs: 2-4 marks.
Mistake: Skipping the second derivative test
CISCE examiners' comments, 2024 and 2025: in optimisation problems candidates found the critical point but did not confirm the maximum or minimum, and could not show that a function has a minimum at a given point.
Wrong working
f′(0) = 0, hence minimum.
Right working
f′(0) = 0 and f″(0) > 0, hence local minimum (or show f′ changes from − to +).
Do this: Always add the test line; it carries a mark.
Usually costs: 1 mark.
Mistake: Confusing increasing with strictly increasing
CISCE examiners' comments, 2025: in a Statement I/II item on a graph, candidates confused 'increasing' with 'strictly increasing' and used incorrect brackets.
Wrong working
Writes strictly increasing on [1, 2].
Right working
Strictly increasing (f′ > 0) on the open interval (1, 2); increasing (f′ ≥ 0) can include flat parts.
Do this: Use ( ) for strictly increasing/decreasing intervals found from f′ > 0 or f′ < 0.
Usually costs: 1 mark.
Mistake: Not differentiating with respect to time
CISCE examiners' comments, 2025: in a related-rates item (popcorn tub), candidates forgot to differentiate with respect to time.
Wrong working
V = πr²h, so h = V/(πr²) = 314/(100π) … treating rates as quantities without a time derivative.
Right working
$\frac{d V}{d t} = π r^{2} \frac{d h}{d t}$ (r fixed), so $\frac{d h}{d t} = \frac{314}{100 π} = 1$ cm per minute.
Do this: Write d/dt on both sides explicitly.
Usually costs: 1 mark.
Mistake: Wrong derivative or not equating it to zero for turning points
CISCE examiners' comments, 2025: finding turning points of a cubic, candidates differentiated incorrectly or did not equate the derivative to zero; in another item they made errors locating the point where the tangent is parallel to the x-axis.
Wrong working
Turning points where f(x) = 0.
Right working
Turning points where f′(x) = 0; then find the y-coordinates from f.
Do this: Roots of f′, not of f.
Usually costs: 1-2 marks.

Practice questions on Applications 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 slope of the normal to $y = x^{2}$ at (1, 1) is:
(a) 2
(b) −2
(c) $\frac{1}{2}$
(d) $- \frac{1}{2}$
Hint
Negative reciprocal of 2.
Answer
(d)
Q2 · 1 mark · MCQBoard level
The rate of change of the area of a circle with respect to its radius when r = 5 cm is:
(a) 5π
(b) 10π
(c) 25π
(d) 2π
Hint
dA/dr = 2πr.
Answer
(b) 10π cm² per cm
Q3 · 1 mark · MCQBoard level
If f′(c) = 0 and f″(c) < 0, then at x = c, f has:
(a) a local minimum
(b) a local maximum
(c) a point of inflection
(d) no conclusion possible
Hint
Frown.
Answer
(b)
Q4 · 1 mark · Very short answerBoard level
Find the critical point of $f (x) = x^{2} - 6 x + 4$ .
Hint
f′ = 0.
Answer
x = 3
Q5 · 2 marks · Short answerBoard level
The radius of a circle increases at 0.5 cm s⁻¹. Find the rate of increase of its area when r = 10 cm.
Hint
dA/dt = 2πr dr/dt.
Answer
10π ≈ 31.4 cm² s⁻¹
Q6 · 2 marks · Short answerHarder board level
Find the intervals in which $f (x) = 2 x^{3} - 9 x^{2} + 12 x + 5$ is strictly increasing and strictly decreasing.
Hint
Factorise f′.
Answer
Increasing on (−∞, 1) ∪ (2, ∞); decreasing on (1, 2).
Q7 · 2 marks · Short answerHarder board level
A 5 m ladder's foot slides away from a wall at 0.3 m s⁻¹. How fast is the top sliding down when the foot is 3 m from the wall?
Hint
x² + y² = 25.
Answer
0.225 m s⁻¹
Q8 · 2 marks · Short answerHOTS / competency
Show that $f (x) = x + \frac{1}{x}$ has a local maximum value that is less than its local minimum value.
Hint
Find both.
Answer
f′ = 1 − 1/x² = 0 at x = ±1; f″ = 2/x³. At x = −1, f″ < 0: local max f(−1) = −2. At x = 1, f″ > 0: local min f(1) = 2. So −2 < 2.
Q9 · 3 marks · Short answerHarder board level
A closed cylinder of volume 128π cm³ is to use the least material. Find its radius and height.
Hint
h = 2r at the optimum.
Answer
r = 4 cm, h = 8 cm (S″ > 0 confirms the minimum).
Q10 · 3 marks · Short answerHarder board level
Show that $f (x) = lo g (1 + x) - \frac{2 x}{2 + x}$ is increasing for x > −1.
Hint
Combine f′ over a common denominator.
Answer
$f^{'} (x) = \frac{1}{1 + x} - \frac{4}{( 2 + x ) ^{2}} = \frac{( 2 + x ) ^{2} - 4 ( 1 + x )}{( 1 + x ) ( 2 + x ) ^{2}} = \frac{x ^{2}}{( 1 + x ) ( 2 + x ) ^{2}} \geq 0$ for x > −1, so f is increasing.
Q11 · 3 marks · Short answerHarder board level
With 100 m of fencing a farmer encloses a rectangle against a straight river (no fence along the river). Find the maximum area.
Hint
2w + l = 100.
Answer
A = w(100 − 2w), A′ = 0 at w = 25, A″ < 0; 25 m × 50 m, area 1250 m².
Q12 · 5 marks · Long answerHarder board level
(a) Squares of side x are cut from the corners of a 24 cm square card to make an open box. Express V in terms of x. [1]
(b) Find x for maximum volume, prove it is a maximum, and find the maximum volume. [4]
Hint
V = x(24 − 2x)².
Answer
(a) $V = x (24 - 2 x)^{2}$ . (b) $V^{'} = (24 - 2 x) (24 - 6 x) = 0$ ⇒ x = 4 (x = 12 gives no box); $V^{''} = 24 x - 192 = - 96 < 0$ at x = 4; $V_{ma x} = 4 \times 1 6^{2} = 1024$ cm³.
Q13 · 2 marks · Short answerHarder board level
Find the equations of the tangent and normal to $y = x^{2} - 2 x + 3$ at x = 2.
Hint
Slope 2 at (2, 3).
Answer
Tangent y = 2x − 1; normal x + 2y = 8.

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.Condition for f to be strictly increasing on an interval?Show answer
f′(x) > 0 throughout the interval.
2.Equation of the normal at $(x_{1}, y_{1})$ ?Show answer
$y - y_{1} = - \frac{1}{m} (x - x_{1})$ .
3.When do two curves cut orthogonally?Show answer
When the product of their slopes at the point of intersection is −1.
4.What is a critical point?Show answer
A point where f′(c) = 0 or f′ does not exist.
5.First derivative test for a local maximum?Show answer
f′ changes sign from + to − at c.
6.Where do absolute extrema on [a, b] occur?Show answer
At critical points in (a, b) or at the end points.
7.Least surface area for a closed cylinder of fixed volume?Show answer
When height = diameter (h = 2r).
8.What if f″(c) = 0?Show answer
The second derivative test fails; use the first derivative test.

Best order to study it

Practise related rates with circles, spheres, cylinders and ladders.
Learn increasing/decreasing tests and sign tables.
Practise tangents and normals, then angles between curves.
Learn local extrema with both tests.
Practise absolute extrema on closed intervals.
Finish with full optimisation problems in real contexts.

Quick revision checklist

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

Can you solve a related-rates problem by differentiating before substituting?
Can you find intervals of increase and decrease with correct brackets?
Can you write the tangent and normal at a point?
Can you find where a tangent is parallel to the x-axis or a given line?
Can you find the angle between two curves?
Can you apply both derivative tests and handle f″ = 0?
Can you find absolute extrema on [a, b]?
Can you set up and solve an optimisation problem with units?

Applications of Derivatives: frequently asked questions

Is the commerce application of calculus (cost, revenue, profit) in ISC 2027?

No. Application of calculus in commerce and economics was part of the old Section C option and is not in the 2027 syllabus. Applications now cover rates, monotonicity, tangents and normals, angle between curves and maxima/minima.

Is the angle between two curves in the syllabus?

Yes. The ISC syllabus lists equations of tangent and normal and the angle between two curves.

Which test should I use for maxima and minima?

The second derivative test is usually quickest, and the syllabus treats it as a provable tool. If f″(c) = 0, or the function is awkward to differentiate twice, use the first derivative test.

What is the difference between increasing and strictly increasing?

Strictly increasing means f(x₁) < f(x₂) whenever x₁ < x₂ (f′ > 0); increasing allows equality (f′ ≥ 0), so flat stretches are allowed.

How many marks does this chapter carry?

It is part of Calculus (35 of 80 marks); optimisation problems are typical 5-mark Section D questions.

Do I need to state units in optimisation answers?

Yes. Give dimensions and the optimum value with units, and answer exactly what the question asks.

Based on: CISCE ISC 2027 Mathematics syllabus; CISCE ISC 2027 Mathematics specimen question paper and answer key; CISCE Analysis of Pupil Performance, ISC Mathematics, 2025; CISCE Analysis of Pupil Performance, ISC Mathematics, 2024