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Application of Derivatives for ISC Class 12 Maths: Methods and 12 Worked Examples
Application of derivatives in ISC Class 12 Maths uses the derivative for four jobs: finding rates of change, deciding where a function is increasing or decreasing, writing tangents and normals (including the angle between two curves), and finding maxima and minima, including real-life optimisation problems. It sits inside the Calculus unit, which carries 35 of the 80 theory marks in the ISC 2027 syllabus. Each question type has a fixed method, and most lost marks come from skipping one step of it.
This guide is part of our ISC home tutors in Gurgaon series for Class 11 and 12. It sets out what the ISC 2027 syllabus includes, then gives a method and worked examples for each question type (every answer checked step by step and by computer algebra), the common mistakes our tutors see, and the link to JEE.
What the ISC 2027 syllabus includes
The official ISC Mathematics (860) syllabus for the 2027 examination lists applications of derivatives as part (ii) of Unit 3, Calculus, in Class XII. It covers rate of change, increasing and decreasing functions, tangents and normals, and maxima and minima. The first derivative test is to be "motivated geometrically" and the second derivative test "given as a provable tool", and the syllabus asks for problems that show basic principles as well as real-life situations.
| Topic | In the ISC 2027 Class XII syllabus? | What to be able to do |
|---|---|---|
| Rate of change ("rate measure") | Yes | Find how fast one quantity changes when another changes with time |
| Increasing and decreasing functions | Yes | Find intervals where a function increases or decreases; find a parameter so a function is increasing |
| Equation of tangent and normal | Yes | Write tangent and normal at a point, for Cartesian, implicit and parametric curves |
| Angle between two curves | Yes | Find the angle between the tangents of two curves where they meet |
| Maxima and minima | Yes | Critical, stationary and turning points; local maxima and minima by the first and second derivative tests; absolute maxima and minima; application problems |
| Approximations and errors using differentials | Not listed | Not in the 2027 Class XII syllabus text |
| Rolle's theorem and the mean value theorem | Not listed | Not in the 2027 Class XII syllabus text |
| Point of inflexion | Only in the suggested projects list | Possible as project work |
The syllabus gives no separate marks for this chapter; it is part of the 35 marks for Calculus. Older books may still teach approximations or the mean value theorems here, but those are not listed in the 2027 Class XII syllabus, so check with your teacher first. For every unit and its marks, see our guide to the ISC Class 12 Maths syllabus for 2027.
Key ideas: what the derivative tells you
Everything in this chapter comes from two meanings of the derivative.
- Rate of change. dy/dx tells you how fast y changes as x changes. If both depend on time t, the chain rule links their rates: dy/dt = (dy/dx) × (dx/dt).
- Slope of the tangent. At the point (x1, y1) on the curve y = f(x), the tangent has slope m = f′(x1). The normal is perpendicular to the tangent, so its slope is −1/m (when m ≠ 0).
From the slope meaning come the other two topics:
- Sign of f′(x). If f′(x) > 0 throughout an interval, f is strictly increasing there; if f′(x) < 0, it is strictly decreasing.
- Where f′(x) = 0. The tangent is horizontal. These stationary points, together with points where f′ does not exist, are the critical points, and every local maximum or minimum inside an interval is at one of them.
A method for each question type
The chapter on one screen; each row is expanded, with worked examples, below.
| Question type | Method in short |
|---|---|
| Rate of change | Write the formula linking the quantities; reduce to one variable if needed; differentiate with respect to t; substitute values last |
| Increasing or decreasing intervals | Find f′(x); factorise; mark the zeros on a number line; test the sign in each interval |
| Parameter for an increasing function | Require f′(x) ≥ 0 for all x; for a quadratic f′, use the discriminant |
| Tangent and normal | Find the point and the slope at it; use y − y1 = m(x − x1); normal slope −1/m |
| Angle between two curves | Find the intersection points; find each curve's slope there; use tan θ = |(m1 − m2)/(1 + m1m2)| |
| Local maxima and minima | Solve f′(x) = 0; use the second derivative test, or the first derivative test if f″ = 0 |
| Absolute maxima and minima on [a, b] | Evaluate f at the critical points inside the interval and at both endpoints; compare |
| Word problems (optimisation) | Name variables; write the quantity to optimise; use the constraint to reduce it to one variable; differentiate; test; answer the question asked |
Rate of change (rate measure)
Rate of change questions give one rate (often dx/dt) and ask for another (dy/dt). The link is a formula connecting x and y, differentiated with respect to time.
Method
- Draw a diagram, name every changing quantity, and write the formula that connects them (area, volume, Pythagoras, similar triangles).
- If the formula has more variables than you have rates for, use a fixed relation (such as a ratio from similar triangles) to remove one.
- Differentiate both sides with respect to t.
- Only now substitute the values at the given instant, and give units. A negative rate means the quantity is decreasing.
Worked example 1: The radius of a circle is increasing at 0.5 cm/s. Find the rate at which its area is increasing when the radius is 4 cm.
Answer: 4π cm2/s (about 12.57 cm2/s). A = πr2, so dA/dt = 2πr × dr/dt. At r = 4 cm with dr/dt = 0.5 cm/s: dA/dt = 2π × 4 × 0.5 = 4π cm2/s. Note that the answer depends on r: the same growth in radius adds area faster when the circle is bigger.
Worked example 2: Water is poured at 4 cm3/s into a conical vessel, vertex down, of base radius 5 cm and height 10 cm. How fast is the water level rising when the water is 4 cm deep?
Answer: 1/π cm/s (about 0.32 cm/s). Let the water have depth h and surface radius r. By similar triangles, r/h = 5/10, so r = h/2. The volume of water is V = (1/3)πr2h = (1/3)π(h/2)2h = πh3/12. Differentiate: dV/dt = (πh2/4) × dh/dt. At h = 4: 4 = (π × 16/4) × dh/dt = 4π × dh/dt, so dh/dt = 1/π cm/s. Trap: keeping both r and h in the volume formula. With two variables and only one known rate, you cannot finish; the similar-triangle ratio removes r.
Increasing and decreasing functions
If f′(x) > 0 at every point of an open interval, f is strictly increasing there; if f′(x) < 0, strictly decreasing. A function can still be strictly increasing if f′(x) = 0 only at isolated points: f(x) = x3 is strictly increasing on all real numbers even though f′(0) = 0.
Method
- Find f′(x) and factorise it fully.
- Find where f′(x) = 0 (and where it does not exist). Mark these on a number line, within the domain.
- Test the sign of f′(x) in each interval, by taking a test point or by the signs of the factors.
- Write the intervals clearly, joining separate intervals with "and" or ∪.
Worked example 3: Find the intervals in which f(x) = 2x3 − 9x2 + 12x + 5 is strictly increasing or strictly decreasing.
Answer: strictly increasing on (−∞, 1) and (2, ∞); strictly decreasing on (1, 2). f′(x) = 6x2 − 18x + 12 = 6(x − 1)(x − 2). The zeros are x = 1 and x = 2. For x < 1, both factors are negative, so f′(x) > 0. For 1 < x < 2, one factor is negative, so f′(x) < 0. For x > 2, both are positive, so f′(x) > 0. Many textbooks also accept the closed forms (−∞, 1] and [2, ∞) for increasing and [1, 2] for decreasing; follow your teacher's convention and be consistent.
Worked example 4: Find the values of k for which f(x) = x3 + kx2 + 3x + 1 is increasing on the set of all real numbers.
Answer: −3 ≤ k ≤ 3. f′(x) = 3x2 + 2kx + 3. We need f′(x) ≥ 0 for every real x. This quadratic has a positive leading coefficient, so it is never negative exactly when its discriminant is not positive: (2k)2 − 4 × 3 × 3 ≤ 0, that is 4k2 − 36 ≤ 0, so k2 ≤ 9 and −3 ≤ k ≤ 3. At the boundary values, for example k = 3, f′(x) = 3(x + 1)2, which is zero only at x = −1, so f is still increasing. Trap: writing the discriminant < 0 and losing k = ±3.
Worked example 5: Find the intervals in [0, 2π] in which f(x) = sin x + cos x is strictly increasing or strictly decreasing.
Answer: strictly increasing on [0, π/4) and (5π/4, 2π]; strictly decreasing on (π/4, 5π/4). f′(x) = cos x − sin x. It is zero where tan x = 1, that is x = π/4 and x = 5π/4 in [0, 2π]. Test a point in each part: at x = 0, f′ = 1 > 0; at x = π/2, f′ = −1 < 0; at x = 3π/2, f′ = 0 − (−1) = 1 > 0. Trap: solving tan x = 1 and keeping only π/4. Always list every solution within the given interval.
From our tutors: the single habit that saves the most marks in this topic is factorising f′(x) before doing anything else, and then drawing a sign chart: a number line with the zeros marked and a + or − in each gap. Students who try to "see" the sign from an unfactorised expression make sign errors far more often. A sign chart also shows the examiner your reasoning, which matters in a board paper where method is marked.
Tangents, normals and the angle between two curves
The tangent at a point has slope equal to the derivative there; the normal is perpendicular to it. The 2027 syllabus also lists the angle between two curves: the angle between their tangents where they meet.
Method
- Find the point of contact (x1, y1). If only x1 is given, find y1 from the curve.
- Find dy/dx and evaluate it at the point: this number is the slope m. For a parametric curve, dy/dx = (dy/dt) ÷ (dx/dt).
- Tangent: y − y1 = m(x − x1). Normal: y − y1 = (−1/m)(x − x1).
- Special cases: if m = 0, the tangent is y = y1 and the normal is x = x1. If dy/dx is undefined (vertical tangent), the tangent is x = x1 and the normal is y = y1.
- For the angle between two curves, find where they intersect, find both slopes at that point, and use tan θ = |(m1 − m2)/(1 + m1m2)|. If m1m2 = −1, the curves cut at right angles (orthogonally).
Worked example 6: Find the equations of the tangent and the normal to the curve y = x3 − 2x + 3 at the point where x = 1.
Answer: tangent y = x + 1 (or x − y + 1 = 0); normal x + y − 3 = 0. At x = 1, y = 1 − 2 + 3 = 2, so the point is (1, 2). dy/dx = 3x2 − 2, which is 1 at x = 1. Tangent: y − 2 = 1(x − 1), so y = x + 1. The normal has slope −1: y − 2 = −(x − 1), so y = 3 − x, that is x + y − 3 = 0.
Worked example 7: Find the equations of the tangent and the normal to the curve x = t2, y = 2t at t = 2.
Answer: tangent x − 2y + 4 = 0; normal 2x + y − 12 = 0. At t = 2 the point is (4, 4). dx/dt = 2t and dy/dt = 2, so dy/dx = 2/(2t) = 1/t, which is 1/2 at t = 2. Tangent: y − 4 = (1/2)(x − 4), so 2y − 8 = x − 4, that is x − 2y + 4 = 0. The normal has slope −2: y − 4 = −2(x − 4), so 2x + y − 12 = 0. (This curve is the parabola y2 = 4x written in parametric form.)
Worked example 8: Find the angle between the curves y2 = 4x and x2 = 4y at each of their points of intersection.
Answer: 90° at the origin, and tan−1(3/4), about 36.9°, at (4, 4). Solve together: from x2 = 4y, y = x2/4; substituting, x4/16 = 4x, so x(x3 − 64) = 0, giving x = 0 or x = 4. The points are (0, 0) and (4, 4). At (4, 4): for y2 = 4x, 2y(dy/dx) = 4, so dy/dx = 2/y = 1/2; for x2 = 4y, dy/dx = x/2 = 2. Then tan θ = |(2 − 1/2)/(1 + 2 × 1/2)| = (3/2)/2 = 3/4. At (0, 0): the tangent to y2 = 4x is vertical (the y-axis) and the tangent to x2 = 4y is horizontal (the x-axis), so the angle is 90°. Trap: stopping at (4, 4) and missing the origin, where the formula cannot be used because one slope is undefined.
Does your child know the methods but still lose marks on long calculus answers? Book a free ISC Maths demo class in Gurgaon. Our tutor will work through a few application of derivatives questions with your child and show exactly where the marks are going.
Book a Free ISC Maths Demo +91 92204 75088Maxima and minima
Maxima and minima has the most sub-topics in this chapter's syllabus list: critical, stationary and turning points, local and absolute maxima and minima, the first and second derivative tests, and application problems.
The two tests
- First derivative test. At a critical point c, if f′(x) changes from positive to negative as x increases through c, f has a local maximum at c. If it changes from negative to positive, a local minimum. If it does not change sign, neither.
- Second derivative test. If f′(c) = 0 and f″(c) < 0, local maximum; if f″(c) > 0, local minimum. If f″(c) = 0, the test fails and you must use the first derivative test. For example, f(x) = x4 and f(x) = x3 both have f″(0) = 0, but x4 has a minimum at 0 and x3 has neither.
Absolute (global) maxima and minima
On a closed interval [a, b], a continuous function has an absolute maximum and minimum, found either at critical points inside the interval or at the endpoints. List all these values and compare; no derivative test is needed.
Method for word problems
- Draw a diagram and name the variables.
- Write the quantity to be maximised or minimised (volume, area, cost, distance).
- Use the given condition (fixed perimeter, fixed volume, fixed sheet size) to write it in terms of one variable. State the allowed range of that variable.
- Differentiate, set the derivative to zero, and solve.
- Show that it is a maximum or minimum with the second derivative test (or the first derivative test).
- Answer the question actually asked, in words, with units.
Worked example 9: Find the local maximum and local minimum values of f(x) = x3 − 6x2 + 9x + 1.
Answer: local maximum value 5 at x = 1; local minimum value 1 at x = 3. f′(x) = 3x2 − 12x + 9 = 3(x − 1)(x − 3), which is zero at x = 1 and x = 3. f″(x) = 6x − 12. At x = 1, f″(1) = −6 < 0, so a local maximum, with value f(1) = 1 − 6 + 9 + 1 = 5. At x = 3, f″(3) = 6 > 0, so a local minimum, with value f(3) = 27 − 54 + 27 + 1 = 1. Trap: giving only the x-values when the question asks for maximum and minimum values.
Worked example 10: Find the absolute maximum and absolute minimum values of f(x) = 2x3 − 15x2 + 36x + 1 on the interval [1, 5].
Answer: absolute maximum 56 at x = 5; absolute minimum 24 at x = 1. f′(x) = 6x2 − 30x + 36 = 6(x − 2)(x − 3), so the critical points are x = 2 and x = 3, both inside [1, 5]. Evaluate: f(1) = 2 − 15 + 36 + 1 = 24; f(2) = 16 − 60 + 72 + 1 = 29; f(3) = 54 − 135 + 108 + 1 = 28; f(5) = 250 − 375 + 180 + 1 = 56. The largest value is 56 and the smallest is 24, both at endpoints. Trap: this is exactly why the endpoints must be checked. The local maximum at x = 2 (value 29) and the local minimum at x = 3 (value 28) are not the absolute ones.
Worked example 11: An open box is made from a square sheet of tin of side 18 cm by cutting equal squares from each corner and folding up the sides. Find the side of the square to be cut off so that the box has the largest possible volume, and find that volume.
Answer: cut squares of side 3 cm; maximum volume 432 cm3. Let the side of each square cut off be x cm, where 0 < x < 9. The box has base (18 − 2x) by (18 − 2x) and height x, so V = x(18 − 2x)2. Differentiate: dV/dx = (18 − 2x)2 − 4x(18 − 2x) = (18 − 2x)(18 − 6x). In 0 < x < 9, dV/dx = 0 only at x = 3. Expanding, V = 4x3 − 72x2 + 324x, so d2V/dx2 = 24x − 144, which is −72 < 0 at x = 3: a maximum. V = 3 × 122 = 432 cm3. Trap: accepting x = 9, which also makes dV/dx zero but gives a box with no base (V = 0), outside the allowed range.
Worked example 12: A closed cylindrical can must hold 128π cm3. Find the radius and height that use the least total surface area of metal, and show that the height then equals the diameter.
Answer: radius 4 cm, height 8 cm (height = diameter); least surface area 96π cm2. Let the radius be r and height h. Volume: πr2h = 128π, so h = 128/r2. Surface area S = 2πr2 + 2πrh = 2πr2 + 256π/r. Then dS/dr = 4πr − 256π/r2 = 0 gives r3 = 64, so r = 4 cm and h = 128/16 = 8 cm = 2r. Check: d2S/dr2 = 4π + 512π/r3 = 12π > 0 at r = 4, so this is a minimum. S = 2π(16) + 256π/4 = 32π + 64π = 96π cm2. The same working with a general volume V shows h = 2r always, a standard result you may be asked to prove.
Common mistakes in ISC answers
Most marks lost in this chapter come from a few repeated slips, not from gaps in knowledge.
| Mistake | Why it costs marks | Fix |
|---|---|---|
| Substituting numbers before differentiating in a rate problem | A value that is changing gets treated as a constant, and its derivative disappears | Differentiate the general formula first; substitute at the end |
| Using the general derivative as the slope | The tangent equation then contains x on both sides and is wrong | Always evaluate dy/dx at the point before writing the line |
| No sign chart for increasing and decreasing | Sign errors, and the examiner cannot see the reasoning | Factorise f′(x) and draw a sign chart every time |
| Declaring a maximum or minimum without a test | Method marks are lost even when the answer is right | Write the second derivative value, or the sign change of f′ |
| Forgetting the endpoints in absolute maxima and minima | The absolute value is often at an endpoint | Always include f(a) and f(b) in the comparison |
| Ignoring the allowed range in word problems | Impossible answers like a zero-volume box are accepted | Write the range of the variable before differentiating |
From our tutors: in optimisation problems, the step students most often skip is writing the constraint as an equation before doing anything else. Without it they try to differentiate an expression in two variables and get stuck. We ask students to write three labelled lines before any calculus: "Quantity to optimise: ...", "Condition: ...", "In one variable: ...". It takes under a minute, and it turns most word problems into routine differentiation.
How application of derivatives links to JEE
Application of derivatives is one of the chapters where ISC Class 12 preparation feeds most directly into JEE. The official JEE (Main) 2026 syllabus lists, under Unit 7, "Applications of derivatives: Rate of change of quantities, monotonic-Increasing and decreasing functions, Maxima and minima of functions of one variable." The JEE (Advanced) 2026 information brochure lists tangents and normals, increasing and decreasing functions, maximum and minimum values of a function, and also Rolle's theorem and Lagrange's mean value theorem with their geometric interpretation.
| Topic | ISC 2027 Class XII | JEE Main 2026 | JEE Advanced 2026 |
|---|---|---|---|
| Rate of change | Yes | Yes | Not named separately |
| Increasing and decreasing functions | Yes | Yes | Yes |
| Tangents and normals | Yes, with the angle between two curves | Not named in this unit | Yes |
| Maxima and minima | Yes | Yes (functions of one variable) | Yes |
| Rolle's theorem and the mean value theorem | Not listed | Not listed | Yes |
The main difference is style. ISC questions reward complete working (the sign chart, the test, the final sentence); JEE questions reward speed and often combine ideas, such as a parameter question needing the discriminant. Learn the chapter the ISC way first, then practise faster JEE-style questions. JEE Advanced aspirants should also learn Rolle's theorem and the mean value theorem from a JEE book.
For the JEE view of the whole of calculus, see our guide to calculus for JEE, and for help with both exams, our JEE Maths home tutors in Gurgaon.
How to study this chapter
Every question here starts with a derivative, so make sure differentiation of implicit, parametric, trigonometric and logarithmic functions is fluent first. Then work in this order:
- Week 1: rate of change; increasing and decreasing functions with sign charts, parameter and trigonometric questions.
- Week 2: tangents and normals for Cartesian, implicit and parametric curves; angle between curves.
- Week 3: local and absolute extrema, then word problems: boxes, cylinders, cones, fences and cost.
- Then: this chapter's questions from past ISC papers and your exam year's specimen paper, under time.
Keep all the standard results (tangent and normal forms, the angle formula, both tests) on one page of your own notes; our ISC Class 12 Maths formula sheet is a good starting point. If you are choosing a practice book for this chapter, see our guide to the best books for ISC Class 12 Maths, and for where this chapter fits in your revision calendar, our ISC board exam 2027 preparation plan. Schools in Gurgaon and across Gurugram teach this chapter at different points in Class 12, so follow your school's order.
Frequently asked questions
What topics are in application of derivatives for ISC Class 12 in 2027?
The ISC 2027 Mathematics syllabus lists rate of change, increasing and decreasing functions, equations of tangent and normal, the angle between two curves, and maxima and minima: critical and stationary points, local and absolute maxima and minima, the first and second derivative tests, and application problems.
Are approximations and errors in the ISC 2027 syllabus?
No. Approximations using differentials are not listed in the ISC 2027 Class XII Mathematics syllabus. Some older textbooks still include them, so check with your teacher before spending time on them.
Is Rolle's theorem in the ISC Class 12 syllabus for 2027?
Rolle's theorem and the mean value theorem are not listed in the ISC 2027 Class XII syllabus text. They are in the JEE (Advanced) 2026 syllabus, so JEE Advanced aspirants should learn them separately.
How many marks is application of derivatives worth in ISC?
The syllabus does not give it separate marks. It is part of Unit 3, Calculus, which carries 35 of the 80 theory marks in Class XII for 2027.
Why do I get the right answer but lose marks in maxima and minima?
Usually because a step is missing: no test to show it is a maximum or minimum, no range for the variable, or no final sentence with units. Show each step.
Is application of derivatives important for JEE?
Yes. Rate of change, increasing and decreasing functions, and maxima and minima are named in the JEE (Main) 2026 syllabus, and JEE Advanced 2026 adds tangents and normals and the mean value theorems.
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