Take a breath. If the words “maxima and minima” have been making your stomach tighten a little, you are in very good company, and you are about to find out that this chapter is far kinder than its reputation. Here is the honest truth: Application of Derivatives asks you to do only one new thing. You already know how to differentiate. That work is done. What this chapter adds is a way of reading the answer — of looking at f′(x) and saying “ah, so the function is climbing here, and levelling off there, and turning around at that point.” That is it. That is the whole idea.
Think of the derivative as a speedometer bolted onto your function. It does not tell you where you are; it tells you how fast you are rising or falling at this exact instant. A positive reading means you are walking uphill. A negative reading means you are walking downhill. A zero reading means the ground has gone flat under your feet for a moment — and flat ground is exactly where hilltops and valley floors live. Once that picture is in your head, every single question in this chapter becomes a variation on the same conversation.
We are going to build this slowly, from zero, and I am going to sit beside you the whole way. We will start with the gentlest idea (rate of change), move to reading the sign of f′(x), then to finding the tops of hills and bottoms of valleys, and finally to the big five-mark word problems where you design a box or a cylinder to be as efficient as possible. Every example below has been worked out completely and checked. Nothing is hand-waved.
What You’ll Learn
- What a Derivative Actually Tells You
- Rate of Change of Quantities
- Related Rates: Two Quantities Changing Together
- Marginal Cost and Marginal Revenue
- Increasing and Decreasing Functions
- Building a Sign Chart for f′(x)
- Proving a Function Is Increasing on an Interval
- Critical Points and What They Mean
- Local Maxima and Minima: The First Derivative Test
- The Second Derivative Test
- Choosing Between the Two Tests
- Absolute Maximum and Minimum on a Closed Interval
- Real-Life Optimisation: The Universal Method
- Optimisation with Areas and Volumes
- Optimisation with Cost, Profit and Distance
- What Was Removed From the Syllabus
- Mistakes That Quietly Cost Marks
Your Game Plan
- Warm up your differentiation. Spend twenty minutes differentiating twenty random functions before you start. This chapter punishes rusty differentiation more than anything else.
- Get comfortable with the sign of f′(x). Do not rush past the increasing/decreasing section. Everything later stands on it.
- Learn one test properly, then the other. Master the second derivative test first (it is faster), then learn the first derivative test for the cases where the second one gives you zero.
- Practise closed-interval problems separately. Absolute maximum and minimum on [a, b] is a mechanical five-step recipe. Drill it until it is automatic.
- Then attack word problems. Give them the most time. Write the constraint, reduce to one variable, differentiate, verify. Same four moves, every single time.
- Finish with the mistakes section. Read it the night before your exam. It is short and it saves marks.
Study Notes
What a Derivative Actually Tells You
Before we apply anything, let us be completely clear about what the symbol dy/dx means, because a fuzzy idea here causes trouble for the rest of the chapter.
If y depends on x, then dy/dx answers this question: when x increases by a tiny amount, how much does y change, per unit of x? It is a rate. Not a total, not a value — a rate. If y is the distance a car has travelled and x is time, then dy/dx is the speed. If y is the area of an oil slick and x is time, then dy/dx is how fast the slick is spreading in square metres per second.
There is a second, equally important reading. Geometrically, f′(a) is the slope of the curve at the point where x = a. Imagine standing on the curve at that point and looking along the direction the curve is heading. If you are tilted upward, the slope is positive. If you are tilted downward, it is negative. If you are standing on level ground, the slope is zero.
Slope reading: f′(x) is the slope (steepness) of the graph of f at the point x.
These are the same fact wearing two different hats. Every question in this chapter uses one hat or the other.
Now here is the sentence that unlocks the chapter. Because f′(x) is a slope, its sign tells you the direction the graph is travelling:
- f′(x) > 0 → the graph is going uphill as you move right. The function is increasing.
- f′(x) < 0 → the graph is going downhill. The function is decreasing.
- f′(x) = 0 → the graph is momentarily flat. Something interesting may be happening here.
Picture yourself hiking. You walk uphill for a while (f′ > 0), reach the summit where for one step the path is level (f′ = 0), then start walking downhill (f′ < 0). That summit is a local maximum. Walk on and you reach the bottom of a valley: downhill, level, then uphill. That is a local minimum. This chapter is, quite literally, the mathematics of hiking.
Do not move on until that hiking picture feels comfortable. Close your eyes and see it: uphill, level, downhill. If you have it, the rest of this chapter is just careful arithmetic.
Rate of Change of Quantities
This is the gentlest part of the chapter and a very reliable source of marks. The whole idea: if a quantity Q depends on a variable u, then dQ/du is the rate at which Q changes with respect to u. Most often u is time t, and we are watching something grow or shrink.
The single most useful tool here is the chain rule, dressed up in a way that looks like common sense. Suppose a balloon’s radius r is growing with time. Its volume V depends on r, and r depends on t. So:
dV dV dr —— = —— × —— dt dr dt
Read it aloud: “the rate the volume changes with time equals how much volume changes per unit radius, times how fast the radius is changing.” It is bookkeeping, not magic. The dr terms behave like they cancel, which is exactly the intuition you want.
Learn these cold (if any of them feel shaky, revise Surface Areas and Volumes first). In a rate-of-change question, half the work is writing the correct formula in line one.
Also watch your units. Rates carry units, and the examiner expects them. Radius in cm and time in seconds gives dA/dt in cm²/s. Writing the number without the unit is a common way to drop half a mark.
A stone is dropped into a still pond. The circular ripple’s radius grows at a constant 3 cm/s. How fast is the enclosed area growing at the instant the radius is 10 cm?
Step 1 — name what you know. dr/dt = 3 cm/s, and we want dA/dt when r = 10 cm.
Step 2 — write the relationship. A = πr².
Step 3 — differentiate with respect to t.
dA/dt = 2πr × (dr/dt)
Step 4 — substitute. dA/dt = 2π × 10 × 3 = 60π cm²/s (about 188.50 cm²/s).
Why it works: notice we substituted r = 10 only after differentiating. If you had put r = 10 into A = πr² first, you would be differentiating the constant 100π and getting zero. Differentiate first, substitute second — always.
The edge of a cube is increasing at 3 cm/s. When the edge is 10 cm, find (i) the rate of increase of the surface area and (ii) the rate of increase of the volume.
Let the edge be a. Given da/dt = 3 cm/s, and a = 10 cm at the instant of interest.
(i) Surface area. S = 6a², so dS/dt = 12a × (da/dt) = 12 × 10 × 3 = 360 cm²/s.
(ii) Volume. V = a³, so dV/dt = 3a² × (da/dt) = 3 × 100 × 3 = 900 cm³/s.
Why it works: one given rate (da/dt) feeds two different formulas. The pattern is identical each time — differentiate the formula, multiply by the rate of the inner variable. Notice too that volume grows much faster than surface area at this size, which matches intuition: cubing amplifies more than squaring.
Air is pumped into a spherical balloon so that its volume increases at 25 cm³/s. How fast is the radius increasing when the radius is 5 cm?
Here the unknown is the inner rate, which is the reverse of Example 1. Given dV/dt = 25 cm³/s; find dr/dt at r = 5.
V = (4/3)πr³ dV/dt = 4πr² × (dr/dt) 25 = 4π(5)² × (dr/dt) 25 = 100π × (dr/dt) dr/dt = 25 / (100π) = 1/(4π) cm/s
So the radius grows at 1/(4π) cm/s, roughly 0.0796 cm/s.
Why it works: the equation dV/dt = 4πr² × dr/dt links three things. Give me any two and I can find the third. That is the entire skill of this section — recognising which of the three is missing.
A particle moves along a line so that its displacement after t seconds is s = t³ − 6t² + 9t metres. Find its velocity and acceleration, and the times at which it is momentarily at rest.
Velocity is the rate of change of displacement: v = ds/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3).
Acceleration is the rate of change of velocity: a = dv/dt = 6t − 12 = 6(t − 2).
At rest means v = 0, so 3(t − 1)(t − 3) = 0, giving t = 1 s and t = 3 s.
At t = 1: s = 1 − 6 + 9 = 4 m and a = −6 m/s² (slowing and about to reverse). At t = 3: s = 27 − 54 + 27 = 0 m and a = 6 m/s².
Why it works: “at rest” never means “at the origin”. It means velocity is zero. The particle at t = 1 s is sitting 4 m from the start, perfectly still for an instant, before turning around. Read the physical words carefully.
Related Rates: Two Quantities Changing Together
Sometimes two quantities are tied together by a geometric relationship, and both are changing at once. A ladder slides down a wall: as the foot moves out, the top must come down. Sand falls into a cone: as the volume grows, so does the height. These are called related rates, and they are just the chain rule applied to an equation rather than a formula.
The method never changes:
- Draw the picture and label every length with a letter, not a number. Numbers go in at the end.
- Write the equation that ties the letters together (Pythagoras, a volume formula, a similar-triangles ratio).
- If there are three letters, use the given relationship to eliminate one, so only two remain.
- Differentiate the whole equation with respect to t.
- Now substitute the instantaneous values and solve for the unknown rate.
A 5 m ladder leans against a vertical wall. Its foot is pulled away from the wall along the ground at 2 cm/s. How fast is the top of the ladder sliding down when the foot is 4 m from the wall?
Let x be the distance of the foot from the wall and y the height of the top above the ground. The ladder length is fixed, so Pythagoras gives the tie:
x² + y² = 25 Differentiate w.r.t. t: 2x(dx/dt) + 2y(dy/dt) = 0 x(dx/dt) + y(dy/dt) = 0
When x = 4, we get y = √(25 − 16) = 3 m. And dx/dt = 2 cm/s = 0.02 m/s. Substituting:
4(0.02) + 3(dy/dt) = 0
0.08 + 3(dy/dt) = 0
dy/dt = −0.08/3 = −2/75 m/s
= −8/3 cm/s
The top is sliding down at 8/3 cm/s (about 2.67 cm/s).
Why it works: the negative sign fell out of the algebra by itself, and it is telling the truth — y is shrinking. Also notice we converted 2 cm/s into metres before substituting, because x and y are in metres. Mixed units are the number-one source of wrong answers in ladder problems.
Sand is poured onto the ground at 12 cm³/s and forms a cone whose height is always one-sixth of the base radius. How fast is the height rising when the height is 4 cm?
Volume of a cone is V = (1/3)πr²h — but that has two changing letters, r and h. We must use the given relationship to remove one. “Height is one-sixth of the radius” means h = r/6, that is r = 6h.
V = (1/3)π(6h)²h = (1/3)π(36h²)h = 12πh³ dV/dt = 36πh² × (dh/dt) At h = 4, dV/dt = 12: 12 = 36π(16)(dh/dt) 12 = 576π (dh/dt) dh/dt = 12/(576π) = 1/(48π) cm/s
The height rises at 1/(48π) cm/s, about 0.0066 cm/s — slow, which is exactly what you would expect for a pile that is much wider than it is tall.
Why it works: step 3 of the method did the heavy lifting. Two changing letters make the problem unsolvable; the given ratio lets you rewrite V in terms of h alone, and then it is a one-line chain rule. Whenever a cone or cylinder problem gives you a ratio between r and h, that ratio exists precisely so you can eliminate a variable.
The length x of a rectangle is decreasing at 5 cm/min while the width y is increasing at 4 cm/min. Find the rate of change of (i) the perimeter and (ii) the area, at the moment when x = 8 cm and y = 6 cm.
Given: dx/dt = −5 cm/min (decreasing, so negative) and dy/dt = +4 cm/min.
(i) Perimeter. P = 2(x + y), so dP/dt = 2(dx/dt + dy/dt) = 2(−5 + 4) = −2 cm/min. The perimeter is shrinking at 2 cm/min.
(ii) Area. A = xy needs the product rule:
dA/dt = x(dy/dt) + y(dx/dt)
= 8(4) + 6(−5)
= 32 − 30 = +2 cm²/min
The area is increasing at 2 cm²/min.
Why it works: this is a lovely example because the perimeter is shrinking while the area is growing at the same instant. There is no contradiction — they are different quantities and each has its own formula. Trust the algebra, not your gut.
Marginal Cost and Marginal Revenue
Economists use derivatives constantly, and CBSE includes a small, very scoreable slice of it. The vocabulary is the only new thing.
If R(x) is the total revenue from selling x units, the marginal revenue is MR = dR/dx — roughly, the revenue from selling one more unit.
Whenever you see the word “marginal”, differentiate and substitute. That is the whole instruction.
(i) The total cost in rupees of producing x units is C(x) = 0.007x³ − 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.
MC = dC/dx = 0.021x² − 0.006x + 15 At x = 17: 0.021(289) = 6.069 0.006(17) = 0.102 MC = 6.069 − 0.102 + 15 = ₹20.967
(ii) The total revenue in rupees from selling x units is R(x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.
MR = dR/dx = 26x + 26 = 26(x + 1) At x = 7: MR = 26(8) = ₹208
Why it works: the constant 4000 in the cost function is the fixed cost — rent, machinery, salaries — and it vanishes on differentiation, which is exactly right, because fixed costs do not change when you make one more unit. The mathematics is quietly modelling something true about the world.
Increasing and Decreasing Functions
Let us make the hiking picture precise. A function f is increasing on an interval if, as x moves right, f(x) never goes down. It is strictly increasing if f(x) genuinely goes up: whenever x₁ < x₂ in that interval, f(x₁) < f(x₂). Decreasing is the mirror image: as x moves right, f(x) comes down.
Checking that definition directly would be exhausting — you would have to test infinitely many pairs. The derivative rescues us completely.
• if f′(x) > 0 for all x, f is strictly increasing;
• if f′(x) < 0 for all x, f is strictly decreasing;
• if f′(x) = 0 for all x, f is constant.
So the question “where is f increasing?” becomes the far easier question “where is f′(x) positive?”
That last sentence is the whole trick. You never have to think about the original function again. You differentiate once, and then you are just solving an inequality — something you have been doing since Class 10.
Find the intervals on which f(x) = 2x³ − 9x² + 12x + 5 is increasing and decreasing.
Step 1 — differentiate and factorise. Factorising is not optional; it is what makes the sign obvious.
f′(x) = 6x² − 18x + 12
= 6(x² − 3x + 2)
= 6(x − 1)(x − 2)
Step 2 — find where f′(x) = 0. x = 1 and x = 2. These two numbers chop the real line into three pieces.
Step 3 — test one convenient point in each piece.
x = 0 : f′(0) = 6(−1)(−2) = +12 positive x = 1.5 : f′(1.5) = 6(0.5)(−0.5) = −1.5 negative x = 3 : f′(3) = 6(2)(1) = +12 positive
Step 4 — state the answer in interval form.
f is increasing on (−∞, 1) and on (2, ∞), and decreasing on (1, 2).
Why it works: a product of factors changes sign only where one of the factors is zero. Between consecutive zeros the sign cannot flip, so one test point per interval is genuinely enough. That is why we bother to factorise.
Find the intervals of increase and decrease for f(x) = x⁴ − 8x³ + 22x² − 24x + 11.
f′(x) = 4x³ − 24x² + 44x − 24
= 4(x³ − 6x² + 11x − 6)
Try x = 1: 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor.
x³ − 6x² + 11x − 6 = (x − 1)(x² − 5x + 6)
= (x − 1)(x − 2)(x − 3)
f′(x) = 4(x − 1)(x − 2)(x − 3)
Zeros at x = 1, 2, 3, giving four intervals. Test points:
x = 0 : 4(−1)(−2)(−3) = −24 negative x = 1.5 : 4(0.5)(−0.5)(−1.5) = +1.5 positive x = 2.5 : 4(1.5)(0.5)(−0.5) = −1.5 negative x = 4 : 4(3)(2)(1) = +24 positive
Increasing on (1, 2) and (3, ∞); decreasing on (−∞, 1) and (2, 3).
Why it works: with three simple (non-repeated) factors the sign alternates as you cross each zero, exactly as it did here: −, +, −, +. Spotting that alternation is a great self-check. If your signs do not alternate across simple roots, you have made an arithmetic slip.
Find the intervals on which f(x) = sin x + cos x is increasing and decreasing, for 0 < x < 2π.
f′(x) = cos x − sin x f′(x) = 0 ⇒ cos x = sin x ⇒ tan x = 1 In (0, 2π): x = π/4 and x = 5π/4
Those two values split (0, 2π) into three intervals. Test one point in each:
x = π/2 : f′ = 0 − 1 = −1 negative x = π : f′ = −1 − 0 = −1 negative x = 7π/4 : f′ = 1/√2 − (−1/√2) = √2 positive
We also need a point in (0, π/4). Take x = π/6: f′ = (√3)/2 − 1/2 ≈ 0.366, positive.
So f is increasing on (0, π/4) and (5π/4, 2π), and decreasing on (π/4, 5π/4).
Why it works: notice that both π/2 and π sit inside the same interval (π/4, 5π/4), so testing both was redundant — but harmless, and they agree, which is reassuring. The real lesson is that with trigonometric functions you must solve f′(x) = 0 within the stated interval only. Outside (0, 2π) there are infinitely many more solutions, and including them would be wrong.
Building a Sign Chart for f′(x)
The sign chart is the single most useful piece of paper-work in this chapter. It is nothing more than a number line with the critical values marked and the sign of f′(x) written in each gap. Draw one every single time, even when you think you can see the answer. It costs thirty seconds and it prevents almost every sign error.
Here is the procedure written out in full, using f′(x) = 6(x − 1)(x − 2) from Example 9 as the running case:
- Solve f′(x) = 0 and also note any x where f′(x) does not exist. Here: x = 1 and x = 2.
- Mark those values on a number line in increasing order. They cut the line into intervals — here, three of them.
- Pick an easy test value inside each interval. Choose whole numbers where possible; 0 is your friend.
- Compute the sign only, not the value. You do not care that f′(0) = 12; you care that it is positive. Counting minus signs in the factorised form is faster than multiplying out.
- Write + or − above each interval, and translate: + means increasing, − means decreasing.
Now look at the table below, which lays out the same information in the form examiners like to see.
| Interval | Test value | Sign of f′(x) | Behaviour of f |
|---|---|---|---|
| (−∞, 1) | x = 0 | +12, positive | Increasing |
| x = 1 | — | 0 | Local maximum, f(1) = 10 |
| (1, 2) | x = 1.5 | −1.5, negative | Decreasing |
| x = 2 | — | 0 | Local minimum, f(2) = 9 |
| (2, ∞) | x = 3 | +12, positive | Increasing |
Notice something beautiful in that table. The sign flipped from + to − at x = 1, and the function stopped climbing and started falling — a summit. It flipped from − to + at x = 2, and the function stopped falling and started climbing — a valley floor. You have just discovered the first derivative test without anyone teaching it to you. We will make it official shortly.
Proving a Function Is Increasing on an Interval
A slightly different flavour of question asks you to prove something, usually: “Show that f(x) is strictly increasing on R” or “Show that f is increasing for all x > 0”. There is no sign chart here. Instead you must argue that f′(x) is positive everywhere on the stated set.
The trick is nearly always to rewrite f′(x) in a form whose sign is obvious. Three patterns cover almost every question:
- A perfect square: anything of the form (something)² is ≥ 0 automatically.
- A completed square: ax² + bx + c can be written as a[(x + p)² + q], and if a > 0 and q > 0 the whole thing is positive.
- A fraction with an obviously positive denominator: if f′(x) = x/(1 + x)², the bottom is a square so it is positive, and the sign of the whole thing is just the sign of the top.
Show that f(x) = 4x³ − 18x² + 27x − 7 is increasing on R.
f′(x) = 12x² − 36x + 27
= 3(4x² − 12x + 9)
= 3(2x − 3)²
A square is never negative, so 3(2x − 3)² ≥ 0 for every real x. Therefore f′(x) ≥ 0 everywhere, and f is increasing on R.
A careful footnote. f′(x) = 0 at the single point x = 3/2. Does that break “strictly increasing”? No. A function is still strictly increasing if its derivative vanishes at isolated points, as long as it is positive everywhere else. Here f′ is positive on both sides of 3/2 and the graph merely flattens for an instant before continuing upward. Say this explicitly in your answer — it shows the examiner you understand the subtlety.
Why it works: spotting 4x² − 12x + 9 as (2x − 3)² is the whole question. If factorising does not jump out, compute the discriminant: b² − 4ac = 144 − 144 = 0, which tells you it is a perfect square. If the discriminant were negative and a > 0, the quadratic would be strictly positive — that is an equally good argument.
Show that f(x) = log(1 + x) − x/(1 + x) is increasing for all x > 0.
Differentiate, using the quotient rule on the second term:
d/dx [log(1 + x)] = 1/(1 + x)
d/dx [x/(1+x)] = [(1+x)(1) − x(1)] / (1+x)²
= 1/(1 + x)²
f′(x) = 1/(1 + x) − 1/(1 + x)²
Common denominator (1 + x)²:
f′(x) = [(1 + x) − 1] / (1 + x)²
= x / (1 + x)²
For x > 0 the numerator x is positive and the denominator (1 + x)² is a square, hence positive. So f′(x) > 0 and f is increasing on (0, ∞).
Why it works: the messy-looking derivative collapsed to something transparent once we put it over a common denominator. Whenever your f′(x) is a difference of two fractions, combine them before you try to judge the sign. Guessing the sign of a difference is how marks get lost.
Critical Points and What They Mean
We keep talking about “the places where f′(x) = 0”. Let us give them their proper name and be precise about what they can and cannot tell us.
• f′(c) = 0 — the graph is momentarily flat (also called a stationary point), or
• f′(c) does not exist — the graph has a sharp corner or a vertical tangent.
Every local maximum and every local minimum of a differentiable function occurs at a critical point. So critical points are your complete list of suspects.
That last line deserves emphasis, because it is what makes the whole method possible. Instead of searching an infinite line for hilltops, you solve one equation and get a short finite list. Everything after this is just interrogating the suspects.
But be careful about the converse. A critical point need not be a maximum or a minimum. Consider f(x) = x³. Then f′(x) = 3x², which is zero at x = 0. Yet f is increasing everywhere — it is negative to the left of 0 and positive to the right. The graph just pauses at the origin and carries on climbing. That is a point of inflection, not a turning point. Being a critical point makes you a suspect, not a convict.
Find the critical points of (i) f(x) = x⁴ − 62x² + 120x + 9 and (ii) g(x) = x/2 + 2/x.
(i)
f′(x) = 4x³ − 124x + 120
= 4(x³ − 31x + 30)
Try x = 1: 1 − 31 + 30 = 0, so (x − 1) is a factor.
x³ − 31x + 30 = (x − 1)(x² + x − 30)
= (x − 1)(x + 6)(x − 5)
f′(x) = 4(x − 1)(x + 6)(x − 5)
Critical points: x = −6, x = 1, x = 5. This is a polynomial, so f′ exists everywhere and there are no others.
(ii) The domain of g excludes x = 0.
g′(x) = 1/2 − 2/x²
= (x² − 4) / (2x²)
= (x − 2)(x + 2) / (2x²)
Setting the numerator to zero gives x = 2 and x = −2. What about x = 0, where g′ fails to exist? It is not a critical point, because 0 is not in the domain of g in the first place. A critical point must belong to the function’s domain.
Why it works: for a fraction, f′(x) = 0 exactly when the numerator is zero and the denominator is not. Always simplify to a single fraction first — then read off the zeros from the top and the danger points from the bottom.
Local Maxima and Minima: The First Derivative Test
First, the vocabulary, because CBSE is fussy about it. A function f has a local maximum at x = c if f(c) is at least as big as f(x) for all x in some small interval around c. “Local” means you only have to beat your immediate neighbours, not the whole world. A mound in a field is a local maximum even if there is a mountain twenty kilometres away. Similarly, a local minimum at c means f(c) is the smallest value in some small neighbourhood of c.
The value f(c) is called the local maximum value (or minimum value); the point x = c is the point of local maximum. Questions often ask for both, so give both.
• sign changes + to − → f has a local maximum at c (uphill, then downhill — a summit);
• sign changes − to + → f has a local minimum at c (downhill, then uphill — a valley);
• no sign change → c is a point of inflection; f has neither a maximum nor a minimum there.
The test is called “motivated geometrically” in the syllabus, and that phrase is a gift: you are allowed to justify it with the hiking picture. If you are climbing and then descending, you must have passed over a top. There is nothing more to it than that.
Find the local maximum and local minimum values of f(x) = x³ − 6x² + 9x + 15.
f′(x) = 3x² − 12x + 9
= 3(x² − 4x + 3)
= 3(x − 1)(x − 3)
Critical points: x = 1, x = 3
Now the sign chart, testing x = 0, x = 2 and x = 4:
f′(0) = 3(−1)(−3) = +9 positive f′(2) = 3(1)(−1) = −3 negative f′(4) = 3(3)(1) = +9 positive
At x = 1: sign goes + → −. Local maximum. Value: f(1) = 1 − 6 + 9 + 15 = 19.
At x = 3: sign goes − → +. Local minimum. Value: f(3) = 27 − 54 + 27 + 15 = 15.
Why it works: and here is a thought worth pausing on — the local maximum value 19 is larger than the local minimum value 15, which feels obvious, but it does not have to be true in general. A local maximum on one hill can easily sit lower than a local minimum in a distant valley, because “local” only compares you to your neighbours. Never assume the local max value exceeds the local min value.
Examine f(x) = (x − 1)³(x + 1)² for local maxima and minima.
Differentiate with the product rule, then factorise ruthlessly:
f′(x) = 3(x−1)²(x+1)² + (x−1)³ × 2(x+1)
Take out the common factor (x−1)²(x+1):
f′(x) = (x−1)²(x+1)[3(x+1) + 2(x−1)]
= (x−1)²(x+1)(3x + 3 + 2x − 2)
= (x−1)²(x+1)(5x + 1)
Critical points: x = −1, x = −1/5, x = 1
The factor (x − 1)² is a square, so it is never negative and never changes the sign. The sign of f′(x) is therefore entirely decided by (x + 1)(5x + 1):
x = −2 : (−1)(−9) = + positive x = −0.5 : (0.5)(−1.5) = − negative x = 0 : (1)(1) = + positive x = 2 : (3)(11) = + positive
x = −1: + → −, so a local maximum, value f(−1) = (−2)³(0)² = 0.
x = −1/5: − → +, so a local minimum, value f(−1/5) = (−6/5)³(4/5)² = (−216/125)(16/25) = −3456/3125 ≈ −1.106.
x = 1: the sign is positive on both sides — no change. So x = 1 is a point of inflection, neither a maximum nor a minimum. The value there is f(1) = 0.
Why it works: repeated factors are the story here. An even power such as (x − 1)² touches zero without crossing it, so the sign is unchanged; an odd power crosses and flips the sign. Learn to look at the exponent of each factor before you test anything — it will often tell you the answer at a glance.
The Second Derivative Test
The first derivative test always works, but it needs a sign chart and several test values. When the function is well behaved there is a much faster route, and the syllabus explicitly describes it as “a provable tool” — meaning you may simply use it.
Here is the intuition first. The second derivative f″(x) measures how the slope is changing — the curvature. If f″(c) < 0 the curve bends downward like the top of an arch, so a flat point there must be a summit. If f″(c) > 0 the curve bends upward like the inside of a bowl, so a flat point there must be a valley floor. Think of a smile and a frown: a positive second derivative gives you a smile shape (minimum at the bottom), a negative one gives you a frown shape (maximum at the top).
• If f″(c) < 0 → local maximum at c (frown shape).
• If f″(c) > 0 → local minimum at c (smile shape).
• If f″(c) = 0 → the test fails. It gives no information at all. Go back and use the first derivative test.
Find the local extrema of f(x) = 2x³ − 21x² + 36x − 20.
f′(x) = 6x² − 42x + 36 = 6(x − 1)(x − 6) f″(x) = 12x − 42 Critical points: x = 1, x = 6 At x = 1: f″(1) = 12 − 42 = −30 < 0 → LOCAL MAXIMUM At x = 6: f″(6) = 72 − 42 = +30 > 0 → LOCAL MINIMUM
Now the values:
f(1) = 2 − 21 + 36 − 20 = −3
f(6) = 2(216) − 21(36) + 36(6) − 20
= 432 − 756 + 216 − 20 = −128
Local maximum value −3 at x = 1; local minimum value −128 at x = 6.
Why it works: compare the effort with Example 15. No sign chart, no choosing test points, just two substitutions into f″. When f″ is easy to compute, this test is faster and less error-prone. Also note that both extreme values are negative — that is completely fine. “Maximum” describes the shape of the graph, not the sign of the number.
Find the local maximum and minimum values of f(x) = x⁴ − 62x² + 120x + 9.
We found the critical points in Example 14: x = −6, 1, 5. Now:
f″(x) = 12x² − 124 f″(−6) = 12(36) − 124 = 432 − 124 = +308 > 0 → MINIMUM f″(1) = 12(1) − 124 = 12 − 124 = −112 < 0 → MAXIMUM f″(5) = 12(25) − 124 = 300 − 124 = +176 > 0 → MINIMUM
The values, computed carefully:
f(−6) = 1296 − 62(36) + 120(−6) + 9
= 1296 − 2232 − 720 + 9 = −1647
f(1) = 1 − 62 + 120 + 9 = 68
f(5) = 625 − 62(25) + 600 + 9
= 625 − 1550 + 600 + 9 = −316
Local minima: −1647 at x = −6 and −316 at x = 5. Local maximum: 68 at x = 1.
Why it works: this is the example that proves the point made in Example 15. The local minimum value at x = 5 is −316, and the local maximum value at x = 1 is 68 — fine. But look at the two minima: −1647 and −316. They are wildly different, because each one only has to beat its own neighbourhood. A quartic with a positive leading coefficient goes up to +∞ on both ends, dips into two valleys, and rises to a local peak between them. Sketch that shape in your head and the three answers stop being a list of numbers and become a picture.
Find the local maximum value of f(x) = (log x)/x for x > 0.
Quotient rule:
f′(x) = [x × (1/x) − (log x)(1)] / x²
= (1 − log x) / x²
f′(x) = 0 ⇒ 1 − log x = 0 ⇒ log x = 1 ⇒ x = e
For the second derivative test, differentiate again (quotient rule once more):
f″(x) = [x²(−1/x) − (1 − log x)(2x)] / x⁴
= [−x − 2x + 2x log x] / x⁴
= (2 log x − 3) / x³
At x = e (where log x = 1):
f″(e) = (2 − 3)/e³ = −1/e³ < 0 → LOCAL MAXIMUM
The local maximum value is f(e) = (log e)/e = 1/e ≈ 0.368.
Why it works: the second derivative looked frightening until we substituted x = e, at which point log x = 1 and everything collapsed. That is typical. Do not simplify f″ more than you need to — you only ever have to evaluate its sign at the critical point, and substituting early often makes the algebra vanish.
Choosing Between the Two Tests
Students often ask which test they are “supposed” to use. The honest answer is: whichever gets you to a correct answer fastest. But there are clear situations where one beats the other, and knowing them saves real time in an exam.
| Aspect | First Derivative Test | Second Derivative Test |
|---|---|---|
| What you compute | Sign of f′(x) on both sides of c | Value of f″(c) |
| Speed | Slower — needs a sign chart and test values | Faster — one substitution per critical point |
| Can it ever fail? | No. It always gives a verdict | Yes. If f″(c) = 0 it says nothing |
| Detects inflection points | Yes — shows up as “no sign change” | No — just returns zero and stops |
| Best used when | f″ is messy, or f″(c) = 0, or repeated factors appear in f′ | f is a polynomial or another function that is easy to differentiate twice |
| Also gives you | The full increasing/decreasing picture for free | Only the classification at c |
My practical advice: reach for the second derivative test first, because it is quick. If f″(c) comes out zero, do not panic and do not guess — simply say “the second derivative test fails here” and switch to a sign chart. Writing that sentence explicitly earns credit; silently guessing does not.
Absolute Maximum and Minimum on a Closed Interval
So far we have hunted for local extrema — hilltops and valleys. Now we ask a bigger question: over a closed interval [a, b], what is the very largest value the function reaches, and the very smallest? These are the absolute (or global) maximum and minimum.
Here is the crucial new idea, and it is easy to picture. Imagine walking a mountain trail that starts at a fenced gate and ends at another fenced gate. The highest point of your walk might be a summit in the middle — or it might be the gate itself, if the trail simply climbs the whole way. Endpoints are candidates too. Beginners forget this constantly, and it is the single biggest source of lost marks in this section.
1. Find f′(x).
2. Solve f′(x) = 0 and keep only the roots that lie inside [a, b]. Discard the rest.
3. Make a candidate list: those critical points plus the two endpoints a and b.
4. Evaluate f at every candidate.
5. The largest value is the absolute maximum; the smallest is the absolute minimum.
You never need the first or second derivative test here. Just compare numbers.
That last line is worth reading twice, because it makes this section genuinely easy. There is no classification to do, no sign chart, no curvature. You build a short list of numbers and pick the biggest and smallest. A continuous function on a closed interval is guaranteed to attain both, so the answer always exists.
Find the absolute maximum and minimum values of f(x) = 2x³ − 15x² + 36x + 1 on [0, 5].
f′(x) = 6x² − 30x + 36 = 6(x − 2)(x − 3) Critical points: x = 2 and x = 3. Both lie inside [0, 5], so both stay. Candidates: x = 0, 2, 3, 5 f(0) = 1 f(2) = 16 − 60 + 72 + 1 = 29 f(3) = 54 − 135 + 108 + 1 = 28 f(5) = 250 − 375 + 180 + 1 = 56
Absolute maximum = 56 at x = 5. Absolute minimum = 1 at x = 0.
Why it works: look what happened. x = 2 is a genuine local maximum (value 29) and x = 3 a genuine local minimum (value 28) — but neither wins. Both absolute extremes occur at the endpoints, because the cubic rises steeply after x = 3 and was low at x = 0. If you had classified the critical points and stopped there, you would have answered 29 and 28 and lost the question entirely. Always evaluate the endpoints.
Find the absolute maximum and minimum of f(x) = x³ − 3x + 3 on [−3, 3/2].
f′(x) = 3x² − 3 = 3(x − 1)(x + 1) Critical points: x = −1 and x = 1. Interval is [−3, 1.5]. Both −1 and 1 lie inside, so both stay. Candidates: x = −3, −1, 1, 3/2 f(−3) = −27 + 9 + 3 = −15 f(−1) = −1 + 3 + 3 = 5 f(1) = 1 − 3 + 3 = 1 f(3/2) = 27/8 − 9/2 + 3 = 15/8 = 1.875
Absolute maximum = 5 at x = −1. Absolute minimum = −15 at x = −3.
Why it works: this time the maximum is at an interior critical point and the minimum is at an endpoint — a mixed case, which is very common. Note also how carefully we handled the fractional endpoint 3/2. Convert to a decimal at the end if you like, but do the arithmetic in fractions: 27/8 − 36/8 + 24/8 = 15/8. Decimals invite rounding errors mid-calculation.
Find the absolute maximum and minimum values of f(x) = sin x + cos x on [0, π].
f′(x) = cos x − sin x = 0 ⇒ tan x = 1 In [0, π] the only solution is x = π/4. (5π/4 is a solution of tan x = 1 but lies OUTSIDE [0, π] — discard it.) Candidates: x = 0, π/4, π f(0) = 0 + 1 = 1 f(π/4) = 1/√2 + 1/√2 = 2/√2 = √2 ≈ 1.414 f(π) = 0 + (−1) = −1
Absolute maximum = √2 at x = π/4. Absolute minimum = −1 at x = π.
Why it works: step 2 of the method did real work here. Trigonometric equations have infinitely many solutions, and only those inside the interval are candidates. Writing “5π/4 lies outside [0, π], so we reject it” is worth stating explicitly — it shows the examiner you understood the restriction rather than getting lucky.
Real-Life Optimisation: The Universal Method
This is where the chapter earns its name, and where the five-mark questions live. It is also the calculus twin of Linear Programming, where you optimise with straight lines instead of derivatives. A word problem hands you a real situation — a sheet of tin, a fixed length of fencing, a delivery cost — and asks you to make something as big, or as cheap, as possible. It feels intimidating because there is prose to read. It is not intimidating once you have a fixed routine.
2. Write the objective. The quantity to be made largest or smallest — volume, area, cost, distance.
3. Write the constraint. The fixed fact — total fencing, given volume, the sheet size, a relationship between letters.
4. Reduce to ONE variable. Use the constraint to eliminate every letter but one. This is the step that decides whether you can finish.
5. Differentiate, set to zero, solve. Note the sensible domain while you are here.
6. Verify with the second derivative and answer the actual question that was asked, with units.
Step 4 is the one students stumble on, so let me say more about it. If your objective still has two letters in it after step 3, you cannot differentiate it — there is nothing to differentiate with respect to. The constraint exists precisely to let you write one letter in terms of the other. Whenever you feel stuck in a word problem, ask yourself: “have I used the constraint yet?” Nine times out of ten, you have not.
And step 6 matters more than it looks. Many students find x = 5 and stop. But the question asked for the volume, or the dimensions, or the cost. Finding the critical value is the middle of the answer, not the end.
Find two positive numbers whose sum is 24 and such that the product of one number and the square of the other is as large as possible.
Move 1 and 2. Let the numbers be x and y. The objective is P = x·y².
Move 3. The constraint is x + y = 24.
Move 4 — reduce to one variable. From the constraint, y = 24 − x. So
P(x) = x(24 − x)², 0 < x < 24
Move 5 — differentiate. Product rule:
P′(x) = (24 − x)² + x × 2(24 − x)(−1)
= (24 − x)[(24 − x) − 2x]
= (24 − x)(24 − 3x)
= 3(24 − x)(8 − x)
P′(x) = 0 ⇒ x = 24 or x = 8
x = 24 would make y = 0, which is not a positive number, so we reject it. That leaves x = 8.
Move 6 — verify and answer. Expand to differentiate again: P(x) = x³ − 48x² + 576x, so P″(x) = 6x − 96. At x = 8, P″(8) = 48 − 96 = −48 < 0, confirming a maximum.
The numbers are 8 and 16, and the largest product is 8 × 16² = 8 × 256 = 2048.
Why it works: notice we factorised P′ instead of expanding it. Expanding gives 3x² − 96x + 576, which factorises to the same thing but with more chances to slip. Also notice we rejected x = 24 using the physical wording (“two positive numbers”). Mathematics gave two roots; the real world eliminated one. That happens in almost every word problem, so always check your roots against the situation.
A rectangular sheet of tin measuring 45 cm by 24 cm has a square of side x cut from each corner. The flaps are folded up to make an open box. Find x so that the volume is a maximum, and state that maximum volume.
Moves 1–4. After cutting squares of side x from each corner, the base measures (45 − 2x) by (24 − 2x), and folding up the flaps gives a height of x. So the objective is already in one variable:
V(x) = x(45 − 2x)(24 − 2x)
The domain. We need x > 0, and both base dimensions positive: 45 − 2x > 0 gives x < 22.5, and 24 − 2x > 0 gives x < 12. The tighter condition wins, so 0 < x < 12.
Move 5 — expand and differentiate.
(45 − 2x)(24 − 2x) = 1080 − 90x − 48x + 4x²
= 4x² − 138x + 1080
V(x) = 4x³ − 138x² + 1080x
V′(x) = 12x² − 276x + 1080
= 12(x² − 23x + 90)
= 12(x − 5)(x − 18)
V′(x) = 0 ⇒ x = 5 or x = 18
x = 18 lies outside 0 < x < 12 — you cannot cut 18 cm squares from a sheet only 24 cm wide. Reject it. So x = 5.
Move 6 — verify and answer. V″(x) = 24x − 276, so V″(5) = 120 − 276 = −156 < 0. Maximum confirmed.
V(5) = 5 × (45 − 10) × (24 − 10)
= 5 × 35 × 14
= 2450 cm³
The squares should have side 5 cm, giving a maximum volume of 2450 cm³ (a box 35 cm by 14 cm by 5 cm).
Why it works: stating the domain was not decoration — it is what let us throw away x = 18 with a proper reason rather than a hunch. In a five-mark question, the marks are typically split across setting up V(x), finding V′ and its roots, rejecting the invalid root with justification, verifying with V″, and computing the final volume. Write all five steps and you collect all five marks.
Optimisation with Areas and Volumes
Geometric optimisation questions are the most common type in the board paper. They all follow the six moves; what changes is which formula supplies the objective and which supplies the constraint. Keep the mensuration formulas from the rate-of-change section within reach.
A closed cylindrical can must hold 250π cm³. Find the radius and height that use the least metal, and the minimum total surface area.
Objective: total surface area S = 2πr² + 2πrh (two circular ends plus the curved side).
Constraint: πr²h = 250π, so h = 250/r².
S(r) = 2πr² + 2πr(250/r²)
= 2πr² + 500π/r (r > 0)
S′(r) = 4πr − 500π/r²
S′(r) = 0 ⇒ 4πr = 500π/r²
⇒ 4r³ = 500
⇒ r³ = 125
⇒ r = 5 cm
Verify: S″(r) = 4π + 1000π/r³. At r = 5, S″(5) = 4π + 1000π/125 = 4π + 8π = 12π > 0, so this is a minimum.
h = 250/r² = 250/25 = 10 cm S = 2π(25) + 2π(5)(10) = 50π + 100π = 150π cm² ≈ 471.24 cm²
Radius 5 cm, height 10 cm, minimum surface area 150π cm².
Why it works — and a fact worth memorising: notice that h = 10 = 2r. This is not a coincidence of the numbers. For any closed cylinder of fixed volume, the surface area is least when the height equals the diameter. If your answer to a closed-cylinder problem does not satisfy h = 2r, you have made a mistake somewhere. That is a free self-check on every question of this type.
Show that the rectangle of maximum area that can be inscribed in a circle of radius 10 cm is a square, and find that area.
Place the rectangle with its centre at the centre of the circle. Its diagonal is then a diameter, of length 20 cm. If the sides are x and y, Pythagoras gives the constraint x² + y² = 400, so y = √(400 − x²).
A = x√(400 − x²), 0 < x < 20 Maximise A² instead. Let Z = A²: Z = x²(400 − x²) = 400x² − x⁴ dZ/dx = 800x − 4x³ = 4x(200 − x²) dZ/dx = 0 ⇒ x² = 200 ⇒ x = 10√2 (x = 0 is rejected — it gives no rectangle)
Verify: d²Z/dx² = 800 − 12x². At x² = 200, this is 800 − 2400 = −1600 < 0, so Z (and hence A) is maximum.
y = √(400 − 200) = √200 = 10√2 = x So the sides are equal — it IS a square. Area = 10√2 × 10√2 = 100 × 2 = 200 cm²
The rectangle is a square of side 10√2 cm, with maximum area 200 cm².
Why it works: the question asked us to show it is a square, so finding x was only half the job — we had to compute y as well and observe that y = x. Read “show that” questions carefully; they tell you exactly what your final line must say.
Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 12 cm.
Let the cone have height h (measured from its base up to the apex, which touches the sphere) and base radius r. Put the centre of the sphere at O. The base circle sits at a distance |h − 12| from O, so by Pythagoras in the right triangle from O to the rim of the base:
r² + (h − 12)² = 12² = 144
r² = 144 − (h − 12)²
= 144 − (h² − 24h + 144)
= 24h − h²
V = (1/3)πr²h
= (1/3)π(24h − h²)h
= (1/3)π(24h² − h³), 0 < h < 24
dV/dh = (1/3)π(48h − 3h²)
= πh(16 − h)
dV/dh = 0 ⇒ h = 16 (h = 0 rejected)
Verify: d²V/dh² = π(16 − 2h). At h = 16 this equals π(16 − 32) = −16π < 0, so the volume is maximum.
r² = 24(16) − 256 = 384 − 256 = 128 V = (1/3)π(128)(16) = 2048π/3 cm³ ≈ 2144.66 cm³
Maximum volume = 2048π/3 cm³, achieved when the height is 16 cm and the base radius is √128 = 8√2 cm.
Why it works — another free check: the optimal height came out as 16 = (4/3) × 12. For a cone inscribed in a sphere of radius R, the maximum volume always occurs at h = 4R/3. Substituting that back gives a maximum volume of (8/27) of the sphere’s volume. Use it to sanity-check your answer, but do show the full derivation in the exam — quoting the shortcut alone will not earn the method marks.
A cylinder is inscribed in a right circular cone of height 12 cm and base radius 6 cm, standing on the same base. Find the maximum volume of the cylinder.
Let the cylinder have radius r and height h. Take a vertical cross-section: the cone becomes a triangle of height 12 and half-base 6, and the top corner of the cylinder lies on the slanted side. Similar triangles give the constraint:
h / 12 = (6 − r) / 6
h = 12(6 − r)/6 = 2(6 − r)
V = πr²h = πr² × 2(6 − r)
= 2π(6r² − r³), 0 < r < 6
dV/dr = 2π(12r − 3r²)
= 6πr(4 − r)
dV/dr = 0 ⇒ r = 4 (r = 0 rejected)
Verify: d²V/dr² = 2π(12 − 6r). At r = 4 this is 2π(12 − 24) = −24π < 0, a maximum.
h = 2(6 − 4) = 4 cm V = π(16)(4) = 64π cm³ ≈ 201.06 cm³
Maximum cylinder volume = 64π cm³, with radius 4 cm and height 4 cm.
Why it works: the entire difficulty of this problem is in one line — getting h/12 = (6 − r)/6 from the cross-section. When the cylinder has radius r, the remaining horizontal distance out to the cone’s slanted edge is 6 − r, and the small triangle sitting above the cylinder is similar to the whole cone. Draw the cross-section every time; do not try to see this in three dimensions.
Optimisation with Cost, Profit and Distance
The syllabus asks for “real-life situations”, and not all of them are about shapes. Money and distance problems use exactly the same six moves; the only extra care needed is in translating the words into an objective function.
Two translation habits will carry you through nearly every one of these:
- “Per” means divide. Cost per kilometre is total cost divided by kilometres. If you can write the total and the count separately, the “per” version follows.
- For distance, minimise the square. The distance between (a, b) and (x, y) is √[(x − a)² + (y − b)²]. Minimise the expression inside the root and take the square root at the end — same reasoning as in the previous section.
An open rectangular tank with a square base must hold 32000 cm³ of water. Find the dimensions that use the least sheet metal, and the area of metal required.
Objective: the tank is open at the top, so the metal covers the square base plus four rectangular sides. With base side x and height h:
S = x² + 4xh
Constraint: x²h = 32000, so h = 32000/x²
S(x) = x² + 4x(32000/x²)
= x² + 128000/x, x > 0
S′(x) = 2x − 128000/x²
S′(x) = 0 ⇒ 2x³ = 128000
⇒ x³ = 64000
⇒ x = 40 cm
Verify: S″(x) = 2 + 256000/x³. At x = 40, S″ = 2 + 256000/64000 = 2 + 4 = 6 > 0, a minimum.
h = 32000/1600 = 20 cm S = 40² + 4(40)(20) = 1600 + 3200 = 4800 cm²
Base 40 cm by 40 cm, height 20 cm, using 4800 cm² of metal.
Why it works — and notice the pattern: here x = 2h, that is, the base side is twice the height. Compare this with the closed cylinder in Example 25, where h = 2r. Removing the lid changes the optimal proportions, which makes physical sense: with no top to pay for, it is worth making the tank wider and shallower. Whenever a question says “open”, check that you have left out the top face — forgetting to do so is the classic error here.
A truck driven at a steady speed of v km/h burns fuel costing v²/25 rupees per hour. The driver’s wages are a further ₹100 per hour. Find the speed that makes the cost of a journey per kilometre as small as possible, and that minimum cost.
Objective — and here is where the “per” habit matters. In one hour the truck costs (v²/25 + 100) rupees and covers v kilometres. So the cost per kilometre is:
C(v) = (v²/25 + 100) / v
= v/25 + 100/v, v > 0
C′(v) = 1/25 − 100/v²
C′(v) = 0 ⇒ 100/v² = 1/25
⇒ v² = 2500
⇒ v = 50 km/h (v > 0)
Verify: C″(v) = 200/v³, which is positive for every v > 0. So v = 50 gives a minimum.
C(50) = 50/25 + 100/50
= 2 + 2 = ₹4 per km
The most economical speed is 50 km/h, costing ₹4 per kilometre.
Why it works: the mathematics is telling a genuine story. Drive too slowly and the wage bill is spread over very few kilometres; drive too fast and the fuel term v/25 dominates. The minimum sits where the two costs balance — and indeed at v = 50 they are exactly equal, ₹2 each. That balance is not an accident and it is a nice remark to include in your answer.
Find the point on the parabola y² = 4x that is nearest to the point (3, 0), and find that shortest distance.
Let (x, y) be a point on the parabola, so y² = 4x. The squared distance to (3, 0) is:
D² = (x − 3)² + y²
= (x − 3)² + 4x (using y² = 4x)
Let Z = D² = x² − 6x + 9 + 4x
= x² − 2x + 9, x ≥ 0
dZ/dx = 2x − 2
dZ/dx = 0 ⇒ x = 1
Verify: d²Z/dx² = 2 > 0, so this is a minimum.
At x = 1: y² = 4(1) = 4, so y = 2 or y = −2 Z = 1 − 2 + 9 = 8 D = √8 = 2√2 ≈ 2.83
The nearest points are (1, 2) and (1, −2), each at a distance of 2√2 units from (3, 0).
Why it works: two things earned their keep. First, substituting y² = 4x turned a two-variable expression into a clean quadratic in x alone — the constraint doing its job. Second, minimising D² rather than D avoided differentiating a square root entirely, and the answer is identical because D is non-negative. Finally, do not forget that y² = 4 has two solutions; the parabola is symmetric about the x-axis, so there are two equally near points. Giving only one costs a mark.
What Was Removed From the Syllabus
This short section exists to save you time, not to teach you anything. Older textbooks, second-hand notes and many YouTube playlists still cover two topics that used to live in this chapter and have since been removed from the rationalised CBSE syllabus. For 2026–27 they are not examinable.
2. Approximations. Using dy ≈ f′(x)·Δx to estimate values such as √36.6 or (25.02)^(1/2). Deleted.
If a practice paper, an old sample paper or a printed guide asks you for “the equation of the tangent” or “find the approximate value of”, it is following the pre-rationalisation syllabus. Skip it.
So what is in scope? The official 2026–27 wording for this chapter under Unit III (Calculus) is short and specific: applications of derivatives covering rate of change of quantities, increasing and decreasing functions, and maxima and minima — with the first derivative test motivated geometrically and the second derivative test given as a provable tool — together with simple problems that illustrate basic principles and real-life situations.
That is exactly the spine of this article, which is why so much of it is devoted to word problems. The phrase “real-life situations” sits in the syllabus text itself, and it is a strong hint about where the heavier marks go.
Mistakes That Quietly Cost Marks
Read this section the night before your exam. Then test yourself on the CBSE Class 12 Maths sample paper. Every item below is something I have seen capable students do while knowing perfectly well how to avoid it. These are not knowledge gaps; they are pressure errors, and naming them is most of the cure.
| The mistake | What to do instead |
|---|---|
| Substituting the given value before differentiating | Differentiate first, keep letters, substitute in the last line only |
| Treating every solution of f′(x) = 0 as an extremum | Apply a test to each critical point; remember f(x) = x³ at x = 0 |
| Forgetting the endpoints in a closed-interval question | Your candidate list is always: interior critical points plus a and b |
| Keeping a root that the physical situation forbids | State the domain early, then reject invalid roots with a reason |
| Reading f″(c) = 0 as “no maximum or minimum” | Write “second derivative test fails”, then use a sign chart |
| Flipping the meaning of f″ > 0 and f″ < 0 | Sketch a tiny bowl (∪, minimum) or arch (∩, maximum) in the margin |
| Giving the point x = c when the value f(c) was asked for | “Point of maximum” is c; “maximum value” is f(c). Give both if unsure |
| Mixing units (cm/s with metres) in a rate problem | Convert everything to one system before substituting |
| Solving a trigonometric f′(x) = 0 over all of R | Keep only the solutions inside the stated interval, and say you rejected the rest |
| Including the top face on an “open” box or tank | Re-read the word “open” and count the faces on your sketch |
| Stopping at x = 5 without answering the question | Write a final sentence with the requested quantity and its units |
| Differentiating a square root when you did not have to | Optimise the square of the distance or area, then take the root at the end |
2. Did I factorise f′(x) rather than leaving it expanded?
3. Did I state the domain and reject invalid roots with a reason?
4. Did I verify with f″ or a sign chart, and write the conclusion in words?
5. Did I answer the question that was actually asked, with units?
Five questions, half a minute, and it catches almost everything. Build the habit now, in practice, so that it happens automatically in the exam hall.
Practice Worksheet
Ten questions, arranged roughly easiest to hardest. Work each one on paper before you open the answer — reading a solution feels like learning, but it is not the same thing. If you get stuck, look back at the worked example that matches the type rather than jumping straight to the answer.
Question 1.
The radius of a spherical balloon is increasing at 2 cm/s. Find the rate at which (i) its volume and (ii) its surface area are increasing when the radius is 7 cm.
Show Answer
(i) V = (4/3)πr³, so dV/dt = 4πr²(dr/dt) = 4π(49)(2) = 392π cm³/s ≈ 1231.50 cm³/s.
(ii) S = 4πr², so dS/dt = 8πr(dr/dt) = 8π(7)(2) = 112π cm²/s ≈ 351.86 cm²/s.
Remember: differentiate first, then substitute r = 7.
Question 2.
Find the intervals on which f(x) = 2x³ − 3x² − 36x + 7 is increasing and on which it is decreasing.
Show Answer
f′(x) = 6x² − 6x − 36
= 6(x² − x − 6)
= 6(x − 3)(x + 2)
Critical points: x = −2, x = 3
x = −3 : 6(−6)(−1) = +36 positive
x = 0 : 6(−3)(2) = −36 negative
x = 4 : 6(1)(6) = +36 positive
Increasing on (−∞, −2) and (3, ∞); decreasing on (−2, 3).
Question 3.
Find the local maximum and local minimum values of f(x) = x³ − 9x² + 24x − 12 using the second derivative test.
Show Answer
f′(x) = 3x² − 18x + 24 = 3(x − 2)(x − 4) f″(x) = 6x − 18 Critical points: x = 2, x = 4 f″(2) = 12 − 18 = −6 < 0 → LOCAL MAXIMUM f″(4) = 24 − 18 = +6 > 0 → LOCAL MINIMUM f(2) = 8 − 36 + 48 − 12 = 8 f(4) = 64 − 144 + 96 − 12 = 4
Local maximum value 8 at x = 2; local minimum value 4 at x = 4.
Question 4.
Examine f(x) = x⁴ − 4x³ + 4x² + 7 for local maxima and minima. State every extreme value.
Show Answer
f′(x) = 4x³ − 12x² + 8x
= 4x(x² − 3x + 2)
= 4x(x − 1)(x − 2)
Critical points: x = 0, 1, 2
f″(x) = 12x² − 24x + 8
f″(0) = 8 > 0 → LOCAL MINIMUM
f″(1) = 12 − 24 + 8 = −4 < 0 → LOCAL MAXIMUM
f″(2) = 48 − 48 + 8 = +8 > 0 → LOCAL MINIMUM
f(0) = 7
f(1) = 1 − 4 + 4 + 7 = 8
f(2) = 16 − 32 + 16 + 7 = 7
Local minimum value 7 at x = 0 and at x = 2; local maximum value 8 at x = 1. The two minima are equal because this quartic is symmetric about x = 1.
Question 5.
Find the absolute maximum and absolute minimum values of f(x) = 2x³ − 24x + 107 on the interval [1, 3].
Show Answer
f′(x) = 6x² − 24 = 6(x − 2)(x + 2) Critical points: x = 2 and x = −2. x = −2 lies OUTSIDE [1, 3] — reject it. Candidates: x = 1, 2, 3 f(1) = 2 − 24 + 107 = 85 f(2) = 16 − 48 + 107 = 75 f(3) = 54 − 72 + 107 = 89
Absolute maximum = 89 at x = 3; absolute minimum = 75 at x = 2. Notice the maximum occurs at an endpoint — exactly the trap warned about earlier.
Question 6.
Show that f(x) = x + 1/x has a local minimum for x > 0, and find that minimum value.
Show Answer
f′(x) = 1 − 1/x² = (x² − 1)/x² = (x − 1)(x + 1)/x² f′(x) = 0 ⇒ x = 1 or x = −1 For x > 0 the only critical point is x = 1. f″(x) = 2/x³ f″(1) = 2 > 0 → LOCAL MINIMUM f(1) = 1 + 1 = 2
Local minimum value 2, at x = 1. (For x < 0 the same function has a local maximum value −2 at x = −1, but that was not asked.)
Question 7.
A man 2 m tall walks away from a lamp post 6 m high at a steady 5 km/h. At what rate is the length of his shadow increasing?
Show Answer
Let x be his distance from the post and s the length of his shadow. The tip of the shadow is x + s from the post. The lamp, the top of his head and the shadow tip are in a line, so by similar triangles:
6 / (x + s) = 2 / s 6s = 2(x + s) 6s = 2x + 2s 4s = 2x s = x/2 Differentiate with respect to t: ds/dt = (1/2)(dx/dt) = (1/2)(5) = 2.5 km/h
The shadow lengthens at 2.5 km/h (that is, 5/2 km/h). Note that the answer does not depend on how far he has already walked — the rate is constant.
Question 8.
Find two positive numbers whose sum is 20 and whose sum of squares is as small as possible.
Show Answer
Let the numbers be x and 20 − x, 0 < x < 20.
S(x) = x² + (20 − x)²
S′(x) = 2x + 2(20 − x)(−1)
= 2x − 40 + 2x
= 4x − 40
S′(x) = 0 ⇒ x = 10
S″(x) = 4 > 0 → MINIMUM
S(10) = 100 + 100 = 200
The numbers are 10 and 10, and the least sum of squares is 200.
Question 9.
A box with a square base and an open top is to hold 4000 cm³. Find the dimensions that require the least material, and the area of material used.
Show Answer
Let the base side be x cm and the height h cm.
Constraint: x²h = 4000 ⇒ h = 4000/x²
Open top, so the material is the base plus four sides:
S = x² + 4xh
= x² + 4x(4000/x²)
= x² + 16000/x, x > 0
S′(x) = 2x − 16000/x²
S′(x) = 0 ⇒ 2x³ = 16000
⇒ x³ = 8000
⇒ x = 20 cm
S″(x) = 2 + 32000/x³
S″(20) = 2 + 32000/8000 = 2 + 4 = 6 > 0 → MINIMUM
h = 4000/400 = 10 cm
S = 400 + 16000/20 = 400 + 800 = 1200 cm²
Base 20 cm by 20 cm, height 10 cm, using 1200 cm² of material. Once again x = 2h, the signature of an open square-based box.
Question 10.
A wire of length 28 cm is cut into two pieces. One piece is bent into a square and the other into a circle. Find the length of each piece so that the combined area of the square and the circle is a minimum, and find that minimum area.
Show Answer
Let one piece have length x cm (bent into the square),
so the other has length 28 − x cm (bent into the circle).
Square: side = x/4, area = x²/16
Circle: 2πr = 28 − x ⇒ r = (28 − x)/(2π)
area = πr² = (28 − x)²/(4π)
A(x) = x²/16 + (28 − x)²/(4π), 0 < x < 28
A′(x) = x/8 − (28 − x)/(2π)
A′(x) = 0 ⇒ x/8 = (28 − x)/(2π)
⇒ 2πx = 8(28 − x)
⇒ 2πx = 224 − 8x
⇒ x(2π + 8) = 224
⇒ x = 224/(2π + 8) = 112/(π + 4)
A″(x) = 1/8 + 1/(2π) > 0 → MINIMUM
So the square piece is 112/(π + 4) cm ≈ 15.68 cm and the circular piece is 28 − 112/(π + 4) = 28π/(π + 4) cm ≈ 12.32 cm.
Minimum area: x²/16 = 784/(π + 4)² (28−x)²/(4π) = 196π/(π + 4)² A = (784 + 196π)/(π + 4)² = 196(4 + π)/(π + 4)² = 196/(π + 4) cm² ≈ 27.44 cm²
Minimum combined area = 196/(π + 4) cm². This is the hardest question on the sheet — if you got the setup right and only stumbled on the algebra, you are in very good shape.
One last thought before you close this page. You do not need to be brilliant at this chapter tomorrow. You need to be a little better than you were yesterday — one more correct question, one fewer sign error, one word problem where you finally remembered to state the domain. Do that daily and by the time the exam arrives you will be genuinely fluent, without ever having had a heroic day. Small, steady, honest practice beats panic every single time. I am proud of you for getting this far down the page.