Application of Derivatives

A speeding car's speedometer, a company's marginal profit, the fastest-draining shape for a water tank, the largest box you can fold from a fixed sheet of tin — every one of these is a question about how one quantity changes as another quantity changes, and Chapter 5 already handed you the tool that answers it: the derivative. This chapter turns that tool outward. Instead of asking "what is $f'(x)$ ," it asks "what does $f'(x)$ tell me about the world." Four ideas carry the whole chapter, and they build on each other in a straight line. First, if two quantities are linked by a rule and both are changing with time, the Chain Rule lets you find the rate of one from the rate of the other — this is how a five-centimetre-per-second wave tells you how fast the area inside it grows. Second, the sign of $f'(x)$ across an interval tells you whether the graph is climbing or falling there, which is the precise, provable meaning of "increasing" and "decreasing" — no more eyeballing a sketch. Third, wherever a graph turns from climbing to falling (or back), it must pass through a point where the tangent is momentarily flat, a critical point, and classifying these points with the first or second derivative test finds every local peak and valley of a function without ever plotting it. Fourth, when a problem restricts you to a closed interval — a rectangle you must fold from an actual sheet, a fixed budget, a fixed length of wire — the closed interval method combines critical points with the two endpoints to guarantee the true, global maximum or minimum. By the end you will be able to convert a word problem — a ladder sliding down a wall, a cone filling with sand, a manufacturer choosing a price — into a function, differentiate it, and read off exactly where it is largest, smallest, rising, or falling, with a proof-backed reason for every claim.

Key takeaways

Related rates: if $x = f(t)$ and $y = g(t)$ , then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ whenever $\dfrac{dx}{dt} \neq 0$ — this is how one measured rate (radius growing, edge shrinking) gives you another (area, volume, surface area).
Marginal cost and marginal revenue are just $C'(x)$ and $R'(x)$ — the instantaneous rate of change of total cost or revenue with respect to the number of units produced or sold.
Increasing/decreasing (Theorem 1): if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then $f'(x) > 0$ throughout $(a,b)$ makes $f$ increasing on $[a,b]$ , $f'(x) < 0$ makes it decreasing, and $f'(x) = 0$ throughout makes it constant — proved with the Mean Value Theorem.
Critical points are where $f'(c) = 0$ or $f'(c)$ fails to exist; every local maximum or minimum of a differentiable-enough function occurs at a critical point, but not every critical point is an extremum (test $f(x) = x^3$ at $x = 0$ ).
First derivative test: if $f'$ switches sign from positive to negative at $c$ , $c$ is a local maximum; negative to positive, a local minimum; no sign change, a point of inflexion.
Second derivative test: at a critical point with $f'(c) = 0$ , if $f''(c) < 0$ it is a local maximum, if $f''(c) > 0$ it is a local minimum, and if $f''(c) = 0$ the test is silent and you must fall back on the first derivative test.
Closed interval (absolute extrema) method: on a closed interval $[a,b]$ , list every critical point inside $(a,b)$ , add the two endpoints $a$ and $b$ , evaluate $f$ at all of them, and the largest and smallest values you get are the absolute maximum and absolute minimum on $[a,b]$ .
Optimization problems — largest area, least cost, shortest distance, greatest volume — all reduce to writing one variable's objective function in terms of a single free variable, then applying the first or second derivative test (often with the domain restricted, so check the restriction before trusting a critical point).
A local extremum is about a small neighbourhood only; an absolute extremum is about the whole interval — a function can have several local maxima yet only one of them be the tallest of all.
Key idea: the sign of the first derivative classifies where a function rises or falls, its critical points locate every candidate turning point, and the first or second derivative test (backed by comparing endpoint values on closed domains) turns those candidates into certified local or absolute maxima and minima.

The core insight

Reading the sign of $f'(x)$ tells you where $f$ rises and falls, the zeros (or non-differentiable points) of $f'$ mark every candidate turning point, and the first or second derivative test — extended by comparing the two endpoint values on a closed interval — turns those candidates into certified local or absolute maxima and minima.

What this chapter covers

The chapter has four connected pieces, and this article follows them in the same order NCERT does. The core ideas section opens with rate of change of quantities, where the Chain Rule converts one measured rate (say, how fast a radius grows) into another (how fast an area or volume grows) — this is the "related rates" machinery used throughout physics and economics, including the special case of marginal cost and marginal revenue. It then builds the precise, provable definitions of increasing and decreasing functions, replacing "the graph goes up" with a sign test on $f'(x)$ that is backed by the Mean Value Theorem. From there it defines local maxima and minima through critical points, and gives you two complementary tools to classify them — the first derivative test (always works, needs a sign chart) and the second derivative test (faster, but can fail). Finally it handles the practically important case of absolute maxima and minima on a closed interval, the closed interval method, which is what every real optimization problem — boxes, cans, ladders, poles, trapeziums — actually needs. Four worked derivations prove the four central theorems in full; four worked examples show board-level and JEE/NEET-level problems solved start to finish; and the cheatsheet at the end collects every formula and the recipe for using it.

The core ideas

Rate of change of quantities

If a quantity $y$ varies with another quantity $x$ according to a rule $y = f(x)$ , the derivative $\dfrac{dy}{dx}$ (equivalently $f'(x)$ ) is by definition the rate of change of $y$ with respect to $x$ , and its value at a particular point $x = x_0$ , written $\left.\dfrac{dy}{dx}\right|_{x=x_0}$ , is the rate of change of $y$ with respect to $x$ at that instant. The genuinely useful extension is when both $x$ and $y$ are themselves functions of a third variable $t$ (typically time): if $x = f(t)$ and $y = g(t)$ , the Chain Rule links their rates directly.

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, \qquad \text{provided } \frac{dx}{dt} \neq 0. \qquad (1)

where: $dy/dt$ and $dx/dt$ are the rates at which $y$ and $x$ change with time, and dividing one by the other cancels the common variable $t$ to leave the rate of $y$ with respect to $x$ directly. This single identity is the engine behind every "related rates" problem in the chapter: a circle's radius grows at a known rate, and you are asked how fast its area grows; a cube's volume is pumped up at a known rate, and you are asked how fast its surface area responds; a cone of sand grows in height at an unknown rate while volume is poured in at a known rate. In every case you write down the geometric relation between the two quantities (say $A = \pi r^2$ ), differentiate both sides with respect to $t$ using the Chain Rule, substitute the known rate, and solve for the wanted one. A useful sign convention accompanies this: $dy/dx$ is positive exactly when $y$ increases as $x$ increases, and negative exactly when $y$ decreases as $x$ increases — a shrinking length is recorded as a negative rate, an inflating volume as a positive one. Two special applications recur throughout economics-flavoured problems in the chapter. If $C(x)$ is the total cost of producing $x$ units, the marginal cost is $MC = C'(x)$ , the instantaneous rate of change of total cost with respect to output — not the cost of literally one more unit, but its calculus idealisation. Likewise if $R(x)$ is the total revenue from selling $x$ units, the marginal revenue is $MR = R'(x)$ . Both are found exactly the way any other rate is found: differentiate, then substitute the given value of $x$ .

Increasing and decreasing functions

Look at the graph of $f(x) = x^2$ . To the right of the origin, moving left to right, the height of the graph keeps climbing; to the left of the origin, moving left to right, the height keeps falling. NCERT turns that visual impression into an algebraic definition that does not depend on a picture at all. Let $I$ be an interval in the domain of a real function $f$ . Then $f$ is increasing on $I$ if $x_1 < x_2$ in $I$ forces $f(x_1) \leq f(x_2)$ for all $x_1, x_2$ in $I$ , and strictly increasing if that inequality is strict, $f(x_1) < f(x_2)$ . The mirror definitions — decreasing and strictly decreasing — reverse the inequality on $f$ . A function is constant on $I$ if $f(x) = c$ for every $x$ in $I$ . A function is monotonic on $I$ if it is either increasing throughout $I$ or decreasing throughout $I$ (not a mixture). These definitions apply equally at a single point $x_0$ : $f$ is increasing (or decreasing) at $x_0$ if there is some open interval around $x_0$ on which $f$ has that behaviour. Checking this definition directly, by picking arbitrary $x_1 < x_2$ and comparing $f(x_1)$ and $f(x_2)$ algebraically, works but is slow. The chapter's central practical result is that the sign of the derivative decides monotonicity outright, and this is proved rigorously (not just illustrated) in the first derivation below using the Mean Value Theorem. Once proved, the working method is short: differentiate $f$ , solve $f'(x) = 0$ to find the points that could divide the real line into pieces, then test the sign of $f'(x)$ on each resulting sub-interval by plugging in one convenient test value. A cubic like $f(x) = 4x^3 - 6x^2 - 72x + 30$ factors as $f'(x) = 12(x-3)(x+2)$ , so $x = -2$ and $x = 3$ split the line into three intervals, and the sign of the product $(x-3)(x+2)$ on each piece — read off instantly from how many of the two factors are negative — tells you at a glance where $f$ climbs and where it falls.

Figure

A sign chart and graph sketch for a cubic function showing it increasing on the interval up to negative two, decreasing between negative two and three, and increasing again beyond three, with the derivative sign marked plus minus plus along the number line

Critical points, local maxima and local minima

A local extremum is a "nearby champion," not necessarily a global one. Formally, an interior point $c$ in the domain of $f$ is a point of local maxima if there is some $h > 0$ such that $f(c) \geq f(x)$ for every $x$ within $h$ of $c$ (that is, in $(c-h, c+h)$ ); $f(c)$ is then the local maximum value. A point of local minima is defined the same way with the inequality reversed. Geometrically, at a local maximum the graph is climbing just to the left of $c$ and falling just to the right, so the tangent line at $c$ must be momentarily horizontal — which is exactly the content of the theorem that follows. If $f$ has a local maximum or minimum at an interior point $c$ , then either $f'(c) = 0$ or $f$ is simply not differentiable at $c$ . A point where either of these happens — $f'(c) = 0$ , or the derivative fails to exist — is called a critical point of $f$ . This gives a finite hunting ground: every local extremum sits at a critical point, so to find all local extrema you first find all critical points and then test each one. The converse is false and this is the single most common source of exam errors in the whole chapter: a critical point need not be an extremum at all. Take $f(x) = x^3$ ; then $f'(x) = 3x^2$ and $f'(0) = 0$ , yet $x = 0$ is neither a local maximum nor a local minimum, because $f$ is strictly increasing right through $x = 0$ . Such a point, where $f'$ vanishes but does not change sign, is called a point of inflexion.

The first and second derivative tests

Having narrowed the search to critical points, the chapter gives two ways to classify each one. The first derivative test watches how the sign of $f'(x)$ behaves on either side of a critical point $c$ at which $f$ is continuous: if $f'(x) > 0$ immediately to the left of $c$ and $f'(x) < 0$ immediately to the right, $c$ is a local maximum; if the sign pattern is negative-then-positive, $c$ is a local minimum; and if $f'(x)$ keeps the same sign on both sides, $c$ is a point of inflexion, neither a maximum nor a minimum. This test always works, even where $f$ is not twice differentiable, but it requires building a full sign chart. The second derivative test is a faster shortcut for the common case where $f$ is twice differentiable at $c$ and $f'(c) = 0$ : if $f''(c) < 0$ , $c$ is a local maximum; if $f''(c) > 0$ , $c$ is a local minimum; and if $f''(c) = 0$ , the test gives no verdict at all and you must return to the first derivative test. This failure case is not a rare edge case to be ignored — $f(x) = x^4$ at $x = 0$ has $f'(0) = f''(0) = 0$ yet $x=0$ is a genuine local (and absolute) minimum, while $f(x) = x^3$ at $x=0$ has $f'(0)=f''(0)=0$ and is a point of inflexion; the second derivative test cannot distinguish these two cases, only the first derivative test (or higher derivatives) can.

Absolute maxima and minima on a closed interval

Local extrema are a purely local statement; they say nothing about which point is tallest across the entire domain. The absolute maximum (or greatest value) of $f$ on an interval $I$ is a value $f(c)$ with $f(c) \geq f(x)$ for every $x$ in $I$ ; the absolute minimum (or least value) reverses the inequality. On an open interval a continuous function may have no absolute extrema at all — $f(x) = x$ on $(0,1)$ never attains a largest or smallest value, because however close you get to the endpoints there is always a point closer still. Restrict the same function to the closed interval $[0,1]$ and it immediately has an absolute minimum $0$ at $x=0$ and absolute maximum $1$ at $x=1$ ; a continuous function on a closed interval is always guaranteed both an absolute maximum and an absolute minimum, and each is attained either at an interior critical point or at one of the two endpoints. This gives the four-step closed interval method, proved as a derivation below: list the critical points inside $(a,b)$ , add the endpoints $a$ and $b$ , evaluate $f$ at every point on that list, and read off the largest and smallest values as the absolute maximum and minimum.

Figure

A graph on a closed interval from a to b showing a continuous curve with a local minimum inside the interval that is not the lowest point overall, the true absolute minimum sitting at the left endpoint and the true absolute maximum sitting at the right endpoint

Derivations

Derivation ATheorem 1 — a positive derivative forces a function to increase

The goal is to prove, not merely illustrate, that the sign of $f'(x)$ across an interval controls whether $f$ is increasing or decreasing there — this is the theorem that upgrades "the graph looks like it is climbing" into an airtight algebraic guarantee. Suppose $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$ . The claim is that $f'(x) > 0$ for every $x$ in $(a,b)$ makes $f$ increasing on $[a,b]$ ; the proofs that $f'(x) < 0$ gives decreasing and $f'(x) = 0$ gives constant follow by an identical argument with the inequality reversed or removed, so only the increasing case needs to be written out in full. The one external tool the proof borrows is the Mean Value Theorem from the previous chapter, which says that for any two points $x_1 < x_2$ in $[a,b]$ there exists some $c$ strictly between them with $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$ .

Step 1. Take any two points $x_1, x_2$ in $[a,b]$ with $x_1 < x_2$ . The definition of "increasing on $[a,b]$ " demands exactly one thing: show that $f(x_1) < f(x_2)$ follows from $x_1 < x_2$ , for every such pair. So fix an arbitrary pair and aim to compare $f(x_1)$ and $f(x_2)$ .

Step 2. Apply the Mean Value Theorem to $f$ on the sub-interval $[x_1, x_2]$ (this sub-interval lies inside $[a,b]$ , so $f$ is continuous on it and differentiable on its interior, exactly what the Mean Value Theorem needs). There exists a point $c$ with $x_1 < c < x_2$ such that

f(x_2) - f(x_1) = f'(c)\,(x_2 - x_1). \qquad (2)

Step 3. Read off the sign of the right-hand side. Since $c$ lies in $(a,b)$ , the hypothesis $f'(x) > 0$ throughout $(a,b)$ applies to it, so $f'(c) > 0$ . Also $x_2 - x_1 > 0$ because $x_1 < x_2$ . A positive number times a positive number is positive, so the right-hand side of equation (2) is positive, which forces the left-hand side to be positive as well: $f(x_2) - f(x_1) > 0$ .

Step 4. Conclude. $f(x_2) - f(x_1) > 0$ means exactly $f(x_2) > f(x_1)$ , i.e. $f(x_1) < f(x_2)$ . Since $x_1$ and $x_2$ were an arbitrary pair with $x_1 < x_2$ in $[a,b]$ , this holds for every such pair, which is precisely the definition of $f$ being increasing on $[a,b]$ . The decreasing case repeats this argument with $f'(c) < 0$ flipping the final inequality, and the constant case repeats it with $f'(c) = 0$ making $f(x_2) - f(x_1) = 0$ for every pair, so $f$ never changes value. A useful strengthening, stated without proof in the chapter, is that the hypothesis only needs to hold in the interior of the interval: if $f'(x) > 0$ everywhere in an interval except possibly at its endpoints, and $f$ stays continuous there, the same increasing conclusion still holds.

Go deeper · Derivation B · 2 minThe first derivative test — turning a sign change into a verified extremum

Derivation BThe first derivative test — turning a sign change into a verified extremum

The goal is to justify, using only Theorem 1 already proved above, why a sign change in $f'$ around a critical point $c$ genuinely certifies a local maximum or minimum, rather than merely suggesting one. The first derivative test is stated as a rule to be applied, but it is not an independent axiom — it falls straight out of the increasing/decreasing theorem once you apply that theorem separately on each side of $c$ . Suppose $f$ is defined on an open interval around $c$ , is continuous at $c$ , and $c$ is a critical point, with $f'(x) > 0$ for all $x$ close to and to the left of $c$ , and $f'(x) < 0$ for all $x$ close to and to the right of $c$ — the "positive-to-negative" case, which the test claims is a local maximum.

Step 1. Apply Theorem 1 on the left piece. On some interval $(c-h, c)$ , $f'(x) > 0$ throughout, so by Theorem 1 (using the strengthened interior version, since $f$ is continuous up to $c$ ), $f$ is increasing on $(c-h, c]$ . In particular, for any $x$ in $(c-h, c)$ , $x < c$ implies $f(x) < f(c)$ .

Step 2. Apply Theorem 1 on the right piece. On some interval $(c, c+h)$ , $f'(x) < 0$ throughout, so $f$ is decreasing on $[c, c+h)$ . In particular, for any $x$ in $(c, c+h)$ , $c < x$ implies $f(c) > f(x)$ , i.e. $f(x) < f(c)$ .

Step 3. Combine the two pieces. For every $x$ in the punctured neighbourhood $(c-h, c+h)$ with $x \neq c$ — whether $x$ is to the left or to the right of $c$ — Step 1 or Step 2 gives $f(x) < f(c)$ . This is exactly Definition 4's requirement for $c$ to be a point of local maxima, with $f(c)$ the local maximum value. The negative-to-positive case is the mirror image: applying Theorem 1's decreasing half on the left and increasing half on the right gives $f(x) > f(c)$ on both sides, which is the definition of a local minimum.

Step 4. Handle the no-sign-change case. If $f'(x)$ keeps the same sign on both sides of $c$ — say positive on both $(c-h,c)$ and $(c,c+h)$ — then Theorem 1 makes $f$ increasing on the whole punctured neighbourhood, gluing to a function that is increasing straight through $c$ (using the continuity of $f$ at $c$ itself). Then points just left of $c$ have smaller values than $f(c)$ and points just right of $c$ have larger values than $f(c)$ , so neither the local-maximum nor the local-minimum inequality holds in every direction: $c$ is a point of inflexion, not an extremum. This is exactly why $f(x)=x^3$ at $x=0$ fails to be an extremum — $f'(x) = 3x^2 \geq 0$ never changes sign at $0$ .

Go deeper · Derivation C · 2 minThe second derivative test — reading a verdict off the second derivative alone

Derivation CThe second derivative test — reading a verdict off the second derivative alone

The goal is to derive the second derivative test from the definition of the derivative as a limit, showing precisely why $f''(c) < 0$ forces the same positive-to-negative sign pattern in $f'$ that the first derivative test needs for a local maximum — turning a single number, $f''(c)$ , into a full local sign-chart verdict without actually building the chart. Assume $f'(c) = 0$ and $f$ is twice differentiable at $c$ , with $f''(c) < 0$ .

Step 1. Write $f''(c)$ as the limiting difference quotient of $f'$ itself, since the second derivative is just the derivative of the first derivative.

f''(c) = \lim_{x \to c} \frac{f'(x) - f'(c)}{x - c}. \qquad (3)

Step 2. Use $f'(c) = 0$ to simplify. Equation (3) collapses to $f''(c) = \lim_{x \to c} \dfrac{f'(x)}{x - c}$ , because subtracting zero changes nothing in the numerator. This is the key rewriting: the second derivative at $c$ is the limit of the ratio of $f'(x)$ to $(x - c)$ .

Step 3. Read the sign implication of a negative limit. Since this limit equals $f''(c)$ , which is negative, the ratio $\dfrac{f'(x)}{x-c}$ must itself be negative for all $x$ sufficiently close to $c$ (a limit that is strictly negative cannot be approached through values that are zero or positive arbitrarily close to $c$ ). A negative ratio means $f'(x)$ and $(x-c)$ have opposite signs at every such nearby $x$ .

Step 4. Translate opposite signs into the sign pattern the first derivative test needs. For $x$ slightly less than $c$ , $(x - c) < 0$ ; opposite sign forces $f'(x) > 0$ . For $x$ slightly greater than $c$ , $(x - c) > 0$ ; opposite sign forces $f'(x) < 0$ . That is exactly " $f'$ positive to the left, negative to the right" — the first derivative test's condition for a local maximum at $c$ . By the previous derivation, this certifies $c$ as a point of local maxima. The mirror argument with $f''(c) > 0$ gives $f'(x) < 0$ to the left and $f'(x) > 0$ to the right, certifying a local minimum. When $f''(c) = 0$ , Step 3's argument breaks — a zero limit is consistent with $f'(x)/(x-c)$ being positive on one side and negative on the other, or keeping one sign throughout, or oscillating — so no verdict can be extracted this way, and the test is correctly declared inconclusive, sending you back to the first derivative test.

Go deeper · Derivation D · 2 minThe closed interval method — why checking critical points and endpoints is enough

Derivation DThe closed interval method — why checking critical points and endpoints is enough

The goal is to justify the four-step working rule for absolute extrema on a closed interval $[a,b]$ : that scanning only the critical points inside $(a,b)$ together with the two endpoints $a$ and $b$ is guaranteed to catch the true absolute maximum and minimum, with no other point ever needing to be checked. The argument rests on two facts taken as given (their proofs belong to real analysis and are stated without proof in the chapter): a continuous function on a closed interval always attains an absolute maximum and an absolute minimum somewhere in that interval, and at any interior point where a differentiable function attains an absolute extremum, the derivative there must vanish.

Step 1. Existence is guaranteed. Because $f$ is continuous on the closed, bounded interval $[a,b]$ , it attains its absolute maximum at some point $c_1$ in $[a,b]$ and its absolute minimum at some point $c_2$ in $[a,b]$ — these are not candidates to be hunted for, they are guaranteed to exist. The only question is where $c_1$ and $c_2$ can be.

Step 2. Split into interior points and endpoints. Every point of $[a,b]$ is either one of the two endpoints $a$ , $b$ , or an interior point strictly between them. So $c_1$ (and likewise $c_2$ ) is either an endpoint, in which case it is already on the list to be checked in Step 2 of the working rule, or an interior point.

Step 3. If the extremum is interior, it must be a critical point. Suppose $c_1$ is interior and $f$ is differentiable at $c_1$ . Since $f$ attains its absolute maximum on the whole interval at $c_1$ , in particular $f(c_1) \geq f(x)$ for all $x$ in some small neighbourhood of $c_1$ inside $(a,b)$ — so $c_1$ is also a point of local maxima. By the necessary condition for local extrema (proved via the same left-and-right sign argument used in the first derivative test derivation, run in reverse: if $f'(c_1)$ were positive, $f$ would be increasing just past $c_1$ and exceed $f(c_1)$ , contradicting the maximum; if negative, $f$ would exceed $f(c_1)$ just before $c_1$ , again a contradiction), it follows that $f'(c_1) = 0$ . If instead $f$ is simply not differentiable at $c_1$ , then by definition $c_1$ is already a critical point on that count alone. Either way, an interior absolute extremum is always a critical point, so it appears on the Step 1 list of the working rule.

Step 4. Assemble the guarantee. Every possible location for the absolute maximum — endpoint or interior critical point — is on the combined list of {critical points in $(a,b)$ } union { $a, b$ }. Evaluating $f$ at every point on this finite list and taking the largest value therefore cannot miss the true absolute maximum wherever it sits; the same argument with the inequality reversed guarantees the smallest value on the list is the true absolute minimum. This is why the working rule needs nothing beyond solving $f'(x) = 0$ , locating non-differentiable points, and plugging numbers into $f$ — no sign chart is even required, only a comparison of the resulting values.

Worked examples

Board level — how fast is the enclosed area growing (NCERT Example 3)

A stone dropped into a quiet lake sends out circular waves at a speed of $4\ \text{cm/s}$ . At the instant the radius of the circular wave is $10\ \text{cm}$ , how fast is the enclosed area increasing?

Step 1. Set up the relation between the two changing quantities. The area enclosed by a circular wave of radius $r$ is $A = \pi r^2$ . Both $A$ and $r$ change with time $t$ , so differentiate with respect to $t$ using the Chain Rule: $\dfrac{dA}{dt} = \dfrac{d}{dr}(\pi r^2) \cdot \dfrac{dr}{dt} = 2\pi r \dfrac{dr}{dt}$ .

Step 2. Substitute the given rate and the given instant. It is given that $\dfrac{dr}{dt} = 4\ \text{cm/s}$ (the wave's radius grows at this constant speed), and the instant asked about is $r = 10\ \text{cm}$ .

Step 3. Compute. $\dfrac{dA}{dt} = 2\pi (10)(4) = 80\pi$ . The enclosed area is increasing at the rate of $80\pi\ \text{cm}^2/\text{s}$ when $r = 10\ \text{cm}$ .

Board level — intervals of increase and decrease for a cubic (NCERT Example 11)

Find the intervals in which the function $f(x) = 4x^3 - 6x^2 - 72x + 30$ is (a) increasing (b) decreasing.

Step 1. Differentiate and factor. $f'(x) = 12x^2 - 12x - 72 = 12(x^2 - x - 6) = 12(x-3)(x+2)$ .

Step 2. Locate the critical points. Setting $f'(x) = 0$ gives $x = -2$ and $x = 3$ . These two points split the real line into three disjoint intervals: $(-\infty, -2)$ , $(-2, 3)$ , and $(3, \infty)$ .

Step 3. Test the sign of $f'(x) = 12(x-3)(x+2)$ on each interval by checking one convenient value. On $(-\infty,-2)$ , take $x = -3$ : both factors $(x-3)$ and $(x+2)$ are negative, so their product is positive, and $f'(-3) > 0$ . On $(-2,3)$ , take $x = 0$ : $(x-3) < 0$ and $(x+2) > 0$ , so the product is negative, and $f'(0) < 0$ . On $(3,\infty)$ , take $x = 4$ : both factors are positive, so $f'(4) > 0$ .

Step 4. Read off the conclusion. $f$ is increasing on $(-\infty, -2)$ and on $(3, \infty)$ , and decreasing on $(-2, 3)$ . Note that $f$ is neither increasing nor decreasing on all of $\mathbb{R}$ — the three pieces genuinely alternate, and no single word describes the whole real line.

Competitive level (JEE / NEET) — the largest open box cut from a rectangular sheet (NCERT Example 36)

An open-topped box is made by removing equal squares of side $x$ metres from each corner of a $3\ \text{m}$ by $8\ \text{m}$ rectangular aluminium sheet and folding up the sides. Find the volume of the largest such box.

Step 1. Express the volume as a function of the single free variable $x$ . After removing squares of side $x$ from each corner and folding up the flaps, the box has height $x$ , length $8 - 2x$ , and breadth $3 - 2x$ . So $V(x) = x(3-2x)(8-2x) = 4x^3 - 22x^2 + 24x$ , valid for $0 < x < 1.5$ (breadth must stay positive).

Step 2. Differentiate and solve for critical points. $V'(x) = 12x^2 - 44x + 24 = 4(x-3)(3x-2)$ , so $V'(x) = 0$ at $x = 3$ or $x = \dfrac{2}{3}$ . Immediately check the domain restriction: $x = 3$ lies outside $0 < x < 1.5$ (you cannot cut a $3\ \text{m}$ square from a sheet whose short side is only $3\ \text{m}$ without eliminating the breadth entirely), so only $x = \dfrac{2}{3}$ is admissible.

Step 3. Classify $x = \dfrac{2}{3}$ with the second derivative test. $V''(x) = 24x - 44$ , so $V''\!\left(\dfrac{2}{3}\right) = 16 - 44 = -28 < 0$ . A negative second derivative certifies a local maximum, and since $x=2/3$ is the only critical point in the admissible domain (with $V\to 0$ at both ends of $(0,1.5)$ ), it is also the absolute maximum on that domain.

Step 4. Compute the maximum volume. $V\!\left(\dfrac{2}{3}\right) = 4\left(\dfrac{2}{3}\right)^3 - 22\left(\dfrac{2}{3}\right)^2 + 24\left(\dfrac{2}{3}\right) = \dfrac{32}{27} - \dfrac{88}{9} + 16 = \dfrac{200\sqrt{3}}{27}$ after simplification is the textbook's stated closed form; numerically this is $V\!\left(\tfrac{2}{3}\right) = \tfrac{200}{27}\ \text{m}^3 \approx 7.41\ \text{m}^3$ . Cutting a square of side $\dfrac{2}{3}\ \text{m}$ from each corner gives the largest box, with volume $\dfrac{200}{27}\ \text{m}^3 \approx 7.41\ \text{m}^3$ . Timing note: about 3 minutes. The domain check in Step 2 is the part JEE most often tests separately — a critical point that solves $f'(x)=0$ algebraically but sits outside the physical domain must be discarded before you even reach the second derivative test; plugging an inadmissible root straight into $V''$ and reporting it as the answer is the single most common slip in box-and-sheet optimization problems.

Competitive level (JEE / NEET) — shortest distance from a soldier to a moving target (NCERT Example 29)

An enemy helicopter flies along the curve $y = x^2 + 7$ . A soldier standing at $(3, 7)$ wants to shoot it down when it is nearest to him. Find the nearest distance.

Step 1. Write the squared distance as a function of $x$ , since minimizing distance and minimizing squared distance happen at the same point (squaring is increasing for non-negative values) and squared distance avoids a square root. For a helicopter position $(x, x^2+7)$ , the squared distance to $(3,7)$ is $f(x) = (x-3)^2 + (x^2 + 7 - 7)^2 = (x-3)^2 + x^4$ .

Step 2. Differentiate and factor. $f'(x) = 2(x-3) + 4x^3 = 2\left(2x^3 + x - 3\right) = 2(x-1)(2x^2+2x+3)$ . (Check: expanding $(x-1)(2x^2+2x+3) = 2x^3+2x^2+3x-2x^2-2x-3 = 2x^3+x-3$ , matching.)

Step 3. Solve $f'(x) = 0$ . The quadratic factor $2x^2+2x+3$ has discriminant $2^2 - 4(2)(3) = 4 - 24 = -20 < 0$ , so it has no real roots and never changes sign (it stays positive, since its leading coefficient is positive). The only real critical point is therefore $x = 1$ , and since there are no domain endpoints to worry about (the curve extends over all real $x$ ), this single critical point is the entire candidate list.

Step 4. Evaluate and confirm it is a minimum, not a maximum. $f(1) = (1-3)^2 + 1^4 = 4 + 1 = 5$ . To confirm this is a minimum rather than a maximum, compare a nearby value: $f(0) = (0-3)^2 + 0^4 = 9 > 5$ , so $f$ is larger away from $x=1$ , confirming $x=1$ gives the minimum squared distance. The minimum distance is $\sqrt{5}$ . Timing note: about 2.5 minutes. The discriminant check in Step 3 is what separates a fast, certain solution from a slow one — factoring a cubic derivative into a linear factor times an unfactorable quadratic, then proving that quadratic has no real roots, is faster and more rigorous than attempting to test sign changes for a cubic with three possible real roots. Always try to factor $f'(x)$ completely before reaching for a sign chart or the second derivative test.

Common traps

Mistaking a critical point for an automatic extremum. $f'(c) = 0$ only nominates $c$ as a candidate; it does not certify a maximum or minimum. Every critical point must be tested — by sign change or by the second derivative — before you call it an extremum, exactly because $f(x) = x^3$ at $x=0$ shows the nomination can fail.

Forgetting non-differentiable points are also critical points. A critical point is where $f'(c) = 0$ or where $f'$ fails to exist, such as at a sharp corner like $f(x) = |x|$ at $x=0$ . Solving $f'(x)=0$ alone and stopping there silently discards these corner cases, which is exactly where the second derivative test cannot even be attempted.

Applying the second derivative test when it is silent. If $f''(c) = 0$ , the test gives no information whatsoever — not "probably a maximum," not "probably a minimum," nothing. The only correct move is to fall back on the first derivative test's sign chart.

Trusting an algebraic critical point without checking the domain. In optimization word problems the variable is almost always physically restricted (a length must be positive, a side cut from a fixed sheet cannot exceed the sheet), and an algebraically valid root of $f'(x)=0$ can fall outside that restriction, as $x=3$ did in the open-box example — it must be discarded before testing.

Confusing "increasing on two separate intervals" with "increasing on their union." A function can be increasing on $(-\infty,-2)$ and separately increasing on $(3,\infty)$ without being increasing on the union of the two, and without being increasing (or monotonic at all) on the whole real line. Report each interval separately; never merge disjoint increasing intervals into one blanket statement.

Skipping endpoints in a closed-interval problem. The absolute maximum or minimum on $[a,b]$ can sit at an endpoint even when that endpoint is not a critical point at all. Evaluating $f$ only at critical points and forgetting $f(a)$ and $f(b)$ is the most common way to get an absolute-extremum question wrong, even when every other step was correct.

Misreading marginal cost or marginal revenue as a difference rather than a derivative. Marginal cost is $C'(x)$ , evaluated at the given $x$ — not $C(x) - C(x-1)$ . The chapter's definition is explicitly the instantaneous rate of change, found by differentiating first and substituting second.

Common trapThe single biggest misconception in this chapter is treating "the derivative vanishes here" as proof of a local maximum or minimum

Nothing in the definition of a critical point guarantees an extremum — it only guarantees a horizontal tangent or a non-differentiable point, and both the first and second derivative tests exist precisely because a verdict still has to be earned. The textbook's own counterexample says it plainly: for $f(x) = x^3$ , $f'(0) = 0$ , yet $f$ is strictly increasing straight through $x=0$ , so $0$ is neither a maximum nor a minimum but a point of inflexion. Before writing "local maximum" or "local minimum" next to any critical point, you must show either a sign change in $f'$ across it (first derivative test) or a strict inequality in $f''$ at it (second derivative test, with the caveat that $f''(c)=0$ proves nothing and forces a return to the first derivative test). A bare " $f'(c) = 0$ , therefore extremum" is not a complete argument and will not earn full marks.

You should be able to…

State the precise definitions of increasing, strictly increasing, decreasing, and monotonic on an interval, and connect each to the sign of $f'(x)$ via Theorem 1.
Set up and solve a related-rates problem: write the geometric relation, differentiate with respect to time using the Chain Rule, substitute the given rate and instant, and solve for the unknown rate.
Compute marginal cost and marginal revenue as $C'(x)$ and $R'(x)$ , correctly distinguishing them from a plain difference in totals.
Find all critical points of a function, including both zeros of $f'$ and points where $f'$ fails to exist.
Apply the first derivative test using a sign chart, and explain in your own words why a sign change certifies an extremum.
Apply the second derivative test, and know exactly when it is silent ( $f''(c)=0$ ) and what to do next.
Distinguish a local extremum from an absolute extremum, and state why an open interval can fail to have either.
Execute the closed-interval method end to end: find critical points inside $(a,b)$ , add both endpoints, evaluate $f$ everywhere on that list, and identify the largest and smallest values.
Translate a real-world optimization scenario (box, tank, wire, poles, trapezium) into a single-variable function with an explicit, physically sensible domain, and discard critical points that fall outside it.
Recognize and avoid the chapter's signature trap: treating $f'(c)=0$ as proof of an extremum without a supporting sign-change or second-derivative argument.

Ask the tutor: if you are stuck on classifying a critical point or setting up an optimization problem from this chapter, ask the tutor to walk through your specific function step by step.

What exactly is the difference between a local maximum and an absolute maximum?

A local maximum only has to beat its immediate neighbours — there exists some small interval around the point where it is the tallest value, but points far away on the graph are allowed to be even taller. An absolute maximum has to beat every single point in the entire domain (or the entire interval under discussion) at once. A function can have several local maxima of different heights, and only the tallest of them is also the absolute maximum; on an open interval, a function might have no absolute maximum at all even while having plenty of local maxima. Whenever a question says "greatest value" or "global maximum," it means absolute; "turning point" or "peak" on its own usually means local, so read the wording carefully.

Why does the first derivative test require f to be continuous at the critical point?

The whole argument rests on gluing together two separate applications of the increasing/decreasing theorem — one on the left piece, one on the right piece — into a single statement about $f(c)$ itself. That gluing step needs $f(c)$ to actually be the limiting value approached from both sides, which is exactly what continuity at $c$ guarantees. If $f$ had a jump discontinuity at $c$ , the values just left and just right of $c$ could both be, say, less than the limiting values but unrelated to the actual value $f(c)$ sitting at the jump, and the comparison used in the proof would simply not go through. In practice, every function you differentiate in this chapter is continuous wherever it is defined, so this hypothesis is rarely the binding constraint, but it is the reason the theorem is stated the way it is.

If f double prime of c is zero, does that mean c is definitely a point of inflexion?

No, and this is a common overcorrection. $f''(c) = 0$ only means the second derivative test is silent; it does not by itself tell you whether $c$ is a maximum, a minimum, or an inflexion point. Compare $f(x) = x^4$ and $f(x) = x^3$ at $x = 0$ : both have $f'(0) = f''(0) = 0$ , yet $x^4$ has a genuine local (and absolute) minimum at $0$ while $x^3$ has a point of inflexion there. The only reliable way to settle the question when $f''(c)=0$ is to go back to the first derivative test and check whether $f'$ actually changes sign around $c$ .

In a related-rates problem, how do I know which rate is given and which one I am solving for?

Read the problem for two clues: the rate stated as a fixed numerical value with units like "cm/s" or "cubic metres per second" is the given rate, and the phrase "how fast is ... increasing/decreasing" points at the unknown rate you must solve for, usually at a specific stated instant (a particular radius, height, or time). Write down the geometric equation linking the two quantities first (area in terms of radius, volume in terms of edge, and so on), differentiate the whole equation with respect to time using the Chain Rule, then substitute both the given rate and the given instantaneous value of the quantity — never substitute the instantaneous value before differentiating, since a constant does not have a meaningful derivative and differentiating too early throws away the relationship you need.

Why do optimization word problems always end with checking the second derivative or a sign chart, even after solving f prime of x equals zero?

Solving $f'(x) = 0$ only produces candidates — it cannot by itself distinguish a maximum from a minimum from an inflexion point, exactly as the chapter's own $x^3$ example shows. A word problem asking for the "maximum volume" or "minimum cost" needs you to certify which type of critical point you have found, not just locate it; skipping the second derivative test (or a first-derivative sign chart) means you have found a stationary value, not necessarily the one the question asked for. It is entirely possible for a critical point to solve $f'(x)=0$ and yet be a minimum when the question wanted a maximum, so the classification step is not optional bookkeeping — it is the step that actually answers the question.