Application of Derivatives

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 6.

Why does a balloon grow slower and slower?

Blow into a balloon at a steady pace. The air goes in at the same rate every second, yet the balloon seems to swell quickly at first and then more and more lazily. Nothing has changed except its size. This chapter explains why, and then goes much further. The derivative tells you how fast something changes, which way it is heading, where it turns around and, finally, what the best possible value is. Those four questions are the four sections of the chapter.

The chapter in one line: rate of change → increasing or decreasing → local maximum and minimum → absolute maximum and minimum and optimisation.

The derivative as a rate of change

If $y = f(x)$, then $\dfrac{dy}{dx}$ is the rate at which $y$ changes for a unit change in $x$. At one particular point $x = x_0$ the rate is the number $f'(x_0)$. A positive value means $y$ grows as $x$ grows. A negative value means $y$ shrinks.

In real situations both quantities depend on time $t$, and the link between their rates is the chain rule:

$$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}, \qquad \dfrac{dx}{dt} \neq 0$$

This is the engine of every "related rates" question. The radius of a balloon and its volume are tied together by a formula, so if you know how fast one changes, differentiating the formula with respect to $t$ tells you how fast the other changes. The table gives the four formulas CUET uses again and again.

Why the sphere result explains the balloon. If the volume grows at a fixed rate, then $\dfrac{dr}{dt} = \dfrac{1}{4\pi r^2}\cdot\dfrac{dV}{dt}$. The factor $r^2$ sits in the denominator, so as the balloon gets bigger the same air adds a thinner and thinner layer to the radius. That is exactly why it feels lazier.

Worked example: the swelling balloon. Air is pumped into a spherical balloon at $100\ \text{cm}^3\ \text{s}^{-1}$. How fast is the radius increasing when $r = 5$ cm?

Common wrong step. Putting $r = 5$ into the formula before differentiating. Once $r$ is a fixed number, its derivative is zero and the chain rule has nothing to work on. Differentiate first, substitute afterwards.

Pause and predict: the balloon is inflated at a constant rate. When its radius has doubled, what happens to $\dfrac{dr}{dt}$?

Check your answer: it becomes one quarter of what it was. Since $\dfrac{dr}{dt} = \dfrac{1}{4\pi r^2}\dfrac{dV}{dt}$, doubling $r$ multiplies $r^2$ by 4.

Worked example: the growing cube. The volume of a cube increases at $9\ \text{cm}^3\ \text{s}^{-1}$. How fast is its surface area increasing when an edge is 10 cm?

Common wrong step. Writing $\dfrac{dS}{dt} = 12x$ and forgetting the factor $\dfrac{dx}{dt}$. A rate with respect to time always carries the "$\times\,\dfrac{d(\text{variable})}{dt}$" tail.

Marginal cost and marginal revenue. If $C(x)$ is the total cost of making $x$ items, the marginal cost is $\dfrac{dC}{dx}$, and the marginal revenue from $R(x)$ is $\dfrac{dR}{dx}$. Each is the instantaneous rate of change of the total with output. For $R(x) = 3x^2 + 36x + 5$, the marginal revenue is $6x + 36$, which equals $126$ at $x = 15$.

Beyond the textbook. For the exam: marginal cost is defined as $\dfrac{dC}{dx}$ at the current output. In reality: shopkeepers read it as "the extra cost of making one more item". That reading works because $C(x+1) - C(x) \approx C'(x)$ when $x$ is large and $C$ changes smoothly. Why it matters: it explains why businesses compare marginal cost with marginal revenue to decide how much to produce.

Increasing and decreasing functions

A graph that climbs as you move right is increasing, and one that falls is decreasing. The precise definitions use two points $x_1 < x_2$ of an interval $I$.

Why the sign of $f'$ decides everything. The derivative is the slope of the tangent. Between any two points of a smooth graph there is a point where the tangent is parallel to the line joining them. If every tangent slopes upward, that line slopes upward too, so the graph must rise. The same reasoning works for downward slopes and for flat ones.

So, for $f$ continuous on $[a, b]$ and differentiable on $(a, b)$:

$$f'(x) > 0 \text{ on } (a,b) \Rightarrow f \text{ increasing}, \qquad f'(x) < 0 \text{ on } (a,b) \Rightarrow f \text{ decreasing}, \qquad f'(x) = 0 \text{ on } (a,b) \Rightarrow f \text{ constant}$$

The method in four steps: differentiate → solve $f'(x) = 0$ → mark the roots on a number line, which splits it into intervals → read the sign of $f'$ in each interval.

Worked example: a cubic with two turns. Find where $f(x) = 4x^3 - 6x^2 - 72x + 30$ is increasing and decreasing.

Common wrong step. Writing "increasing on $(-\infty,-2)\cup(3,\infty)$" as one set. A function is increasing on an interval, so the two pieces are stated separately. A union of two intervals is not an interval, and a function can be higher on the left piece than on the right one.

Sign of a quadratic in one line. For $a(x-p)(x-q)$ with $a > 0$ and $p < q$, the product is positive outside the roots and negative between them. If $a < 0$ the pattern flips.

Beyond the textbook. For the exam: $f'(x) > 0$ gives an increasing function. In reality: the condition is sufficient but not necessary. The function $f(x) = x^3$ has $f'(0) = 0$, yet it is strictly increasing on all of $\mathbb{R}$, because the slope is zero at only one point and positive everywhere else. Why it matters: a statement like "if $f$ is increasing then $f'>0$ everywhere" is false, and NTA uses this in statement-based questions.

Maxima and minima: the turning points

A function has a maximum at $c$ if $f(c) \ge f(x)$ for every $x$ in its interval, and a minimum if $f(c) \le f(x)$. The word local shrinks the demand to a small neighbourhood: $c$ is a local maximum if $f(c) \ge f(x)$ for all $x$ in some interval $(c - h, c + h)$. A local peak need not be the highest point overall.

Why $f'(c) = 0$ at a smooth turning point. Just before a peak the graph rises (positive slope) and just after it falls (negative slope). A smooth slope cannot jump from positive to negative without passing through zero. So at a smooth peak or valley the tangent is horizontal.

A critical point is a point $c$ in the domain where $f'(c) = 0$ or $f$ is not differentiable at $c$. Every local extremum is a critical point, but the converse is false: $f(x) = x^3$ has $f'(0) = 0$ and no extremum. Such a point, where $f'$ is zero but does not change sign, is a point of inflection.

Two tests classify a critical point.

Worked example: three critical points. Find the local extrema of $f(x) = 3x^4 + 4x^3 - 12x^2 + 12$.

Common wrong step. Stopping at "$x = 0$ is the maximum" and quoting $f''(0)$ as the answer. The question usually asks for the extreme value, which means substituting the point back into $f$ itself.

Beyond the textbook. For the exam: if $f'(c) = 0$ and $f''(c) = 0$ the second derivative test gives no answer. In reality: "no answer" does not mean "no extremum". The function $f(x) = x^4$ has $f'(0) = 0$ and $f''(0) = 0$, and $0$ is clearly a minimum, because $x^4 \ge 0$ everywhere. The first derivative test settles it, since $f'(x) = 4x^3$ changes from negative to positive at $0$. Why it matters: a failed test sends you back to the sign of $f'$, never to a conclusion.

Why the second derivative test works. $f''$ is the rate of change of the slope. If $f'(c) = 0$ and $f''(c) > 0$, the slope is zero at $c$ and rising, so it must be negative just before and positive just after. That is the "$-$ to $+$" pattern of a valley. If $f''(c) < 0$ the slope is falling through zero, which gives a hilltop.

Pause and predict: is $x = 0$ a local maximum, a local minimum or an inflection point for $f(x) = x^3$? And for $f(x) = x^4$?

Check your answer: inflection for $x^3$, because $f'(x) = 3x^2$ is positive on both sides. Local minimum for $x^4$, because $f'(x) = 4x^3$ is negative on the left and positive on the right. Both have $f'(0) = 0$ and $f''(0) = 0$.

Absolute maximum and minimum on a closed interval

A continuous function on $[a, b]$ always attains an absolute maximum and an absolute minimum somewhere on that interval. They can sit at a turning point or at an end, and at an end the derivative need not be zero. That is why checking only $f'(x) = 0$ is not enough.

The working rule: (1) find the critical points inside $(a, b)$; (2) add the two endpoints $a$ and $b$; (3) evaluate $f$ at every one of these points; (4) the largest value is the absolute maximum and the smallest is the absolute minimum.

A function that is purely increasing or purely decreasing has no turning points at all, so its extremes sit at the endpoints.

Worked example: a cubic on $[1, 5]$. Find the absolute maximum and minimum of $f(x) = 2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$.

Common wrong step. Comparing only the critical-point values $29$ and $28$ and calling $29$ the maximum. The true maximum is at the endpoint $x = 5$, where $f'(5) = 6 \cdot 3 \cdot 2 = 36 \neq 0$.

Optimisation: the best possible value

Word problems that ask for "the greatest area", "the least cost" or "the maximum volume" all follow the same recipe. Write the quantity to be optimised as a function of one variable, find the domain, find the critical points and check which one gives the maximum or minimum.

Worked example: the open box. A square tin sheet of side $24$ cm is made into an open box by cutting a square of side $x$ cm from each corner and folding up the sides. For what $x$ is the volume greatest, and what is that volume?

Common wrong step. Keeping both roots. At $x = 12$ the flaps eat the whole sheet and the volume is zero, which is a minimum, not a maximum. The domain filter in step 3 of the recipe is there for this reason.

Pause and predict: a wire $40$ cm long is bent into a rectangle. Which shape has the greatest area?

Check your answer: a square of side $10$ cm. If one side is $x$, the other is $20 - x$, so $A = x(20-x)$, $A' = 20 - 2x = 0$ gives $x = 10$, and $A'' = -2 < 0$ confirms the maximum area $100\ \text{cm}^2$.

Beyond the textbook. For the exam: optimisation problems are solved by $\dfrac{dQ}{dx} = 0$ followed by a check. In reality: the same idea runs the training of machine learning models. A model's error is a function of many numbers, and the computer repeatedly moves them in the direction in which the derivative says the error falls, until the slope is close to zero. Why it matters: "set the derivative to zero" is the single most reused idea in applied mathematics.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Continuity and Differentiability gives the derivative rules, the chain rule and the second derivative used throughout.
• Next: Integrals reverses the process, recovering a function from its rate of change.
• Earlier idea reused: Limits and Derivatives introduced the derivative as a slope and as a limit of average rates.
• Same family of ideas: Application of Integrals uses the same curve sketching habits, and Linear Programming is another optimisation topic.