Application Of Derivatives — Complete Guide with Solved Examples

What Application of Derivatives Actually Means

Here’s the thing most textbooks miss: derivatives were invented to solve real problems — finding how fast something changes, where it peaks, and what shape a curve takes. When we move from “finding the derivative” to “applying” it, we’re finally doing the interesting part.

Class 12 boards give this chapter roughly 8-10 marks every year. JEE Main? Expect 2-3 questions, with at least one from maxima-minima that looks simple but has a hidden catch. This is a chapter where students who understand the why behind each method comfortably score full marks, while students who memorise steps lose marks on application-based twists.

We’ll cover four major applications: rate of change, tangents and normals, increasing/decreasing functions, and maxima-minima. Every section has the worked examples and the exact mistakes that cost students marks.

Key Terms & Definitions

Derivative as Rate of Change: If $y = f(x)$ , then $\frac{dy}{dx}$ tells us how fast $y$ changes with respect to $x$ . If $x$ is time, $\frac{dy}{dt}$ is the rate of change over time.

Tangent at a Point: The straight line that just touches a curve at a point, with slope equal to $f'(x_0)$ at that point $x_0$ .

Normal at a Point: The line perpendicular to the tangent at the same point. If tangent slope is $m$ , normal slope is $-\frac{1}{m}$ .

Increasing Function: $f$ is increasing on $(a, b)$ if $f'(x) > 0$ for all $x \in (a, b)$ . The curve goes upward as $x$ increases.

Decreasing Function: $f'(x) < 0$ for all $x$ in the interval. Curve goes downward.

Critical Point: A point where $f'(x) = 0$ or $f'(x)$ does not exist. These are candidates for maxima or minima — not guaranteed.

Local Maximum: $f(c)$ is a local maximum if $f(c) \geq f(x)$ for all $x$ near $c$ .

Local Minimum: $f(c)$ is a local minimum if $f(c) \leq f(x)$ for all $x$ near $c$ .

Absolute Maximum/Minimum: The largest/smallest value of $f$ over a closed interval $[a, b]$ .

Methods & Concepts

1. Rate of Change

The idea: express one quantity as a function of another, then differentiate.

Standard setup: If the area of a circle is $A = \pi r^2$ , then $\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$ . We multiplied by $\frac{dr}{dt}$ using chain rule — this connects two rates.

Always write down what you’re differentiating with respect to before you start. Most errors in rate-of-change problems happen because students forget to apply the chain rule when the variable isn’t $x$ .

Worked Example (CBSE Level): A balloon is being inflated so its radius increases at 2 cm/s. Find the rate of increase of its volume when radius = 5 cm.

V = \frac{4}{3}\pi r^3

\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} = 4\pi (5)^2 \cdot 2 = 200\pi \text{ cm}^3/\text{s}

2. Tangents and Normals

Slope of tangent at $(x_1, y_1)$ : $m_T = f'(x_1)$

Equation of tangent: $y - y_1 = m_T(x - x_1)$

Slope of normal: $m_N = -\frac{1}{m_T}$ (when $m_T \neq 0$ )

Equation of normal: $y - y_1 = m_N(x - x_1)$

CBSE marking scheme: 1 mark for finding slope, 1 mark for substituting the point, 1 mark for the final equation. Even if your final answer is wrong, you get 2/3 if the method is right.

Special cases to know:

If $f'(x_1) = 0$ : tangent is horizontal ( $y = y_1$ ), normal is vertical ( $x = x_1$ )
If $f'(x_1)$ is undefined (vertical tangent): tangent is $x = x_1$ , normal is $y = y_1$

3. Increasing and Decreasing Functions

Method:

Find $f'(x)$
Set $f'(x) = 0$ , find critical points
Make a sign chart: test $f'(x)$ in each interval
Where $f'(x) > 0$ → increasing; where $f'(x) < 0$ → decreasing

Worked Example: Find intervals where $f(x) = x^3 - 6x^2 + 9x + 5$ is increasing.

f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)

Sign chart:

$x < 1$ : both factors negative → product positive → $f'(x) > 0$ → increasing
$1 < x < 3$ : $(x-1) > 0$ , $(x-3) < 0$ → product negative → $f'(x) < 0$ → decreasing
$x > 3$ : both factors positive → $f'(x) > 0$ → increasing

Answer: Increasing on $(-\infty, 1) \cup (3, \infty)$ , decreasing on $(1, 3)$ .

4. Maxima and Minima

We have three tests. Know when to use which.

At a critical point $x = c$ where $f'(c) = 0$ :

$f'$ changes positive → negative: local maximum
$f'$ changes negative → positive: local minimum
$f'$ doesn’t change sign: neither (point of inflection)

At critical point $c$ where $f'(c) = 0$ :

f''(c) < 0 \Rightarrow \text{local maximum}

f''(c) > 0 \Rightarrow \text{local minimum}

$f''(c) = 0 \Rightarrow \text{test fails --- use first derivative test}$

Absolute extrema on closed interval $[a, b]$ :

Find all critical points inside $(a, b)$
Evaluate $f$ at critical points AND at endpoints $a$ , $b$
Largest value = absolute maximum; smallest = absolute minimum

Solved Examples

Example 1 — Easy (CBSE Board Pattern)

Problem: Find the point on the curve $y = x^2$ where the tangent is parallel to the line $y = 4x - 3$ .

Solution: Parallel lines have equal slopes. Slope of given line = 4.

\frac{dy}{dx} = 2x = 4 \implies x = 2

When $x = 2$ : $y = 4$ . Point is $(2, 4)$ .

Example 2 — Medium (JEE Main Pattern)

Problem: Find the maximum and minimum values of $f(x) = \sin x + \frac{1}{2}\cos 2x$ on $[0, \frac{\pi}{2}]$ .

Solution:

f'(x) = \cos x - \sin 2x = \cos x - 2\sin x \cos x = \cos x(1 - 2\sin x)

Setting $f'(x) = 0$ : either $\cos x = 0 \Rightarrow x = \frac{\pi}{2}$ , or $\sin x = \frac{1}{2} \Rightarrow x = \frac{\pi}{6}$ .

Now evaluate at critical points and endpoints:

Point	$f(x)$
$x = 0$	$0 + \frac{1}{2} = \frac{1}{2}$
$x = \frac{\pi}{6}$	$\frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} = \frac{3}{4}$
$x = \frac{\pi}{2}$	$1 + \frac{1}{2}(-1) = \frac{1}{2}$

Absolute maximum = $\frac{3}{4}$ at $x = \frac{\pi}{6}$ . Absolute minimum = $\frac{1}{2}$ at $x = 0$ and $x = \frac{\pi}{2}$ .

Example 3 — Hard (JEE Advanced Style)

Problem: A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter is 10 m, find the dimensions that maximise the area.

Solution: Let the rectangle have width $2r$ (so semicircle has radius $r$ ) and height $h$ .

Perimeter: $2h + 2r + \pi r = 10 \Rightarrow h = \frac{10 - 2r - \pi r}{2} = 5 - r - \frac{\pi r}{2}$

Area: $A = 2rh + \frac{\pi r^2}{2}$

Substituting $h$ :

A = 2r\left(5 - r - \frac{\pi r}{2}\right) + \frac{\pi r^2}{2} = 10r - 2r^2 - \pi r^2 + \frac{\pi r^2}{2} = 10r - 2r^2 - \frac{\pi r^2}{2}

\frac{dA}{dr} = 10 - 4r - \pi r = 10 - r(4 + \pi) = 0

r = \frac{10}{4 + \pi}

\frac{d^2A}{dr^2} = -(4 + \pi) < 0 \Rightarrow \text{maximum confirmed.}

Width $= \frac{20}{4+\pi}$ m, Height $= \frac{10}{4+\pi}$ m.

Optimisation word problems follow a fixed pattern: set up two equations (one constraint, one to optimise), eliminate one variable, differentiate, solve. Practice recognising which equation is which — that’s the actual skill being tested.

Exam-Specific Tips

CBSE Class 12 Boards

The chapter carries 8 marks typically distributed as:

1 short question (2 marks): tangent/normal or rate of change
1 long question (5-6 marks): maxima-minima or increasing/decreasing with proof

Always state the second derivative test result explicitly — “since $f''(c) < 0$ , the function has a local maximum at $x = c$ ”. Skipping this costs 1 mark.

JEE Main

JEE Main 2024 and 2023 both had a maxima-minima question where the critical point was a fraction. The trap: students set up the function correctly but made arithmetic errors solving $f'(x) = 0$ . Slow down at that step.

Weightage: roughly 2-3 questions per paper, mix of tangent-normal and optimisation. Rate of change appears less frequently but is easy marks when it does appear.

JEE Advanced

Expect the unexpected: functions defined piecewise, checking differentiability before applying tests, or optimisation with two variables and Lagrange-style constraints. The second derivative test failing ( $f''(c) = 0$ ) and needing higher-order analysis appears at this level.

Common Mistakes to Avoid

Mistake 1: Forgetting to check endpoints in absolute extrema problems.
The absolute maximum/minimum may not be at a critical point — it could be at $x = a$ or $x = b$ . Always evaluate all candidates.

Mistake 2: Confusing local and absolute extrema.
A local maximum is the highest point nearby, not necessarily the highest overall. When the problem says “maximum value on $[a, b]$ ”, it wants absolute maximum.

Mistake 3: Applying second derivative test when $f''(c) = 0$ .
When $f''(c) = 0$ , the second derivative test is inconclusive. You must use the first derivative test. Students who blindly write “neither maximum nor minimum” here are wrong — it could still be a maximum or minimum.

Mistake 4: Getting the normal slope wrong.
Normal slope is $-\frac{1}{m}$ , not $\frac{1}{m}$ . The negative sign is missed under time pressure. Write it out explicitly every time.

Mistake 5: Using degrees instead of radians in trigonometric rate-of-change problems.
If $\theta$ is in degrees, $\frac{d}{d\theta}(\sin\theta) \neq \cos\theta$ . Calculus only works cleanly with radians. Check the unit in the problem.

Practice Questions

Q1. Find the rate of change of the area of a circle with respect to its radius when $r = 6$ cm.

$A = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r$ . At $r = 6$ : $\frac{dA}{dr} = 12\pi$ cm²/cm.

Q2. Find the equation of the tangent to $y = \sqrt{3x - 2}$ at the point where $x = 3$ .

At $x = 3$ : $y = \sqrt{7}$ . $\frac{dy}{dx} = \frac{3}{2\sqrt{3x-2}}$ . At $x=3$ : slope $= \frac{3}{2\sqrt{7}}$ .

Tangent: $y - \sqrt{7} = \frac{3}{2\sqrt{7}}(x - 3)$ , which simplifies to $2\sqrt{7}y - 14 = 3x - 9$ , i.e., $3x - 2\sqrt{7}y + 5 = 0$ .

Q3. Find the intervals where $f(x) = 2x^3 - 9x^2 + 12x - 5$ is increasing.

$f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$ .

$f'(x) > 0$ when $x < 1$ or $x > 2$ . Increasing on $(-\infty, 1) \cup (2, \infty)$ .

Q4. Find local maxima and minima of $f(x) = x^3 - 3x$ .

$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$ .

Critical points: $x = 1, -1$ .

$f''(x) = 6x$ .

$f''(-1) = -6 < 0$ : local maximum at $x = -1$ , value $= 2$ .

$f''(1) = 6 > 0$ : local minimum at $x = 1$ , value $= -2$ .

Q5. A wire of length 28 m is cut into two pieces. One piece is bent into a square, the other into a circle. Find the lengths that minimise the total area.

Let side of square $= a$ , so $4a + 2\pi r = 28 \Rightarrow r = \frac{14 - 2a}{\pi}$ .

Total area $= a^2 + \pi r^2 = a^2 + \frac{(14-2a)^2}{\pi}$ .

$\frac{dA}{da} = 2a - \frac{4(14-2a)}{\pi} = 0 \Rightarrow a = \frac{28}{\pi + 4}$ .

Length for square $= 4a = \frac{112}{\pi + 4}$ m, remainder for circle $= 28 - \frac{112}{\pi+4} = \frac{28\pi}{\pi+4}$ m.

Q6. Verify Rolle’s theorem for $f(x) = x^2 - 4$ on $[-2, 2]$ .

$f(-2) = 0 = f(2)$ . $f$ is continuous on $[-2, 2]$ , differentiable on $(-2, 2)$ .

$f'(x) = 2x = 0 \Rightarrow x = 0 \in (-2, 2)$ . Verified.

Q7. The sum of two positive numbers is 16. Find the numbers so their product is maximum.

Let numbers be $x$ and $16 - x$ . Product $P = x(16-x) = 16x - x^2$ .

$\frac{dP}{dx} = 16 - 2x = 0 \Rightarrow x = 8$ .

$\frac{d^2P}{dx^2} = -2 < 0$ : maximum at $x = 8$ .

Both numbers are 8.

Q8. Find the point on the curve $y^2 = 4x$ closest to the point $(2, 1)$ .

Any point on curve: $(t^2, 2t)$ . Distance² $= D = (t^2 - 2)^2 + (2t - 1)^2$ .

$\frac{dD}{dt} = 2(t^2 - 2)(2t) + 2(2t-1)(2) = 4t^3 - 8t + 8t - 4 = 4t^3 - 4 = 0$ .

$t = 1$ . Point: $(1, 2)$ .

FAQs

What is the difference between local maxima and absolute maxima?

Local maxima is the highest value in a neighbourhood — points nearby are lower. Absolute maxima is the highest value over the entire domain or interval. On a closed interval $[a, b]$ , the absolute maximum could be at a local maximum or at one of the endpoints.

How do I know when to use first vs second derivative test?

Use the second derivative test when $f''(c)$ is easy to compute and clearly non-zero. Use the first derivative test when $f''(c) = 0$ (second test fails), when $f''$ is hard to compute, or when you need to show sign change explicitly. For board exams, either is accepted.

Why does the second derivative test fail when $f''(c) = 0$ ?

Because the second derivative’s sign tells us about concavity at that point. When $f''(c) = 0$ , the concavity is ambiguous — the function could still curve up, curve down, or have a point of inflection. We need to check actual function values nearby to decide.

What is a point of inflection and how is it different from a critical point?

A critical point has $f'(x) = 0$ or undefined — it’s a candidate for maxima/minima. A point of inflection is where the curve changes concavity (from concave up to concave down or vice versa) — it’s where $f''(x) = 0$ and the sign of $f''$ actually changes. A point can be both, but usually isn’t.

How do I set up optimisation problems from word problems?

Identify: (1) what to maximise/minimise (your objective function), (2) what constraint you’re given. Write both as equations, use the constraint to reduce your objective to one variable, then differentiate. The hardest part is usually the setup — the calculus is straightforward.

Do I need to verify that a critical point is actually a maximum/minimum?

Yes — always. In CBSE papers this is required for full marks. In JEE, skipping this verification costs you in multi-part questions. The minimum you need: state whether $f''(c) < 0$ (maximum) or $f''(c) > 0$ (minimum).

What happens when $f'(x)$ doesn’t exist at a point?

Points where $f'(x)$ is undefined are also critical points and must be checked. For example, $f(x) = |x|$ has a minimum at $x = 0$ even though $f'(0)$ doesn’t exist. Use the first derivative test (checking sign change from left to right) to classify these.

Which topic within Application of Derivatives has the highest weightage in JEE Main?

Maxima-minima consistently carries the highest weightage — expect one direct question every paper. Increasing/decreasing functions and tangent-normal are the next most frequent. Rate of change appears less often but is typically straightforward marks when it does appear.