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NDA Applications of Derivatives

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This page covers NDA Applications of Derivatives with complete concept notes, 43 graded practice MCQs, key points and exam-specific tips. Free to study.

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Concept Notes

Applications of Derivatives— Rules & Concept

1.Core Concept

Read this first — the foundation of the topic

APPLICATIONS OF DERIVATIVES --- CORE CONCEPT ---

A derivative tells you the RATE OF CHANGE of a function. When we apply this idea to real problems, we get powerful tools: finding maximum and minimum values, checking if a function is increasing or decreasing, finding tangent lines, and solving rate-of-change problems. NDA regularly tests all of these. --- KEY RULES / PROPERTIES ---

1. INCREASING / DECREASING FUNCTIONS If f'(x) > 0 on an interval, the function is INCREASING there.

If f'(x) < 0 on an interval, the function is DECREASING there. If f'(x) = 0 at a point, that point is called a CRITICAL POINT.

2. MAXIMA AND MINIMA At a maximum or minimum, f'(x) = 0. This is called a stationary point.

First Derivative Test: - If f'(x) changes from + to - at point c → LOCAL MAXIMUM at c

- If f'(x) changes from - to + at point c → LOCAL MINIMUM at c - If f'(x) does NOT change sign → it is a POINT OF INFLECTION

Second Derivative Test: - If f'(c) = 0 and f''(c) < 0 → LOCAL MAXIMUM

- If f'(c) = 0 and f''(c) > 0 → LOCAL MINIMUM - If f'(c) = 0 and f''(c) = 0 → Test FAILS, use first derivative test

3. EQUATION OF TANGENT AND NORMAL Tangent at point (x1, y1): y - y1 = f'(x1) × (x - x1)

Normal is perpendicular to tangent. Slope of normal = -1 / f'(x1)

---

2.Formula Block

Memorise — at least one formula appears in every paper
Slope of tangent = dy/dx at the given point
Slope of normal = -1 / (dy/dx)
For maximum/minimum: set dy/dx = 0, solve for x

--

Use d²y/dx² to confirm max or min

---

3.Exam Patterns

What examiners ask — read before attempting PYQs

--- NDA asks: - Find local max/min of a polynomial or trigonometric function - Check if a function is increasing/decreasing in a given interval - Find the equation of tangent/normal at a given point - Word problems: maximize area, volume, or minimize cost --- SHORTCUT / TRICK --- TRICK 1 (Quick Sign Check): For f(x) = ax^n type functions, just check sign of f'(x) around the critical point. No need to fully solve if options are given — plug in values just left and right of the critical point. TRICK 2 (Sum/Product Optimization): If sum of two numbers is fixed (say S), their PRODUCT is maximum when both numbers are EQUAL (each = S/2). If product is fixed, their sum is minimum when equal. This directly solves many NDA word problems in seconds. ---

4.Worked Example

Solve this step-by-step before moving on
1
Step 1

Find f'(x) f'(x) = 6x² - 18x + 12

2
Step 2

Set f'(x) = 0 6x² - 18x + 12 = 0 x² - 3x + 2 = 0 (x - 1)(x - 2) = 0 So x = 1 and x = 2 are critical points.

3
Step 3

Find f''(x) f''(x) = 12x - 18

4
Step 4

Apply Second Derivative Test At x = 1: f''(1) = 12 - 18 = -6 < 0 → LOCAL MAXIMUM At x = 2: f''(2) = 24 - 18 = +6 > 0 → LOCAL MINIMUM

5
Step 5

Find values f(1) = 2 - 9 + 12 + 1 = 6 → Local Maximum value = 6 f(2) = 16 - 36 + 24 + 1 = 5 → Local Minimum value = 5 Answer: Local max = 6 at x = 1; Local min = 5 at x = 2 ---

5.Exam Traps

Common mistakes students make — avoid these

--- Students confuse LOCAL maxima/minima with GLOBAL maxima/minima. A local maximum is not always the biggest value of the whole function. Always re-read the question — it may ask for the absolute (global) maximum over a closed interval.

In that case, also CHECK THE ENDPOINTS of the interval.

Key Points to Remember

  • If f'(x) > 0 in an interval, the function is increasing in that interval.
  • If f'(x) < 0 in an interval, the function is decreasing in that interval.
  • Critical points are found by setting f'(x) = 0 and solving for x.
  • Second Derivative Test: f''(c) < 0 means local max; f''(c) > 0 means local min at critical point c.
  • Slope of tangent to curve y = f(x) at point (x1, y1) is f'(x1).
  • Slope of normal = -1 divided by slope of tangent (perpendicular relationship).
  • If product of two numbers is fixed, their sum is minimum when both numbers are equal.
  • If f'(c) = 0 and f''(c) = 0, the second derivative test fails — use the first derivative test instead.

Exam-Specific Tips

  • At a local maximum or minimum, the first derivative f'(x) equals zero — this is the necessary condition.
  • f''(c) < 0 at a critical point confirms a LOCAL MAXIMUM; f''(c) > 0 confirms a LOCAL MINIMUM.
  • The slope of the normal to a curve is -1/f'(x), which is the negative reciprocal of the tangent slope.
  • A point where f'(x) = 0 but f'(x) does NOT change sign is called a Point of Inflection, not a max or min.
  • For a closed interval [a, b], global maximum/minimum must be checked at critical points AND at endpoints x = a and x = b.
  • If the sum of two positive numbers is S, their product is maximised when each number equals S/2, giving maximum product = S²/4.
  • The equation of tangent at (x1, y1) is: y - y1 = f'(x1)(x - x1).
Practice MCQs

Applications of Derivatives — Practice Questions

43graded MCQs · easy to hard · full solution & trap analysis · showing 20 of 43

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Practice 1easy

A rectangular box with an open top is to be constructed from a square sheet of cardboard of side 12 cm by cutting equal squares of side x cm from each corner and folding up the sides. The volume of the box as a function of x is V(x) = x(12 - 2x)². For what value of x is the volume maximum?

Practice 2easy

A ladder of length 10 m leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 m away from the wall?

Practice 3easy

A function f(x) = x⁴ - 8x² + 5 has local extrema at x = a and x = b. The sum a + b equals:

Practice 4easy

The equation of the tangent line to the curve y = x³ - 3x² + 2 at the point where x = 1 is:

Practice 5easy

A rectangular box with an open top is to be constructed from a square sheet of cardboard of side 12 cm by cutting equal squares of side x cm from each corner and folding up the sides. The volume of the box is maximized when x equals:

Practice 6easy

A particle moves along a straight line such that its position at time t is given by s(t) = t³ - 6t² + 9t + 2 metres. At what time(s) is the particle momentarily at rest?

Practice 7easy

A rectangular box with an open top is to be constructed from a square piece of cardboard of side 12 cm by cutting equal squares of side x cm from each corner and folding up the sides. The volume V of the resulting box as a function of x is V(x) = x(12 - 2x)². For what value of x is the volume maximum?

Practice 8easy

The equation of a curve is y = x³ - 3x² + 2x + 5. At what point on the curve is the tangent line parallel to the x-axis?

Practice 9easy

A function f(x) = x⁴ - 4x³ + 6x² - 4x + 1 has a critical point at x = 1. What is the nature of this critical point?

Practice 10easy

A ladder of length 10 m leans against a vertical wall. The bottom of the ladder is being pulled away from the wall at a constant rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 m away from the wall?

Practice 11easy

A ladder of length 10 m leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 1 m/s. When the bottom is 6 m away from the wall, at what rate is the top of the ladder sliding down the wall?

Practice 12easy

A function f(x) = x³ - 3x² - 9x + 5 has local extrema. Which of the following correctly identifies the nature of the critical points?

Practice 13easy

The radius of a circular disc is increasing at a constant rate of 2 cm/s. At the instant when the radius is 5 cm, what is the rate of change of the area of the disc?

Practice 14medium

A rectangular box with an open top is to be made from a square sheet of cardboard of side 12 cm by cutting equal squares of side x cm from each corner and folding up the sides. The volume V of the box as a function of x is V(x) = x(12 − 2x)². For what value of x is the volume maximum?

Practice 15medium

A particle moves along a curve y = x³ − 3x² + 2x. At what point(s) on the curve is the tangent line horizontal?

Practice 16medium

A ladder of length 10 m leans against a vertical wall. The bottom of the ladder slides away from the wall at a constant rate of 2 m/s. How fast is the top of the ladder sliding down the wall when the bottom is 6 m away from the wall?

Practice 17medium

The function f(x) = x³ − 6x² + 9x + 1 has a local maximum at x = a and a local minimum at x = b. What is the value of a + b?

Practice 18medium

A function f(x) = x⁴ − 4x³ + 6x² − 4x + 1 is defined on the interval [0, 2]. What is the absolute maximum value of f on this interval?

Practice 19medium

A particle moves along a curve given by y = x³ − 3x² + 2x. At what point(s) on the curve is the tangent line horizontal?

Practice 20medium

A cylindrical tank with radius r and height h has volume V = πr²h. If the radius increases at 2 cm/s and the height decreases at 3 cm/s, what is the rate of change of volume (in cm³/s) when r = 5 cm and h = 10 cm?

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60-Second Revision — Applications of Derivatives

  • Remember: f'(x) = 0 gives critical points — always the FIRST step for max/min problems.
  • Formula: Second Derivative Test — f''(c) < 0 means MAX, f''(c) > 0 means MIN at critical point c.
  • Formula: Tangent slope = dy/dx at point; Normal slope = -1/(dy/dx) at that point.
  • Trap: Do NOT forget to check endpoints when question says 'on the interval [a, b]' — local max may not be global max.
  • Trick: Fixed sum → max product when both numbers are equal. Fixed product → min sum when both are equal.
  • Remember: If f''(c) = 0, the second derivative test fails — switch to the first derivative sign change test.
  • Trap: A point of inflection has f'(x) = 0 but NO sign change in f'(x) — it is neither max nor min.
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