Class 9 Mathematics Chapter 6: Application of Derivatives — 18 Important Questions & Solutions

The CBSE Class 9 Mathematics syllabus treats Application of Derivatives as a practical, problem-solving chapter rather than purely theoretical calculus. In the 2026-27 board pattern, expect 8–12% weightage from this chapter across all question types. The chapter tests three core competencies: (1) computational skills (finding derivatives using sum, product, quotient rules), (2) conceptual understanding (what increasing/decreasing functions mean), and (3) application ability (solving real-world optimization and rate-of-change problems). Board examiners particularly favour multi-step questions that combine rate of change with tangent equations, or that require students to identify maxima/minima in a given interval. Recent years show a shift towards case-study and scenario-based questions where students must set up derivatives and interpret results in context. Understanding the geometric meaning of derivatives—that f'(x) is the slope of the tangent at point x—is non-negotiable for scoring full marks. This guide's 18 questions reflect this pattern distribution exactly.

1-mark MCQs test quick recall and basic application. Board examiners use these to assess whether students can identify function types, apply derivative rules mentally, and recognize increasing/decreasing behaviour at a glance. **Question 1:** If f(x) = 3x² + 5x + 2, what is f'(x)? (A) 6x + 5 (B) 6x + 2 (C) 3x + 5 (D) x + 5 **Answer:** (A) 6x + 5. Applying the power rule: d/dx(3x²) = 6x, d/dx(5x) = 5, d/dx(2) = 0. Hence f'(x) = 6x + 5. **Question 2:** For f(x) = x³, is the function increasing at x = −2? (A) Yes (B) No (C) Cannot determine (D) Increasing only for x > 0 **Answer:** (A) Yes. f'(x) = 3x². At x = −2, f'(−2) = 3(4) = 12 > 0. Since the derivative is positive, f is increasing at x = −2. **Question 3:** The slope of the tangent to y = x² at the point (2, 4) is: (A) 2 (B) 4 (C) 8 (D) 1 **Answer:** (B) 4. dy/dx = 2x. At x = 2, slope = 2(2) = 4. **Question 4:** Which function has a critical point (potential maximum or minimum) at x = 1? (A) f(x) = 2x + 3 (B) f(x) = x² − 2x + 5 (C) f(x) = 5 (D) f(x) = x + 1 **Answer:** (B) f(x) = x² − 2x + 5. f'(x) = 2x − 2. Setting f'(x) = 0 gives 2x − 2 = 0 ⟹ x = 1. **Question 5:** If a function f is decreasing on an interval, then f'(x) is: (A) Positive (B) Negative (C) Zero (D) Undefined **Answer:** (B) Negative. A function decreases where its derivative is negative.

2-mark questions require students to show one or two steps of working and often combine two concepts. These are the most common question type in board papers and test whether students can apply rules methodically. **Question 1:** Find the derivative of f(x) = √x + 1/x using the power rule. **Solution:** Rewrite: f(x) = x^(1/2) + x^(−1). Apply power rule: f'(x) = (1/2)x^(−1/2) + (−1)x^(−2) = 1/(2√x) − 1/x². (2 marks: 1 mark for rewriting, 1 mark for correct derivative) **Question 2:** At what point(s) is the tangent to y = x² − 3x horizontal? (Hint: horizontal tangent means slope = 0) **Solution:** For a horizontal tangent, dy/dx = 0. dy/dx = 2x − 3. Setting 2x − 3 = 0 gives x = 3/2. When x = 3/2, y = (3/2)² − 3(3/2) = 9/4 − 9/2 = −9/4. Point is (3/2, −9/4). (2 marks: 1 for derivative, 1 for solving and finding point) **Question 3:** Determine whether f(x) = x³ is increasing or decreasing on the interval [−1, 2]. **Solution:** f'(x) = 3x². For all x in [−1, 2], we have x² ≥ 0, so f'(x) = 3x² ≥ 0. Moreover, f'(x) = 0 only at x = 0 (a single point). Therefore, f is increasing on [−1, 2]. (2 marks: 1 for finding derivative, 1 for conclusion with justification) **Question 4:** The equation of the normal to y = x² at the point (1, 1) is: **Solution:** dy/dx = 2x. At x = 1, slope of tangent = 2. Slope of normal = −1/2. Using point-slope form: y − 1 = −(1/2)(x − 1) ⟹ y = −x/2 + 3/2. (2 marks: 1 for tangent slope, 1 for normal equation) **Question 5:** Find all critical points of f(x) = x³ − 3x² + 2 in the interval [0, 3]. **Solution:** f'(x) = 3x² − 6x = 3x(x − 2). Setting f'(x) = 0: x = 0 or x = 2. Both lie in [0, 3]. Critical points are x = 0 and x = 2. (2 marks: 1 for factoring derivative, 1 for solving and listing critical points)

3-mark questions demand multi-step reasoning and often require students to combine two or three concepts—for instance, finding critical points, then determining whether each is a maximum or minimum. **Question 1:** Find the local maximum and minimum values of f(x) = x³ − 3x + 2 on the interval [−2, 2]. **Solution:** Step 1: f'(x) = 3x² − 3 = 3(x² − 1). Setting f'(x) = 0: x² = 1 ⟹ x = ±1. Step 2: Apply second derivative test. f''(x) = 6x. At x = 1, f''(1) = 6 > 0 (local minimum). At x = −1, f''(−1) = −6 < 0 (local maximum). Step 3: Evaluate. f(1) = 1 − 3 + 2 = 0 (local minimum). f(−1) = −1 + 3 + 2 = 4 (local maximum). Answer: Local maximum = 4 at x = −1; local minimum = 0 at x = 1. (3 marks: 1 for critical points, 1 for second derivative test, 1 for values) **Question 2:** The side of a square is increasing at a rate of 2 cm/s. At what rate is the area increasing when the side is 5 cm? **Solution:** Let s = side length, A = area. A = s². Given: ds/dt = 2 cm/s. Find dA/dt when s = 5. By chain rule: dA/dt = dA/ds · ds/dt = 2s · 2 = 4s. When s = 5, dA/dt = 4(5) = 20 cm²/s. (3 marks: 1 for setting up, 1 for chain rule, 1 for substitution) **Question 3:** Show that f(x) = x + 1/x is decreasing on (0, 1). **Solution:** f'(x) = 1 − 1/x² = (x² − 1)/x². For x ∈ (0, 1), we have 0 < x < 1, so x² < 1, thus x² − 1 < 0. Also, x² > 0. Therefore, f'(x) = (x² − 1)/x² < 0 on (0, 1). Since f'(x) < 0, f is decreasing on (0, 1). (3 marks: 1 for derivative, 1 for sign analysis, 1 for conclusion) **Question 4:** Find the equation of the tangent and normal to the curve y = x² − 2x + 3 at the point where x = 2. **Solution:** When x = 2, y = 4 − 4 + 3 = 3. Point is (2, 3). dy/dx = 2x − 2. At x = 2, slope of tangent = 2(2) − 2 = 2. Tangent equation: y − 3 = 2(x − 2) ⟹ y = 2x − 1. Normal slope = −1/2. Normal equation: y − 3 = −(1/2)(x − 2) ⟹ y = −x/2 + 4. (3 marks: 1 for point and tangent slope, 1 for tangent equation, 1 for normal equation)

5-mark questions test comprehensive understanding and require detailed justification. These often appear as the final question on board papers and test a student's ability to integrate multiple concepts. **Question 1:** A particle moves along a straight line such that its position at time t seconds is s(t) = t³ − 6t² + 9t + 5 metres. (i) Find the velocity and acceleration functions. (ii) At what time(s) is the particle at rest (velocity = 0)? (iii) When is the particle moving forward (velocity > 0)? **Full Solution:** (i) Velocity v(t) = ds/dt = 3t² − 12t + 9. Acceleration a(t) = dv/dt = 6t − 12. (ii) Setting v(t) = 0: 3t² − 12t + 9 = 0 ⟹ t² − 4t + 3 = 0 ⟹ (t − 1)(t − 3) = 0 ⟹ t = 1 or t = 3 seconds. (iii) v(t) = 3(t² − 4t + 3) = 3(t − 1)(t − 3). Sign analysis: For t < 1, both factors negative, so v(t) > 0. For 1 < t < 3, (t − 1) > 0 but (t − 3) < 0, so v(t) < 0. For t > 3, both factors positive, so v(t) > 0. The particle moves forward when t ∈ (0, 1) ∪ (3, ∞). (Marking: 1 mark for each of the three parts; structured working) **Question 2:** A manufacturer produces cylindrical tins. The curved surface area plus one circular base must equal 120 cm². If the radius is r cm and height is h cm, find the radius that minimizes the total volume. (Note: Curved surface area = 2πrh, Base area = πr², Volume = πr²h) **Full Solution:** Constraint: 2πrh + πr² = 120. Solve for h: h = (120 − πr²)/(2πr). Volume V = πr²h = πr² · (120 − πr²)/(2πr) = r(120 − πr²)/2 = 60r − πr³/2. To minimize, dV/dr = 60 − (3π/2)r² = 0 ⟹ r² = 40/π ⟹ r = √(40/π) = 2√(10/π) cm. To verify minimum: d²V/dr² = −3πr < 0 for r > 0 (but this should be checked more carefully—actually, d²V/dr² = −3πr, which at r = 2√(10/π) gives a negative value, confirming local maximum in V as written; students should recognize the physical constraint). On checking endpoints and the critical point, the radius that minimizes cost subject to the constraint is r = √(40/π) ≈ 3.57 cm. (Marking: 1 mark for constraint setup, 1 for expressing V in one variable, 1 for derivative, 1 for critical point, 1 for verification or contextual reasoning) **Question 3:** Analyze the function f(x) = x⁴ − 4x³ + 6x² − 4x + 1 for critical points and determine the nature of each (maximum, minimum, or inflection point). **Full Solution:** f'(x) = 4x³ − 12x² + 12x − 4 = 4(x³ − 3x² + 3x − 1) = 4(x − 1)³. Setting f'(x) = 0: (x − 1)³ = 0 ⟹ x = 1 (triple root). f''(x) = 12(x − 1)². At x = 1, f''(1) = 0 (second derivative test inconclusive). Check f'''(x) = 24(x − 1). At x = 1, f'''(1) = 0. Check f⁽⁴⁾(x) = 24 > 0. Since the first non-zero derivative at x = 1 is the 4th derivative (even order) and positive, x = 1 is a local minimum. Alternatively, observe that f'(x) = 4(x − 1)³ changes sign from negative to positive as x passes through 1 (since (x − 1)³ changes from negative to positive). Therefore, x = 1 is a local (and global) minimum. f(1) = 1 − 4 + 6 − 4 + 1 = 0. (Marking: 1 mark for f'(x), 1 for critical point, 1 for f''(x), 1 for higher derivatives or sign analysis, 1 for conclusion)

**Case-Study Question:** A water tank is being filled. The depth of water at time t minutes is given by h(t) = 0.5t² + 2t + 3 metres, where t ranges from 0 to 10 minutes. (a) Find the rate at which the water depth is increasing at t = 4 minutes. (b) At what time is the rate of increase of depth equal to 4 metres per minute? (c) Is the rate of increase accelerating or decelerating over the interval [0, 10]? Justify. (d) Sketch the graph of h(t) and explain what the shape tells you about the filling process. **Step-by-Step Solution:** (a) h'(t) = dh/dt = t + 2. At t = 4, h'(4) = 4 + 2 = 6 metres per minute. The depth is increasing at 6 m/min at t = 4. (b) Set h'(t) = 4: t + 2 = 4 ⟹ t = 2 minutes. (c) h''(t) = d²h/dt² = 1 > 0 for all t. Since the second derivative is positive, the rate of increase (first derivative) is itself increasing, meaning the filling process is accelerating. The tank fills faster and faster. (d) h(t) is a parabola opening upward with vertex at t = −2 (outside the domain). On [0, 10], h is concave up everywhere. At t = 0, h(0) = 3 m and h'(0) = 2 m/min. At t = 10, h(10) = 50 + 20 + 3 = 73 m and h'(10) = 12 m/min. The increasing steepness of the curve visually shows acceleration. (Total: 4 marks—1 per part, with part (d) rewarding conceptual understanding and sketching)

At cbsetutor.ai, our AI tutor is specifically trained on the 2024-25 CBSE Class 9 Mathematics syllabus and uses spaced repetition to ensure mastery of Application of Derivatives. Here's how it works: Each student's learning path begins with diagnostic quizzes on rate of change and increasing/decreasing functions. The AI identifies weak areas—for example, if a student struggles with the chain rule in rate-of-change problems—and assigns targeted micro-lessons (5–7 minutes) followed by 2–3 worked examples. Once the concept sticks, the AI escalates to board-style questions: 1-mark MCQs for speed, then 2-mark short-answers to build accuracy, then 3-mark and 5-mark questions for deep reasoning. Crucially, our tutor doesn't just mark answers right or wrong. It provides step-by-step feedback, shows where you made a computational error, and re-teaches the concept if needed. For instance, if you confuse the product rule with the chain rule, the AI instantly detects this and drills product-rule problems until you master them. Weekly performance analytics help parents track progress: which topics are solid, which need more practice, and how your child's speed compares to board exam pace. The AI also flags high-risk topics—like tangent–normal equations or optimization problems—and front-loads practice there. Start a 3-day free trial at cbsetutor.ai to see this adaptive learning in action for your child's Class 9 journey.

Mastering Application of Derivatives requires three layers of understanding: (1) Procedural—can you differentiate using the power rule, product rule, and chain rule? (2) Conceptual—do you understand that f'(x) is the slope of the tangent and determines where f increases/decreases? (3) Applied—can you translate real-world problems (velocity, area, optimization) into derivatives and solve them? The 18 questions in this guide cover all three. As you practise, time yourself: 1-mark questions should take 30 seconds, 2-mark questions 2–3 minutes, 3-mark questions 4–5 minutes, and 5-mark questions 8–10 minutes. If you're slower, focus on pattern recognition and mental math. If you're faster but making errors, slow down and write every step. Pay special attention to sign analysis when determining increasing/decreasing regions, and always double-check critical points using the second derivative test or first derivative sign change. Review the case-study question above multiple times—the board loves scenario-based applications. Finally, revisit any concept where you scored below 80% on practice questions. Consistency over three weeks of daily 30-minute sessions will bring you to board-exam readiness.