IB Mathematics: Applications and Interpretation Quiz: Differentiation Rules
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Differentiation RulesQuestion 1 of 20

A curve is described by the equation f(x) = ax³ + bx². The curve has a stationary point at (4, -32).

Find the gradient of the curve at x = 1.

-27
-9
3
9
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Differentiation Rules

Practice Differentiation Rules in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Differentiation Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A curve is described by the equation f(x) = ax³ + bx². The curve has a stationary point at (4, -32).

Find the gradient of the curve at x = 1.

  1. -27
  2. -9 (correct answer)
  3. 3
  4. 9
Explanation: This problem requires setting up and solving a system of equations to find the parameters a and b.
  1. Use the point (4, -32) to form an equation with f(x). f(4) = a(4)³ + b(4)² = -32 64a + 16b = -32. Divide by 16: 4a + b = -2 (Equation 1)
  2. Use the fact that (4, -32) is a stationary point. This means the derivative f'(x) is zero at x = 4. First, find f'(x): f'(x) = 3ax² + 2bx. Now, set f'(4) = 0: f'(4) = 3a(4)² + 2b(4) = 0 48a + 8b = 0. Divide by 8: 6a + b = 0 (Equation 2)
  3. Solve the system of equations. From Eq 2, b = -6a. Substitute this into Eq 1: 4a + (-6a) = -2 -2a = -2 => a = 1. Now find b: b = -6(1) = -6.
  4. The function is f(x) = x³ - 6x², and its derivative is f'(x) = 3x² - 12x.
  5. Find the gradient at x = 1 by evaluating f'(1). f'(1) = 3(1)² - 12(1) = 3 - 12 = -9.

Question 2

The altitude of a drone, A(t) in metres, at time t minutes is given by A(t) = -t³ + 15t² - 48t + 100, for 0 ≤ t ≤ 10.

For what value of t does the drone momentarily stop ascending and begin to descend?

  1. 2.5 minutes
  2. 3.0 minutes
  3. 5.0 minutes
  4. 8.0 minutes (correct answer)
Explanation: The drone stops ascending and begins descending at a local maximum of its altitude function. This occurs when the rate of change of altitude (the vertical velocity) is zero. We need to find the derivative A'(t) and set it to zero. A(t) = -t³ + 15t² - 48t + 100 A'(t) = -3t² + 30t - 48. Set A'(t) = 0 and solve for t: -3t² + 30t - 48 = 0 Divide by -3 to simplify: t² - 10t + 16 = 0 Factor the quadratic equation: (t - 2)(t - 8) = 0. The stationary points are t = 2 and t = 8. A'(t) is a downward-opening parabola, so it is negative before t=2, positive between t=2 and t=8, and negative after t=8. Ascending means A'(t) > 0, descending means A'(t) < 0. The drone stops ascending and starts descending at the point where the derivative changes from positive to negative, which is the local maximum at t = 8.

Question 3

The displacement, s, in metres, of a particle from a fixed point O at time t seconds is given by s(t) = t³ - 12t² + 36t + 5, for t ≥ 0.

Find the instantaneous velocity of the particle when its acceleration is zero.

  1. -12 m/s (correct answer)
  2. 0 m/s
  3. 4 m/s
  4. 37 m/s
Explanation: This is a multi-step problem. Velocity v(t) is the first derivative of displacement s(t), and acceleration a(t) is the second derivative of s(t) (or the first derivative of v(t)).
  1. Find the velocity function, v(t) = s'(t): s(t) = t³ - 12t² + 36t + 5 v(t) = s'(t) = 3t² - 24t + 36.
  2. Find the acceleration function, a(t) = v'(t): a(t) = v'(t) = 6t - 24.
  3. Find the time t when acceleration is zero: Set a(t) = 0: 6t - 24 = 0 => 6t = 24 => t = 4 seconds.
  4. Find the velocity at that time by substituting t = 4 into the velocity function: v(4) = 3(4)² - 24(4) + 36 v(4) = 3(16) - 96 + 36 v(4) = 48 - 96 + 36 = -12 m/s.

Question 4

The height, h, in metres, of the tide in a harbour is modelled by the function h(t) = 6.5 + 2.1cos(π/6 * t), where t is the number of hours after midnight.

At what rate is the tide level changing at 4:00 a.m., to the nearest 0.01 metres per hour?

  1. -0.95 m/hr (correct answer)
  2. -0.55 m/hr
  3. 0.95 m/hr
  4. 5.45 m/hr
Explanation: The rate of change of the tide level is the derivative of the height function, h'(t). The function is h(t) = 6.5 + 2.1cos(π/6 * t). Using the chain rule for the cosine term, the derivative is: h'(t) = -2.1sin(π/6 * t) * (π/6) h'(t) = -0.35πsin(π/6 * t). We need to find the rate at 4:00 a.m., so we evaluate h'(4): h'(4) = -0.35πsin(π/6 * 4) h'(4) = -0.35πsin(2π/3). Since sin(2π/3) = √3/2, we have: h'(4) = -0.35π * (√3/2) h'(4) ≈ -1.0995 * 0.866 ≈ -0.952 m/hr. Rounded to the nearest 0.01, the rate is -0.95 m/hr.

Question 5

The motion of a spring is described by the equation x(t) = 10cos(2t) - 5sin(2t), where x is the displacement in cm and t is the time in seconds.

What is the acceleration of the spring at t = π/2 seconds?

  1. -40 cm/s²
  2. -20 cm/s²
  3. 20 cm/s²
  4. 40 cm/s² (correct answer)
Explanation: Acceleration is the second derivative of the displacement function x(t).
  1. Find the velocity function, v(t) = x'(t). x(t) = 10cos(2t) - 5sin(2t) v(t) = 10(-2sin(2t)) - 5(2cos(2t)) v(t) = -20sin(2t) - 10cos(2t).
  2. Find the acceleration function, a(t) = v'(t). a(t) = -20(2cos(2t)) - 10(-2sin(2t)) a(t) = -40cos(2t) + 20sin(2t).
  3. Evaluate the acceleration at t = π/2. a(π/2) = -40cos(2 * π/2) + 20sin(2 * π/2) a(π/2) = -40cos(π) + 20sin(π). We know that cos(π) = -1 and sin(π) = 0. a(π/2) = -40(-1) + 20(0) = 40 + 0 = 40. The acceleration is 40 cm/s².

Question 6

The equation of a curve is given by y = x² - 3e⁻²ˣ.

Find the gradient of the normal to the curve at x = 0.

  1. 1/6
  2. -1/2
  3. -1/6 (correct answer)
  4. 6
Explanation: When you encounter a question about the gradient of a normal to a curve, remember that the normal is perpendicular to the tangent, so their gradients are negative reciprocals of each other. First, you need to find the gradient of the tangent by differentiating the curve equation. Given y=x2−3e−2xy = x^2 - 3e^{-2x}, apply the differentiation rules: dydx=2x−3⋅(−2)e−2x=2x+6e−2x\frac{dy}{dx} = 2x - 3 \cdot (-2)e^{-2x} = 2x + 6e^{-2x} At x=0x = 0: dydx=2(0)+6e−2(0)=0+6e0=6\frac{dy}{dx} = 2(0) + 6e^{-2(0)} = 0 + 6e^0 = 6 Since the gradient of the tangent is 6, the gradient of the normal is −16-\frac{1}{6}. Looking at the wrong answers: Answer A (16\frac{1}{6}) gives you the reciprocal but forgets the negative sign that makes the normal perpendicular to the tangent. Answer B (−12-\frac{1}{2}) likely comes from incorrectly differentiating the exponential term, getting −3e−2x-3e^{-2x} instead of 6e−2x6e^{-2x}, leading to a tangent gradient of -3 and normal gradient of 13\frac{1}{3} (though this doesn't match exactly). Answer D (6) is simply the gradient of the tangent itself, not the normal. The correct answer is C: −16-\frac{1}{6}. Study tip: Always remember that tangent and normal gradients multiply to give -1. If you find the tangent gradient is mm, the normal gradient is always −1m-\frac{1}{m}. Double-check your exponential differentiation using the chain rule.

Question 7

A company's revenue from selling x units of a product is R(x) = 45x - 0.1x². The cost to produce x units is C(x) = 200 + 5x.

What is the rate of change of profit when 150 units are produced and sold?

  1. 10 (correct answer)
  2. 15
  3. 40
  4. 3550
Explanation: Profit P(x) is Revenue R(x) minus Cost C(x). P(x) = R(x) - C(x) P(x) = (45x - 0.1x²) - (200 + 5x) P(x) = -0.1x² + 40x - 200. The rate of change of profit is the derivative, P'(x), also known as marginal profit. P'(x) = d/dx (-0.1x² + 40x - 200) P'(x) = -0.2x + 40. Now, evaluate P'(x) at x = 150: P'(150) = -0.2(150) + 40 P'(150) = -30 + 40 = 10. The rate of change of profit is 10 (e.g., $10 per unit).

Question 8

The displacement, s, in metres, of a particle from a fixed point O at time t seconds is given by s(t) = t³ - 12t² + 36t + 5, for t ≥ 0.

Find the instantaneous velocity of the particle when its acceleration is zero.

  1. -12 m/s (correct answer)
  2. 0 m/s
  3. 4 m/s
  4. 37 m/s
Explanation: This is a multi-step problem. Velocity v(t) is the first derivative of displacement s(t), and acceleration a(t) is the second derivative of s(t) (or the first derivative of v(t)).
  1. Find the velocity function, v(t) = s'(t): s(t) = t³ - 12t² + 36t + 5 v(t) = s'(t) = 3t² - 24t + 36.
  2. Find the acceleration function, a(t) = v'(t): a(t) = v'(t) = 6t - 24.
  3. Find the time t when acceleration is zero: Set a(t) = 0: 6t - 24 = 0 => 6t = 24 => t = 4 seconds.
  4. Find the velocity at that time by substituting t = 4 into the velocity function: v(4) = 3(4)² - 24(4) + 36 v(4) = 3(16) - 96 + 36 v(4) = 48 - 96 + 36 = -12 m/s.

Question 9

Two different models predict the number of users for a new app. Model F predicts F(t) = 500e⁰.¹ᵗ users. Model G predicts G(t) = 500 + 100t + 2t² users. In both models, t is the number of weeks after launch.

After approximately how many weeks are the two models predicting the same rate of user growth?

  1. 3.2 weeks
  2. 5.9 weeks
  3. 10.0 weeks
  4. 13.9 weeks (correct answer)
Explanation: The rate of user growth for each model is given by its derivative. For Model F: F(t) = 500e⁰.¹ᵗ F'(t) = 500 * (0.1)e⁰.¹ᵗ = 50e⁰.¹ᵗ. For Model G: G(t) = 500 + 100t + 2t² G'(t) = 100 + 4t. To find when the rates are the same, set F'(t) = G'(t): 50e⁰.¹ᵗ = 100 + 4t. This equation cannot be solved algebraically. We must use a graphing calculator or a numerical solver. Graph y₁ = 50e⁰.¹ᵗ and y₂ = 100 + 4t and find their intersection point for t > 0. The calculator shows the intersection occurs at t ≈ 13.92 weeks.

Question 10

The motion of a spring is described by the equation x(t) = 10cos(2t) - 5sin(2t), where x is the displacement in cm and t is the time in seconds.

What is the acceleration of the spring at t = π/2 seconds?

  1. -40 cm/s²
  2. -20 cm/s²
  3. 20 cm/s²
  4. 40 cm/s² (correct answer)
Explanation: Acceleration is the second derivative of the displacement function x(t).
  1. Find the velocity function, v(t) = x'(t). x(t) = 10cos(2t) - 5sin(2t) v(t) = 10(-2sin(2t)) - 5(2cos(2t)) v(t) = -20sin(2t) - 10cos(2t).
  2. Find the acceleration function, a(t) = v'(t). a(t) = -20(2cos(2t)) - 10(-2sin(2t)) a(t) = -40cos(2t) + 20sin(2t).
  3. Evaluate the acceleration at t = π/2. a(π/2) = -40cos(2 * π/2) + 20sin(2 * π/2) a(π/2) = -40cos(π) + 20sin(π). We know that cos(π) = -1 and sin(π) = 0. a(π/2) = -40(-1) + 20(0) = 40 + 0 = 40. The acceleration is 40 cm/s².

Question 11

The cost, C, in euros, of producing x metres of a specialized fabric is C(x) = 1500 + 12x + 0.01x².

The marginal cost of production is found to be 20 euros per metre. How many metres of fabric are being produced?

  1. 340 metres
  2. 400 metres (correct answer)
  3. 800 metres
  4. 1600 metres
Explanation: Marginal cost is the derivative of the cost function, C'(x). We are given the value of the marginal cost and need to find x.
  1. Find the marginal cost function, C'(x). C(x) = 1500 + 12x + 0.01x² C'(x) = 0 + 12 + 0.01(2x) C'(x) = 12 + 0.02x.
  2. Set the marginal cost equal to the given value, 20, and solve for x. 12 + 0.02x = 20 0.02x = 20 - 12 0.02x = 8 x = 8 / 0.02 x = 400. Therefore, 400 metres of fabric are being produced.

Question 12

A curve is described by the equation f(x) = ax³ + bx². The curve has a stationary point at (4, -32).

Find the gradient of the curve at x = 1.

  1. -27
  2. -9 (correct answer)
  3. 3
  4. 9
Explanation: This problem requires setting up and solving a system of equations to find the parameters a and b.
  1. Use the point (4, -32) to form an equation with f(x). f(4) = a(4)³ + b(4)² = -32 64a + 16b = -32. Divide by 16: 4a + b = -2 (Equation 1)
  2. Use the fact that (4, -32) is a stationary point. This means the derivative f'(x) is zero at x = 4. First, find f'(x): f'(x) = 3ax² + 2bx. Now, set f'(4) = 0: f'(4) = 3a(4)² + 2b(4) = 0 48a + 8b = 0. Divide by 8: 6a + b = 0 (Equation 2)
  3. Solve the system of equations. From Eq 2, b = -6a. Substitute this into Eq 1: 4a + (-6a) = -2 -2a = -2 => a = 1. Now find b: b = -6(1) = -6.
  4. The function is f(x) = x³ - 6x², and its derivative is f'(x) = 3x² - 12x.
  5. Find the gradient at x = 1 by evaluating f'(1). f'(1) = 3(1)² - 12(1) = 3 - 12 = -9.

Question 13

The temperature T, in degrees Celsius, of a cup of coffee left to cool is given by T(t) = 22 + 70e⁻⁰.⁰⁵ᵗ, where t is the time in minutes after it was poured.

At what time t is the coffee cooling at a rate of 2 degrees Celsius per minute?

  1. 5.75 minutes
  2. 11.19 minutes (correct answer)
  3. 15.75 minutes
  4. 24.10 minutes
Explanation: The rate of cooling is the derivative of the temperature function, T'(t). A 'cooling rate of 2' means the rate of change is -2. First, find the derivative of T(t): T(t) = 22 + 70e⁻⁰.⁰⁵ᵗ T'(t) = 0 + 70 × (-0.05)e⁻⁰.⁰⁵ᵗ T'(t) = -3.5e⁻⁰.⁰⁵ᵗ. We need to find the time t when the rate of cooling is 2°C per minute, which means T'(t) = -2. Set the derivative equal to -2 and solve for t: -3.5e⁻⁰.⁰⁵ᵗ = -2 e⁻⁰.⁰⁵ᵗ = 2/3.5 = 4/7 Take the natural logarithm of both sides: -0.05t = ln(4/7) t = -ln(4/7)/0.05 ≈ 11.19 minutes.

Question 14

A company's revenue from selling x units of a product is R(x) = 45x - 0.1x². The cost to produce x units is C(x) = 200 + 5x.

What is the rate of change of profit when 150 units are produced and sold?

  1. 10 (correct answer)
  2. 15
  3. 40
  4. 3550
Explanation: Profit P(x) is Revenue R(x) minus Cost C(x). P(x) = R(x) - C(x) P(x) = (45x - 0.1x²) - (200 + 5x) P(x) = -0.1x² + 40x - 200. The rate of change of profit is the derivative, P'(x), also known as marginal profit. P'(x) = d/dx (-0.1x² + 40x - 200) P'(x) = -0.2x + 40. Now, evaluate P'(x) at x = 150: P'(150) = -0.2(150) + 40 P'(150) = -30 + 40 = 10. The rate of change of profit is 10 (e.g., $10 per unit).

Question 15

The value of a certain collectible item, V, in dollars, is modelled by V(t) = 50ln(t + 1) + 100, where t is the number of years since it was purchased.

What is the rate of change of the item's value 9 years after it was purchased?

  1. $115.13 per year
  2. $11.51 per year
  3. $5.00 per year (correct answer)
  4. $215.13 per year
Explanation: When you see a question asking for the "rate of change" of a function at a specific time, you're being asked to find the derivative and evaluate it at that point. The derivative tells you how fast the value is changing at any given moment. Given V(t)=50ln⁡(t+1)+100V(t) = 50\ln(t + 1) + 100, you need to find V′(t)V'(t) first. Using the chain rule, the derivative of ln⁡(t+1)\ln(t + 1) is 1t+1\frac{1}{t + 1}, and the derivative of a constant is zero. So: V′(t)=50⋅1t+1=50t+1V'(t) = 50 \cdot \frac{1}{t + 1} = \frac{50}{t + 1} At t=9t = 9 years: V′(9)=509+1=5010=5V'(9) = \frac{50}{9 + 1} = \frac{50}{10} = 5 The rate of change is $5.00 per year, making C correct. Looking at the wrong answers: A (115.13)appearstobethetotalvaluechangeoversomeperiodratherthantheinstantaneousrate.B(115.13) appears to be the total value change over some period rather than the instantaneous rate. B (11.51) might come from incorrectly calculating 509\frac{50}{9} instead of 5010\frac{50}{10}, forgetting to add 1 in the denominator. D ($215.13) could result from evaluating the original function $V(9)V(9) $ instead of its derivative, or from some other computational error involving the function value rather than its rate of change. Remember: "rate of change" always means derivative. Don't confuse this with the function's actual value or with average rates of change over intervals. Practice identifying derivative keywords like "rate of change," "instantaneous rate," or "how fast" to avoid these common traps.

Question 16

Two different models predict the number of users for a new app. Model F predicts F(t) = 500e⁰.¹ᵗ users. Model G predicts G(t) = 500 + 100t + 2t² users. In both models, t is the number of weeks after launch.

After approximately how many weeks are the two models predicting the same rate of user growth?

  1. 3.2 weeks
  2. 5.9 weeks
  3. 10.0 weeks
  4. 13.9 weeks (correct answer)
Explanation: The rate of user growth for each model is given by its derivative. For Model F: F(t) = 500e⁰.¹ᵗ F'(t) = 500 * (0.1)e⁰.¹ᵗ = 50e⁰.¹ᵗ. For Model G: G(t) = 500 + 100t + 2t² G'(t) = 100 + 4t. To find when the rates are the same, set F'(t) = G'(t): 50e⁰.¹ᵗ = 100 + 4t. This equation cannot be solved algebraically. We must use a graphing calculator or a numerical solver. Graph y₁ = 50e⁰.¹ᵗ and y₂ = 100 + 4t and find their intersection point for t > 0. The calculator shows the intersection occurs at t ≈ 13.92 weeks.

Question 17

The cost, C, in euros, of producing x metres of a specialized fabric is C(x) = 1500 + 12x + 0.01x².

The marginal cost of production is found to be 20 euros per metre. How many metres of fabric are being produced?

  1. 340 metres
  2. 400 metres (correct answer)
  3. 800 metres
  4. 1600 metres
Explanation: Marginal cost is the derivative of the cost function, C'(x). We are given the value of the marginal cost and need to find x.
  1. Find the marginal cost function, C'(x). C(x) = 1500 + 12x + 0.01x² C'(x) = 0 + 12 + 0.01(2x) C'(x) = 12 + 0.02x.
  2. Set the marginal cost equal to the given value, 20, and solve for x. 12 + 0.02x = 20 0.02x = 20 - 12 0.02x = 8 x = 8 / 0.02 x = 400. Therefore, 400 metres of fabric are being produced.

Question 18

The value of a certain collectible item, V, in dollars, is modelled by V(t) = 50ln(t + 1) + 100, where t is the number of years since it was purchased.

What is the rate of change of the item's value 9 years after it was purchased?

  1. $115.13 per year
  2. $11.51 per year
  3. $5.00 per year (correct answer)
  4. $215.13 per year
Explanation: When you see a question asking for the "rate of change" of a function at a specific time, you're being asked to find the derivative and evaluate it at that point. The derivative tells you how fast the value is changing at any given moment. Given V(t)=50ln⁡(t+1)+100V(t) = 50\ln(t + 1) + 100, you need to find V′(t)V'(t) first. Using the chain rule, the derivative of ln⁡(t+1)\ln(t + 1) is 1t+1\frac{1}{t + 1}, and the derivative of a constant is zero. So: V′(t)=50⋅1t+1=50t+1V'(t) = 50 \cdot \frac{1}{t + 1} = \frac{50}{t + 1} At t=9t = 9 years: V′(9)=509+1=5010=5V'(9) = \frac{50}{9 + 1} = \frac{50}{10} = 5 The rate of change is $5.00 per year, making C correct. Looking at the wrong answers: A (115.13)appearstobethetotalvaluechangeoversomeperiodratherthantheinstantaneousrate.B(115.13) appears to be the total value change over some period rather than the instantaneous rate. B (11.51) might come from incorrectly calculating 509\frac{50}{9} instead of 5010\frac{50}{10}, forgetting to add 1 in the denominator. D ($215.13) could result from evaluating the original function $V(9)V(9) $ instead of its derivative, or from some other computational error involving the function value rather than its rate of change. Remember: "rate of change" always means derivative. Don't confuse this with the function's actual value or with average rates of change over intervals. Practice identifying derivative keywords like "rate of change," "instantaneous rate," or "how fast" to avoid these common traps.

Question 19

The altitude of a drone, A(t) in metres, at time t minutes is given by A(t) = -t³ + 15t² - 48t + 100, for 0 ≤ t ≤ 10.

For what value of t does the drone momentarily stop ascending and begin to descend?

  1. 2.5 minutes
  2. 3.0 minutes
  3. 5.0 minutes
  4. 8.0 minutes (correct answer)
Explanation: The drone stops ascending and begins descending at a local maximum of its altitude function. This occurs when the rate of change of altitude (the vertical velocity) is zero. We need to find the derivative A'(t) and set it to zero. A(t) = -t³ + 15t² - 48t + 100 A'(t) = -3t² + 30t - 48. Set A'(t) = 0 and solve for t: -3t² + 30t - 48 = 0 Divide by -3 to simplify: t² - 10t + 16 = 0 Factor the quadratic equation: (t - 2)(t - 8) = 0. The stationary points are t = 2 and t = 8. A'(t) is a downward-opening parabola, so it is negative before t=2, positive between t=2 and t=8, and negative after t=8. Ascending means A'(t) > 0, descending means A'(t) < 0. The drone stops ascending and starts descending at the point where the derivative changes from positive to negative, which is the local maximum at t = 8.

Question 20

The temperature T, in degrees Celsius, of a cup of coffee left to cool is given by T(t) = 22 + 70e⁻⁰.⁰⁵ᵗ, where t is the time in minutes after it was poured.

At what time t is the coffee cooling at a rate of 2 degrees Celsius per minute?

  1. 5.75 minutes
  2. 11.19 minutes (correct answer)
  3. 15.75 minutes
  4. 24.10 minutes
Explanation: The rate of cooling is the derivative of the temperature function, T'(t). A 'cooling rate of 2' means the rate of change is -2. First, find the derivative of T(t): T(t) = 22 + 70e⁻⁰.⁰⁵ᵗ T'(t) = 0 + 70 × (-0.05)e⁻⁰.⁰⁵ᵗ T'(t) = -3.5e⁻⁰.⁰⁵ᵗ. We need to find the time t when the rate of cooling is 2°C per minute, which means T'(t) = -2. Set the derivative equal to -2 and solve for t: -3.5e⁻⁰.⁰⁵ᵗ = -2 e⁻⁰.⁰⁵ᵗ = 2/3.5 = 4/7 Take the natural logarithm of both sides: -0.05t = ln(4/7) t = -ln(4/7)/0.05 ≈ 11.19 minutes.