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MCV4U Derivatives
Grade 12 Unit 2 — practice questions with full solutions

Power, product, quotient and chain rules, tangent lines and implicit differentiation — the toolkit the rest of the course runs on.

Free MCV4U derivatives practice for Grade 12 students in Ontario. These 6 questions cover the same material as a typical unit test or exam review on this topic, and every one comes with a complete worked solution — not just an answer key. Useful whether you are preparing for a MCV4U unit test, catching up on a lesson, or reviewing before the final exam.
MCV4U · Grade 12 Derivatives and their Applications 5 core 1 challenge
1Core
Differentiate: f(x) = 3x4 − 5x2 + 7x − 2.
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  1. Apply the power rule term by term: the derivative of axn is anxn−1.
  2. 3x4 → 12x3.
  3. −5x2 → −10x.
  4. 7x → 7, and the constant −2 differentiates to 0.
Final answerf′(x) = 12x3 − 10x + 7
2Core
Use the product rule to differentiate y = (2x2 + 1)(x3 − 4).
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  1. Product rule: if y = uv then y′ = u′v + uv′.
  2. Let u = 2x2 + 1 so u′ = 4x; let v = x3 − 4 so v′ = 3x2.
  3. y′ = 4x(x3 − 4) + (2x2 + 1)(3x2).
  4. Expand: 4x4 − 16x + 6x4 + 3x2.
  5. Collect like terms: 10x4 + 3x2 − 16x.
Final answery′ = 10x4 + 3x2 − 16x
3Core
Use the quotient rule to differentiate y = (3x − 1) ÷ (x2 + 2).
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  1. Quotient rule: if y = u ÷ v then y′ = (u′v − uv′) ÷ v2.
  2. Here u = 3x − 1 with u′ = 3, and v = x2 + 2 with v′ = 2x.
  3. Numerator: 3(x2 + 2) − (3x − 1)(2x) = 3x2 + 6 − 6x2 + 2x.
  4. Simplify the numerator to −3x2 + 2x + 6.
Final answery′ = (−3x2 + 2x + 6) ÷ (x2 + 2)2
4Core
Use the chain rule to differentiate y = (5x2 − 3)4.
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  1. Chain rule: differentiate the outer function, then multiply by the derivative of the inner one.
  2. Outer: the fourth power gives 4(5x2 − 3)3.
  3. Inner: the derivative of 5x2 − 3 is 10x.
  4. Multiply: 4(5x2 − 3)3 × 10x = 40x(5x2 − 3)3.
Final answery′ = 40x(5x2 − 3)3
5Core
Find the equation of the tangent line to y = x3 − 2x at x = 2.
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  1. First find the point of tangency: y(2) = 8 − 4 = 4, so the point is (2, 4).
  2. The derivative gives the slope: y′ = 3x2 − 2.
  3. At x = 2: y′(2) = 3(4) − 2 = 10.
  4. Point-slope form: y − 4 = 10(x − 2), so y = 10x − 16.
Final answery = 10x − 16
6Challenge
The circle x2 + y2 = 25 passes through (3, 4). Use implicit differentiation to find dy⁄dx at that point, and explain how the answer relates to the radius drawn to (3, 4).
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  1. Differentiate both sides with respect to x, treating y as a function of x: 2x + 2y(dy⁄dx) = 0.
  2. Solve: dy⁄dx = −x ÷ y.
  3. At (3, 4): dy⁄dx = −3 ÷ 4.
  4. The radius from the origin to (3, 4) has slope 4 ÷ 3, and (−3⁄4)(4⁄3) = −1, so the tangent is perpendicular to the radius — exactly as circle geometry predicts.
Final answerdy⁄dx = −3⁄4

MCV4U Derivatives — common questions

Short answers to the things students ask most about this unit.

When do I use the chain rule instead of the product rule?

Use the product rule when two functions are multiplied. Use the chain rule when one function is inside another, such as (5x² − 3)⁴, where a polynomial sits inside a power.

How do I find the equation of a tangent line?

Find the point by substituting into the original function, find the slope by substituting into the derivative, then put both into point-slope form.

More MCV4U practice — grade 12 calculus and vectors

MCV4U Rates of Change & LimitsMCV4U Applications of Derivatives & Curve SketchingMCV4U Derivatives of Exponential & Trigonometric FunctionsMCV4U Geometric VectorsAll MCV4U units
← Unit 1: Rates of Change & LimitsUnit 3: Applications of Derivatives & Curve Sketching →
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