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Quizzes > Mathematics

Product Rule Practice Problems: Grade 12 Quiz

Quick, free product rule quiz with 20 questions. Instant feedback.

Editorial: Review CompletedCreated By: Shradha SatheeshUpdated Aug 28, 2025
Difficulty: Moderate
Grade: Grade 12
Study OutcomesCheat Sheet
Paper art promoting a calculus quiz focused on mastering the product rule.

Use this product rule quiz to practice Grade 12 differentiation with 20 quick questions and build speed and accuracy. You will get instant results and simple steps along the way. Afterward, try the derivative rules quiz, explore the exponent rules quiz, or review polynomial practice problems.

What is the product rule formula for differentiation?
(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
(f(x)g(x))' = f'(x)g'(x) + f(x)g(x)
(f(x)g(x))' = f'(x)g'(x)
(f(x)g(x))' = f'(x) + g'(x)
The product rule states that the derivative of a product of two functions is f'(x)g(x) + f(x)g'(x). This formula ensures that both functions' rates of change are accounted for.
Find the derivative of f(x) = x·sin(x) using the product rule.
sin(x) - x·cos(x)
cos(x) - x·sin(x)
sin(x) + x·cos(x)
x·sin(x) + cos(x)
By applying the product rule to f(x) = x·sin(x), we differentiate x to get 1 and sin(x) to get cos(x). Thus, the derivative is 1·sin(x) + x·cos(x).
In the product rule formula f'(x)g(x) + f(x)g'(x), which term represents the derivative of the first function?
f(x)g'(x)
f(x) + g(x)
f'(x)g(x)
f'(x) + g'(x)
The term f'(x)g(x) comes directly from differentiating the first function, f(x), while keeping g(x) unchanged. This correctly represents how the product rule splits the differentiation process.
What is the derivative of f(x) = 3x · x² using the product rule?
3x²
6x²
9x³
9x²
Although f(x) = 3x · x² simplifies to 3x³, applying the product rule gives: 3·x² + 3x·2x = 3x² + 6x², which simplifies to 9x². This confirms the proper use of the rule.
Identify the mistake in the following differentiation: d/dx [u(x)v(x)] = u'(x)v'(x).
It erroneously multiplies the derivatives instead of applying the product rule.
It adds the functions instead of differentiating them.
It differentiates only one function.
It correctly applies the product rule.
The correct product rule requires adding the product of the derivative of the first function with the second function, and vice versa. Multiplying the two derivatives, as shown, is a common error.
Find the derivative of f(x) = x²·cos(x) using the product rule.
-2x·sin(x) - x²·cos(x)
2x·cos(x) - x²·sin(x)
x·cos(x) - 2x·sin(x)
2x·cos(x) + x²·sin(x)
Differentiate x² to get 2x and cos(x) to get -sin(x). The product rule then combines these as 2x·cos(x) + x²·(-sin(x)), resulting in 2x·cos(x) - x²·sin(x).
Differentiate f(x) = (3x + 1)·eˣ using the product rule.
eˣ·(3x + 4)
eˣ·(3x + 1)
3eˣ
e^(3x + 1)
Differentiate (3x + 1) to obtain 3, and the derivative of eˣ is eˣ. Applying the product rule gives 3eˣ + (3x + 1)eˣ, which simplifies to eˣ·(3x + 4).
Compute the derivative of f(x) = x·ln(x) using the product rule.
ln(x)
ln(x) + 1
1 + x
1/x + ln(x)
Differentiate x to get 1 and ln(x) to get 1/x. Using the product rule, the derivative is 1·ln(x) + x·(1/x), which simplifies to ln(x) + 1.
Find the derivative of f(x) = sin(x)·cos(x) using the product rule.
cos²(x) - sin²(x)
2·sin(x)·cos(x)
sin(x)·cos(x)
-cos²(x) + sin²(x)
Differentiate sin(x) to get cos(x) and cos(x) to get -sin(x). Applying the product rule leads to cos(x)·cos(x) + sin(x)·(-sin(x)), which simplifies to cos²(x) - sin²(x).
Let f(x) = u(x)·v(x). If u(2) = 3, u'(2) = 4, v(2) = 5, and v'(2) = 6, what is f'(2)?
32
26
38
18
Using the product rule, f'(2) = u'(2)v(2) + u(2)v'(2) = 4·5 + 3·6, which equals 20 + 18, resulting in 38.
Differentiate f(x) = (x³ + 2x)(x - 1) using the product rule.
(3x² + 2)(x - 1) + (x³ + 2x)
(3x² + 2)(x + 1) + (x³ + 2x)
3x²(x - 1) + (x³ + 2x)
(3x² + 2)(x - 1) - (x³ + 2x)
Differentiate each function separately: the derivative of x³ + 2x is 3x² + 2, and the derivative of x - 1 is 1. Applying the product rule gives (3x² + 2)(x - 1) + (x³ + 2x)·1.
Differentiate f(x) = eˣ·(x² - 1) using the product rule.
eˣ·(x² + 2x - 1)
eˣ·(x² - 1 - 2x)
(x² - 1)eˣ
eˣ·(2x)
Differentiate eˣ to get eˣ and (x² - 1) to get 2x, then apply the product rule: eˣ·(x² - 1) + eˣ·(2x) which combines to eˣ·(x² + 2x - 1).
Which of the following best explains the product rule's necessity in differentiation?
It accounts for the variation in both factors in a product function.
It only applies when one function is constant.
It is equivalent to the quotient rule.
It simplifies the product into a sum of derivatives.
The product rule is essential because it captures how both functions in a product are changing. It ensures that the derivative reflects the combined effect of the variations in each function.
Determine the derivative of f(x) = (5x - 2)·√x.
5√x + ((5x - 2)/(2√x))
5√x - ((5x - 2)/(2√x))
(5x - 2)√x + 5/(2√x)
10√x
Express √x as x^(1/2) and differentiate: the derivative of (5x - 2) is 5 and that of x^(1/2) is (1/2)x^(-1/2). Applying the product rule results in 5√x + (5x - 2)/(2√x).
Differentiate f(x) = (ln x)·(x²) using the product rule.
2x + ln x
x·(1 + 2 ln x)
x²/(1 + 2 ln x)
1 + 2 ln x
Differentiate ln x to get 1/x and x² to get 2x. The product rule gives (1/x)·x² + ln x·(2x) which simplifies to x + 2x ln x, or x·(1 + 2 ln x).
Differentiate f(x) = (x² + 1)·cos(x³) using both the product and chain rules.
2x·cos(x³) - 3x²·(x² + 1)·sin(x³)
2x·cos(x³) + 3x²·(x² + 1)·sin(x³)
(x² + 1)·sin(x³) - 2x·cos(x³)
2x·cos(x³) - (x² + 1)·sin(x³)
Differentiate (x² + 1) to obtain 2x, and apply the chain rule to differentiate cos(x³) yielding -sin(x³)·3x². Using the product rule then gives 2x·cos(x³) - 3x²·(x² + 1)·sin(x³).
Find the derivative of f(x) = √x·sin(x) using the product rule.
√x·cos(x) - ½·x^(-1/2)·sin(x)
cos(x)/√x + ½·sin(x)·√x
½·x^(-1/2)·sin(x) + √x·cos(x)
sin(x)/√x + cos(x)/√x
Express √x as x^(1/2), whose derivative is ½·x^(-1/2), and differentiate sin(x) normally to get cos(x). The product rule then produces ½·x^(-1/2)·sin(x) + √x·cos(x).
Which of these is an incorrect application of the product rule for f(x) = u(x)v(x)?
f'(x) = u'(x)v(x)
f'(x) = u'(x)v'(x)
f'(x) = u'(x)v(x) + u(x)v'(x)
f'(x) = u(x)v'(x)
The correct product rule adds u'(x)v(x) and u(x)v'(x), but simply multiplying the derivatives, as in u'(x)v'(x), is incorrect. This misstep fails to account for the contribution of each function separately.
Given f(x) = (x²)(eˣ), if one mistakenly differentiates it as f'(x) = 2x·eˣ, what step was omitted?
The derivative of eˣ was omitted.
No mistake was made.
The derivative of x² was omitted.
The product of the derivatives was subtracted.
When applying the product rule to f(x) = x²·eˣ, one must differentiate both x² and eˣ. The error here is neglecting the derivative of eˣ, which leads to an incomplete result.
Differentiate f(x) = x·eˣ·sin(x) by applying the product rule iteratively.
eˣ·(x·sin(x) + cos(x))
eˣ·(sin(x) + x·sin(x) + x·cos(x))
eˣ·(sin(x) + x·cos(x))
x·eˣ·(sin(x) + cos(x))
Treat f(x) as the product of three functions: x, eˣ, and sin(x). Differentiating each while keeping the others constant and summing the results gives eˣ·sin(x) + x·eˣ·sin(x) + x·eˣ·cos(x), which can be factored to eˣ·(sin(x) + x·sin(x) + x·cos(x)).
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Study Outcomes

  1. Understand the concept and definition of the product rule in differentiation.
  2. Apply the product rule to differentiate functions involving products of two differentiable functions.
  3. Analyze expressions to correctly identify component functions for the application of the product rule.
  4. Synthesize different problem-solving approaches when using the product rule in complex scenarios.
  5. Evaluate the effectiveness of the product rule in solving various calculus problems to build confidence for assessments.

Product Rule Worksheet Cheat Sheet

  1. Understand the product rule formula - Get cozy with how the derivative of two functions f(x) and g(x) multiplies out to f'(x)g(x) + f(x)g'(x). This is your golden ticket to tackling products in calculus with confidence.
  2. Practice with different function types - Flex your muscles by applying the product rule to polynomials, trigonometric functions, exponentials, and more to see it in action. Repetition makes perfect, so mix and match functions until it feels like second nature.
  3. Learn the proof from first principles - Dive into the derivation straight from the definition of a derivative to really own your understanding. Seeing the mechanics helps you connect the dots instead of just memorizing the formula.
  4. Extend the rule to multiple functions - Discover how (fgh)' expands to f'gh + fg'h + fgh' and beyond when you have three or more factors. This powerful generalization shows you that calculus can scale to any crease of complexity.
  5. Spot the difference from the chain rule - Remember: product rule is for multiplying functions, while the chain rule handles nested ones like f(g(x)). Knowing when to use each rule is a must-have skill in your calculus toolkit.
  6. Know its limitations - The product rule only works if each function is differentiable at the point you're examining - no exceptions. Spotting non-differentiable cases early saves you from wild goose chases.
  7. Work through practice problems - Reinforce your skills by tackling a variety of exercises, from straightforward to brain-busting. Regular practice helps you anticipate common pitfalls and build unshakeable confidence.
  8. Link to the quotient rule - Notice how the quotient rule comes from applying the product rule to f(x) and [g(x)]❻¹ - it's all connected! This insight makes both rules feel less like separate hoops and more like a unified strategy.
  9. Use fun mnemonics - Try "first times the derivative of the second plus second times the derivative of the first" to keep it stuck in your brain. A catchy phrase goes a long way when exam day stress hits.
  10. Apply it in real-world scenarios - From physics to economics, see how the product rule helps analyze changing quantities in everything from motion to profit. Real examples make abstract formulas feel alive and kickstart your curiosity.
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