Product Rule with Polynomial Products

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Quizzes Created: 7682 | Total Attempts: 9,547,133
| Questions: 15 | Updated: Dec 16, 2025
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1) Find the derivative of f(x) = (2x³ - 5x)(4x² + 3x)

Explanation

Using the Product Rule: u(x) = 2x³ - 5x, u'(x) = 6x² - 5; v(x) = 4x² + 3x, v'(x) = 8x + 3
f'(x) = u'(x)v(x) + u(x)v'(x) = (6x² - 5)(4x² + 3x) + (2x³ - 5x)(8x + 3)
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About This Quiz
Product Rule With Polynomial Products - Quiz

Think you’ve mastered the basics? This quiz builds on your Product Rule skills with more complex polynomial expressions and longer algebraic steps. You’ll work through higher-degree functions, carefully track each derivative, and strengthen your accuracy. It’s a great way to sharpen your technique and reduce common sign and setup errors.

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2) If g(x) = (x² + 6)(3x⁴ - 2x²), what is g'(x)?

Explanation

Using the Product Rule: u(x) = x² + 6, u'(x) = 2x; v(x) = 3x⁴ - 2x², v'(x) = 12x³ - 4x
g'(x) = u'(x)v(x) + u(x)v'(x) = 2x(3x⁴ - 2x²) + (x² + 6)(12x³ - 4x)
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3) What is the derivative of h(x) = (5x + 2)(x³ - 7x)?

Explanation

Using the Product Rule: u(x) = 5x + 2, u'(x) = 5; v(x) = x³ - 7x, v'(x) = 3x² - 7
h'(x) = u'(x)v(x) + u(x)v'(x) = 5(x³ - 7x) + (5x + 2)(3x² - 7)
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4) Find the derivative of f(x) = (x⁴ + 3x²)(2x³ + 4x² - 1)

Explanation

Using the Product Rule: u(x) = x⁴ + 3x², u'(x) = 4x³ + 6x; v(x) = 2x³ + 4x² - 1, v'(x) = 6x² + 8x
f'(x) = u'(x)v(x) + u(x)v'(x) = (4x³ + 6x)(2x³ + 4x² - 1) + (x⁴ + 3x²)(6x² + 8x)
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5) If y = (3x⁵ - x²)(6x + 1), what is dy/dx?

Explanation

Using the Product Rule: u(x) = 3x⁵ - x², u'(x) = 15x⁴ - 2x; v(x) = 6x + 1, v'(x) = 6
dy/dx = u'(x)v(x) + u(x)v'(x) = (15x⁴ - 2x)(6x + 1) + (3x⁵ - x²)(6)
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6) Calculate the derivative of f(x) = (x² - 3x + 2)(x⁴ + 2x³)

Explanation

Using the Product Rule: u(x) = x² - 3x + 2, u'(x) = 2x - 3; v(x) = x⁴ + 2x³, v'(x) = 4x³ + 6x²
f'(x) = u'(x)v(x) + u(x)v'(x) = (2x - 3)(x⁴ + 2x³) + (x² - 3x + 2)(4x³ + 6x²)
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7) Find h'(x) for h(x) = (4x⁶ - 2x⁴ + x)(3x² - 5)

Explanation

Using the Product Rule: u(x) = 4x⁶ - 2x⁴ + x, u'(x) = 24x⁵ - 8x³ + 1; v(x) = 3x² - 5, v'(x) = 6x
h'(x) = u'(x)v(x) + u(x)v'(x) = (24x⁵ - 8x³ + 1)(3x² - 5) + (4x⁶ - 2x⁴ + x)(6x)
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8) What is the derivative of g(x) = (7x - 4)(x⁵ + 3x³ - 2x)?

Explanation

Using the Product Rule: u(x) = 7x - 4, u'(x) = 7; v(x) = x⁵ + 3x³ - 2x, v'(x) = 5x⁴ + 9x² - 2
g'(x) = u'(x)v(x) + u(x)v'(x) = 7(x⁵ + 3x³ - 2x) + (7x - 4)(5x⁴ + 9x² - 2)
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9) Find the derivative of f(x) = (x³ + 2x² - x)(4x² + 5x + 1)

Explanation

Using the Product Rule: u(x) = x³ + 2x² - x, u'(x) = 3x² + 4x - 1; v(x) = 4x² + 5x + 1, v'(x) = 8x + 5
f'(x) = u'(x)v(x) + u(x)v'(x) = (3x² + 4x - 1)(4x² + 5x + 1) + (x³ + 2x² - x)(8x + 5)
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10) If h(x) = (2x⁴ - 6x³ + 5)(x⁵ - 3x²), what is h'(x)?

Explanation

Using the Product Rule: u(x) = 2x⁴ - 6x³ + 5, u'(x) = 8x³ - 18x²; v(x) = x⁵ - 3x², v'(x) = 5x³ - 6x
h'(x) = u'(x)v(x) + u(x)v'(x) = (8x³ - 18x²)(x⁵ - 3x²) + (2x⁴ - 6x³ + 5)(5x³ - 6x)
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11) Some students think they can differentiate (x + 1)(x + 2) by expanding. Why might they be correct?

Explanation

For polynomials, expanding first can be valid and often simpler. After expansion: f(x) = (x + 1)(x + 2) = x² + 3x + 2, then f'(x) = 2x + 3. Using Product Rule directly: u(x) = x + 1, v(x) = x + 2, u'(x) = 1, v'(x) = 1, f'(x) = 1(x + 2) + (x + 1)1 = x + 2 + x + 1 = 2x + 3. Both methods yield the same result. For simple, low-degree polynomials like this one, expanding first can be more efficient. However, for polynomials with higher degrees, the Product Rule is almost always the more efficient and less error-prone method.
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12) A particle's position is s(t) = (t² + 1)(t³ - 2t). What is velocity at t=1?

Explanation

Velocity is the derivative of position: v(t) = s'(t)
Using Product Rule: u(t) = t² + 1, u'(t) = 2t; v(t) = t³ - 2t, v'(t) = 3t² - 2
v(t) = 2t(t³ - 2t) + (t² + 1)(3t² - 2) = 2t⁴ - 4t² + 3t⁴ - 2t² + 3t² - 2 = 5t⁴ - 3t² - 2
v(1) = 5(1) - 3(1) - 2 = 5 - 3 - 2 = 0 m/s
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13) For which scenario is Product Rule NOT most efficient?

Explanation

Function C represents a composite function (a function of a function). For f(x) = (x² + 1)4, the Chain Rule is more efficient than expanding or using the Product Rule.  The other functions explicitly involve products of different functions where the Product Rule is appropriate.
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14) Which function requires the Product Rule?

Explanation

Function A: (x² + 1)³ requires the Chain Rule (power of a function).
Function B: x² + 1 is a sum, not a product, so use Power Rule and Sum Rule.
Function C: x³ - 5x is also a sum, not a product.
Function D: x²sin(x) is a product of x² and sin(x), so it requires the Product Rule. This is the only option that is explicitly a product of two different functions.
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15) The Product Rule can be extended to four functions. What is d/dx[uvwz]?

Explanation

The extended Product Rule for n functions states that d/dx[u₁u₂...uₙ] = u₁'u₂...uₙ + u₁u₂'u₃...uₙ + ... + u₁u₂...uₙ'ₙ. For four functions: d/dx[uvwz] = u'vwz + uv'wz + uvw'z + uvwz', where each term differentiates exactly one of the four functions while keeping the others constant. 

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Find the derivative of f(x) = (2x³ - 5x)(4x² + 3x)
If g(x) = (x² + 6)(3x⁴ - 2x²), what is g'(x)?
What is the derivative of h(x) = (5x + 2)(x³ - 7x)?
Find the derivative of f(x) = (x⁴ + 3x²)(2x³ + 4x² - 1)
If y = (3x⁵ - x²)(6x + 1), what is dy/dx?
Calculate the derivative of f(x) = (x² - 3x + 2)(x⁴ + 2x³)
Find h'(x) for h(x) = (4x⁶ - 2x⁴ + x)(3x² - 5)
What is the derivative of g(x) = (7x - 4)(x⁵ + 3x³ - 2x)?
Find the derivative of f(x) = (x³ + 2x² - x)(4x² + 5x + 1)
If h(x) = (2x⁴ - 6x³ + 5)(x⁵ - 3x²), what is h'(x)?
Some students think they can differentiate (x + 1)(x + 2) by...
A particle's position is s(t) = (t² + 1)(t³ - 2t). What is velocity...
For which scenario is Product Rule NOT most efficient?
Which function requires the Product Rule?
The Product Rule can be extended to four functions. What is...
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