Product Rule Basics

1) The function f(x) = (x + 1)(x - 2)(x + 3) requires an extension of the Product Rule for three functions. What is the correct derivative?

F'(x) = (1)(x - 2)(x + 3) + (x + 1)(1)(x + 3) + (x + 1)(x - 2)(1)

F'(x) = (x + 1)(x - 2) + (x + 1)(x + 3) + (x - 2)(x + 3)

F'(x) = 3(x + 1)(x - 2)(x + 3)

F'(x) = (1)(1)(1) = 1

The extended Product Rule for three functions states: d/dx[uvw] = u'vw + uv'w + uvw'

With u(x) = x + 1, v(x) = x - 2, w(x) = x + 3:

u'(x) = 1, v'(x) = 1, w'(x) = 1

f'(x) = u'vw + uv'w + uvw' = 1(x - 2)(x + 3) + (x + 1)1(x + 3) + (x + 1)(x - 2)1

This can be simplified to f'(x) = (x - 2)(x + 3) + (x + 1)(x + 3) + (x + 1)(x - 2)

Explanation

The extended Product Rule for three functions states: d/dx[uvw] = u'vw + uv'w + uvw'

With u(x) = x + 1, v(x) = x - 2, w(x) = x + 3:

u'(x) = 1, v'(x) = 1, w'(x) = 1

f'(x) = u'vw + uv'w + uvw' = 1(x - 2)(x + 3) + (x + 1)1(x + 3) + (x + 1)(x - 2)1

This can be simplified to f'(x) = (x - 2)(x + 3) + (x + 1)(x + 3) + (x + 1)(x - 2)

2) Find the derivative of f(x) = (3x² + 2x)(x⁴ - 5x)

(6x + 2)(x⁴ - 5x) + (3x² + 2x)(4x³ - 5)

(6x + 2)(x⁴ - 5x) - (3x² + 2x)(4x³ - 5)

(6x + 2)(x⁴ - 5x) + (3x² + 2x)(4x³ + 5)

(6x + 2)(4x³ - 5)

Using the Product Rule: d/dx[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).

Let u(x) = 3x² + 2x, so u'(x) = 6x + 2. Let v(x) = x⁴ - 5x, so v'(x) = 4x³ - 5

Therefore: f'(x) = (6x + 2)(x⁴ - 5x) + (3x² + 2x)(4x³ - 5)

Explanation

Using the Product Rule: d/dx[u(x)v(x)] = u'(x)v(x) + u(x)v'(x).

Let u(x) = 3x² + 2x, so u'(x) = 6x + 2. Let v(x) = x⁴ - 5x, so v'(x) = 4x³ - 5

Therefore: f'(x) = (6x + 2)(x⁴ - 5x) + (3x² + 2x)(4x³ - 5)

3) If g(x) = (2x + 1)(3x² - 4x + 2), what is g'(x)?

2(3x² - 4x + 2) + (2x + 1)(6x - 4)

(2x + 1)(3x² - 4x + 2) × (6x - 4)

2(3x² - 4x + 2) - (2x + 1)(6x - 4)

6x² - 8x + 4 + 12x² - 8x + 4

Using the Product Rule with u(x) = 2x + 1 and v(x) = 3x² - 4x + 2

u'(x) = 2, v'(x) = 6x - 4

g'(x) = u'(x)v(x) + u(x)v'(x) = 2(3x² - 4x + 2) + (2x + 1)(6x - 4)

Explanation

Using the Product Rule with u(x) = 2x + 1 and v(x) = 3x² - 4x + 2

u'(x) = 2, v'(x) = 6x - 4

g'(x) = u'(x)v(x) + u(x)v'(x) = 2(3x² - 4x + 2) + (2x + 1)(6x - 4)

4) Which rule should be used to differentiate f(x) = x²eˣ?

Product Rule

Chain Rule

Power Rule

Quotient Rule

The function f(x) = x²eˣ is a product of two functions: u(x) = x² and v(x) = eˣ. Since both functions are multiplied together and neither is a composition of functions, the Product Rule is the appropriate rule to use.

Explanation

The function f(x) = x²eˣ is a product of two functions: u(x) = x² and v(x) = eˣ. Since both functions are multiplied together and neither is a composition of functions, the Product Rule is the appropriate rule to use.

5) Find the derivative of h(x) = (x³ + 2x)(4x² + 3)

(3x² + 2)(4x² + 3) + (x³ + 2x)(8x)

(3x² + 2)(4x² + 3) - (x³ + 2x)(8x)

3x²(4x² + 3) + 8x(x³ + 2x)

Both A and C are correct

Using the Product Rule where u(x) = x³ + 2x, u'(x) = 3x² + 2; v(x) = 4x² + 3, v'(x) = 8x, we have h'(x) = u'(x)v(x) + u(x)v'(x) = (3x² + 2)(4x² + 3) + (x³ + 2x)(8x).

Explanation

Using the Product Rule where u(x) = x³ + 2x, u'(x) = 3x² + 2; v(x) = 4x² + 3, v'(x) = 8x, we have h'(x) = u'(x)v(x) + u(x)v'(x) = (3x² + 2)(4x² + 3) + (x³ + 2x)(8x).

6) If f(x) = (5x - 3)(2x² + 7x - 1), what is f'(x)?

5(2x² + 7x - 1) + (5x - 3)(4x + 7)

5(2x² + 7x - 1) - (5x - 3)(4x + 7)

10x² + 35x - 5 + 20x² + 35x - 12

(5x - 3)'(2x² + 7x - 1) + (5x - 3)(2x² + 7x - 1)'

Using the Product Rule: u(x) = 5x - 3, u'(x) = 5; v(x) = 2x² + 7x - 1, v'(x) = 4x + 7

f'(x) = u'(x)v(x) + u(x)v'(x) = 5(2x² + 7x - 1) + (5x - 3)(4x + 7)

Explanation

Using the Product Rule: u(x) = 5x - 3, u'(x) = 5; v(x) = 2x² + 7x - 1, v'(x) = 4x + 7

f'(x) = u'(x)v(x) + u(x)v'(x) = 5(2x² + 7x - 1) + (5x - 3)(4x + 7)

7) What is the derivative of f(x) = (x⁴ - 6x²)(3x³ + 8x)?

(4x³ - 12x)(3x³ + 8x) + (x⁴ - 6x²)(9x² + 8)

(4x³ - 12x)(3x³ + 8x) - (x⁴ - 6x²)(9x² + 8)

12x⁶ + 32x⁴ - 36x⁴ - 72x² + 9x⁶ + 8x⁴ - 54x⁴ - 48x²

(4x³ - 12x)(9x² + 8) + (x⁴ - 6x²)(3x³ + 8x)

Using the Product Rule with u(x) = x⁴ - 6x² (u'(x) = 4x³ - 12x) and v(x) = 3x³ + 8x (v'(x) = 9x² + 8), we have f'(x) = u'(x)v(x) + u(x)v'(x) = (4x³ - 12x)(3x³ + 8x) + (x⁴ - 6x²)(9x² + 8).

Explanation

Using the Product Rule with u(x) = x⁴ - 6x² (u'(x) = 4x³ - 12x) and v(x) = 3x³ + 8x (v'(x) = 9x² + 8), we have f'(x) = u'(x)v(x) + u(x)v'(x) = (4x³ - 12x)(3x³ + 8x) + (x⁴ - 6x²)(9x² + 8).

8) Find the derivative of f(x) = (6x⁵ - 2x³ + 4)(x² - 3x)

(30x⁴ - 6x²)(x² - 3x) + (6x⁵ - 2x³ + 4)(2x - 3)

(30x⁴ - 6x²)(2x - 3) + (6x⁵ - 2x³ + 4)(x² - 3x)

(30x⁴ - 6x² + 4)(x² - 3x) + (6x⁵ - 2x³)(2x - 3)

60x⁷ - 180x⁵ + 12x⁵ - 36x³ + 8x² - 24x

Using the Product Rule: u(x) = x⁴ - 6x², u'(x) = 4x³ - 12x; v(x) = 3x³ + 8x, v'(x) = 9x² + 8

f'(x) = u'(x)v(x) + u(x)v'(x) = (4x³ - 12x)(3x³ + 8x) + (x⁴ - 6x²)(9x² + 8)

Explanation

Using the Product Rule: u(x) = x⁴ - 6x², u'(x) = 4x³ - 12x; v(x) = 3x³ + 8x, v'(x) = 9x² + 8

f'(x) = u'(x)v(x) + u(x)v'(x) = (4x³ - 12x)(3x³ + 8x) + (x⁴ - 6x²)(9x² + 8)

9) Calculate g'(x) for g(x) = (x⁴ + 4x² + 1)(5x - 2)

(4x³ + 8x)(5x - 2) + (x⁴ + 4x² + 1)(5)

(4x³ + 8x)(5) + (x⁴ + 4x² + 1)(5x - 2)

20x⁴ - 8x³ + 40x³ - 32x + 5x⁴ - 8x² + 5x² - 2

(4x³ + 8x + 1)(5x - 2) + (x⁴ + 4x²)(5)

Using the Product Rule: u(x) = 6x⁵ - 2x³ + 4, u'(x) = 30x⁴ - 6x²; v(x) = x² - 3x, v'(x) = 2x - 3

f'(x) = u'(x)v(x) + u(x)v'(x) = (30x⁴ - 6x²)(x² - 3x) + (6x⁵ - 2x³ + 4)(2x - 3)

Explanation

Using the Product Rule: u(x) = 6x⁵ - 2x³ + 4, u'(x) = 30x⁴ - 6x²; v(x) = x² - 3x, v'(x) = 2x - 3

f'(x) = u'(x)v(x) + u(x)v'(x) = (30x⁴ - 6x²)(x² - 3x) + (6x⁵ - 2x³ + 4)(2x - 3)

10) What is the derivative of h(x) = (x² - 4x + 3)(2x³ + 6x²)?

(2x - 4)(2x³ + 6x²) + (x² - 4x + 3)(6x² + 12x)

(2x - 4)(6x² + 12x) + (x² - 4x + 3)(2x³ + 6x²)

(2x - 4)(2x³ + 6x²) - (x² - 4x + 3)(6x² + 12x)

4x⁵ + 12x⁴ - 8x⁴ - 24x³ + 6x⁴ + 18x³ - 24x³ - 72x² + 6x³ + 18x²

Using the Product Rule: u(x) = x⁴ + 4x² + 1, u'(x) = 4x³ + 8x; v(x) = 5x - 2, v'(x) = 5

g'(x) = u'(x)v(x) + u(x)v'(x) = (4x³ + 8x)(5x - 2) + (x⁴ + 4x² + 1)(5)

Explanation

Using the Product Rule: u(x) = x⁴ + 4x² + 1, u'(x) = 4x³ + 8x; v(x) = 5x - 2, v'(x) = 5

g'(x) = u'(x)v(x) + u(x)v'(x) = (4x³ + 8x)(5x - 2) + (x⁴ + 4x² + 1)(5)

11) A function is defined as f(x) = (2x - 1)(3x² + 4x). Find f'(2) and interpret this result.

F'(2) = 88, meaning the function has a value of 88 at x = 2

F'(2) = 88, meaning the function's maximum value in the vicinity of x = 2 is 88

F'(2) = 88, meaning the slope of the tangent line is 88 at x = 2

Using the Product Rule: u(x) = x² - 4x + 3, u'(x) = 2x - 4; v(x) = 2x³ + 6x², v'(x) = 6x² + 12x

h'(x) = u'(x)v(x) + u(x)v'(x) = (2x - 4)(2x³ + 6x²) + (x² - 4x + 3)(6x² + 12x)

Explanation

Using the Product Rule: u(x) = x² - 4x + 3, u'(x) = 2x - 4; v(x) = 2x³ + 6x², v'(x) = 6x² + 12x

h'(x) = u'(x)v(x) + u(x)v'(x) = (2x - 4)(2x³ + 6x²) + (x² - 4x + 3)(6x² + 12x)

12) Why is the Product Rule necessary when differentiating products of functions?

Because d/dx[u(x)v(x)] = d/dx[u(x)] × d/dx[v(x)] is not always true

Because the derivative of a product is always the sum of the derivatives

Because constants can be factored out of derivatives

Because the Product Rule applies to all functions, not just polynomials

First, find f'(x) using the Product Rule:

u(x) = 2x - 1, u'(x) = 2; v(x) = 3x² + 4x, v'(x) = 6x + 4

f'(x) = 2(3x² + 4x) + (2x - 1)(6x + 4) = 6x² + 8x + 12x² + 8x - 6x - 4 = 18x² + 10x - 4

f'(2) = 18(4) + 10(2) - 4 = 72 + 20 - 4 = 88

The derivative at x = 2 represents the slope of the tangent line to the function at that point. While A is partially correct about rate of change, D more precisely describes what the derivative represents mathematically.

Explanation

First, find f'(x) using the Product Rule:

u(x) = 2x - 1, u'(x) = 2; v(x) = 3x² + 4x, v'(x) = 6x + 4

f'(x) = 2(3x² + 4x) + (2x - 1)(6x + 4) = 6x² + 8x + 12x² + 8x - 6x - 4 = 18x² + 10x - 4

f'(2) = 18(4) + 10(2) - 4 = 72 + 20 - 4 = 88

The derivative at x = 2 represents the slope of the tangent line to the function at that point. While A is partially correct about rate of change, D more precisely describes what the derivative represents mathematically.

13) Given the following values for differentiable functions f and g at x = 3: f(3) = 4, f'(3) = -2, g(3) = 7, and g'(3) = 5. If h(x) = f(x)g(x), what is h'(3)?

12/20

1/6

14

2/3

The Product Rule is necessary because the derivative of a product is NOT simply the product of the derivatives. For example, if u(x) = x and v(x) = x, then d/dx[x · x] = d/dx[x²] = 2x, but d/dx[x] × d∕dx [x] = 1 × 1 = 1, which is incorrect. The Product Rule states that d/dx[u(x)v(x)] = u'(x)v(x) + u(x)v'(x), which correctly accounts for how both factors change simultaneously.

Explanation

The Product Rule is necessary because the derivative of a product is NOT simply the product of the derivatives. For example, if u(x) = x and v(x) = x, then d/dx[x · x] = d/dx[x²] = 2x, but d/dx[x] × d∕dx [x] = 1 × 1 = 1, which is incorrect. The Product Rule states that d/dx[u(x)v(x)] = u'(x)v(x) + u(x)v'(x), which correctly accounts for how both factors change simultaneously.

14) A company's revenue is given by R(x) = p(x)q(x), where p(x) is the price per unit and q(x) is the quantity sold. If p(5) = 20, p'(5) = -2, q(5) = 100, and q'(5) = 5, what is R'(5)?

R'(5) = -100

R'(5) = -440

R'(5) = 300

R'(5) = 440

Using the Product Rule: h'(x) = f'(x)g(x) + f(x)g'(x). Substituting the given values at x = 3: h'(3) = (-2)(7) + (4)(5) = -14 + 20 = 6.

Explanation

Using the Product Rule: h'(x) = f'(x)g(x) + f(x)g'(x). Substituting the given values at x = 3: h'(3) = (-2)(7) + (4)(5) = -14 + 20 = 6.

15) The function f(x) = (x + 1)(x - 2)(x + 3) requires an extension of the Product Rule for three functions. What is the correct derivative?

F'(x) = (1)(x - 2)(x + 3) + (x + 1)(1)(x + 3) + (x + 1)(x - 2)(1)

F'(x) = (x + 1)(x - 2) + (x + 1)(x + 3) + (x - 2)(x + 3)

F'(x) = 3(x + 1)(x - 2)(x + 3)

F'(x) = (1)(1)(1) = 1

The revenue function R(x) is defined as the product of two other functions, p(x) and q(x). Therefore, to find the derivative R'(x), we must apply the Product Rule of calculus. Using the Product Rule: R'(x) = p'(x)q(x) + p(x)q'(x), we have R'(5) = p(5) * q'(5) + q(5) * p'(5) =(20 * 5) + (100 * -2) = 100 - 200 = -100.

Explanation

The revenue function R(x) is defined as the product of two other functions, p(x) and q(x). Therefore, to find the derivative R'(x), we must apply the Product Rule of calculus. Using the Product Rule: R'(x) = p'(x)q(x) + p(x)q'(x), we have R'(5) = p(5) * q'(5) + q(5) * p'(5) =(20 * 5) + (100 * -2) = 100 - 200 = -100.