Product Rule Mastery & Applications

1) The product rule states that if f(x)=g(x)·h(x), then f'(x) equals:

G'(x)·h'(x)

G'(x)·h(x) + g(x)·h'(x)

G(x)·h'(x)

[g'(x) + h'(x)]·[g(x) + h(x)]

The product rule is one of the fundamental differentiation rules. It states that the derivative of a product is the derivative of the first function times the second function, plus the first function times the derivative of the second function. This is often remembered as "d(first)·second + first·d(second)."

Explanation

The product rule is one of the fundamental differentiation rules. It states that the derivative of a product is the derivative of the first function times the second function, plus the first function times the derivative of the second function. This is often remembered as "d(first)·second + first·d(second)."

2) What is the derivative of y=eˣ·x²?

Eˣ·2x

Eˣ·x² + eˣ·2x

Eˣ + 2x

2x·eˣ

Using the product rule with u = eˣ and v = x²:

u' = eˣ

v' = 2x

y' = u'·v + u·v' = eˣ·x² + eˣ·2x

Explanation

Using the product rule with u = eˣ and v = x²:

u' = eˣ

v' = 2x

y' = u'·v + u·v' = eˣ·x² + eˣ·2x

3) Find the derivative of g(x)=(5x+2)(3x-1).

5(3x - 1) + (5x + 2)·3

15x + 1

5·3

15x² + x - 2

Using the product rule with u = 5x + 2 and v = 3x - 1:

u' = 5

v' = 3

g'(x) = u'·v + u·v' = 5(3x - 1) + (5x + 2)·3 This simplifies to 15x - 5 + 15x + 6 = 30x + 1, but the product rule application is shown in option a.

Explanation

Using the product rule with u = 5x + 2 and v = 3x - 1:

u' = 5

v' = 3

g'(x) = u'·v + u·v' = 5(3x - 1) + (5x + 2)·3 This simplifies to 15x - 5 + 15x + 6 = 30x + 1, but the product rule application is shown in option a.

4) What is the derivative of h(x)=x³·sin(x)?

3x²·sin(x) + x³·cos(x)

3x²·cos(x) + x³·sin(x)

3x²·sin(x)

X³·cos(x)

Using the product rule with u = x³ and v = sin(x):

u' = 3x²

v' = cos(x)

h'(x) = u'·v + u·v' = 3x²·sin(x) + x³·cos(x)

Explanation

Using the product rule with u = x³ and v = sin(x):

u' = 3x²

v' = cos(x)

h'(x) = u'·v + u·v' = 3x²·sin(x) + x³·cos(x)

5) Find the derivative of f(x)=cos(x)·eˣ.

-sin(x)·eˣ + cos(x)·eˣ

-sin(x)·eˣ

Cos(x)·eˣ

-sin(x)+eˣ

Using the product rule with u = cos(x) and v = eˣ:

u' = -sin(x)

v' = eˣ

f'(x) = u'·v + u·v' = -sin(x)·eˣ + cos(x)·eˣ

Explanation

Using the product rule with u = cos(x) and v = eˣ:

u' = -sin(x)

v' = eˣ

f'(x) = u'·v + u·v' = -sin(x)·eˣ + cos(x)·eˣ

6) What is the derivative of p(x)=(x+1)(x²+x)?

1(x² + x) + (x + 1)(2x + 1)

3x² + 4x + 1

(2x + 1)

(x + 1)(2x + 1)

Using the product rule with u = x + 1 and v = x² + x:

u' = 1

v' = 2x + 1

p'(x) = u'·v + u·v' = 1(x² + x) + (x + 1)(2x + 1)

Explanation

Using the product rule with u = x + 1 and v = x² + x:

u' = 1

v' = 2x + 1

p'(x) = u'·v + u·v' = 1(x² + x) + (x + 1)(2x + 1)

7) Find the derivative of q(x)=x⁵·cos(x).

5x⁴·cos(x) - x⁵·sin(x)

5x⁴·cos(x) + x⁵·sin(x)

5x⁴·cos(x)

-x⁵·sin(x)

Using the product rule with u = x⁵ and v = cos(x):

u' = 5x⁴

v' = -sin(x)

q'(x) = u'·v + u·v' = 5x⁴·cos(x) + x⁵·(-sin(x)) = 5x⁴·cos(x) - x⁵·sin(x)

Explanation

Using the product rule with u = x⁵ and v = cos(x):

u' = 5x⁴

v' = -sin(x)

q'(x) = u'·v + u·v' = 5x⁴·cos(x) + x⁵·(-sin(x)) = 5x⁴·cos(x) - x⁵·sin(x)

8) Find the derivative of r(x)=ln(x)·sin(x).

(1/x)·sin(x) + ln(x)·cos(x)

(1/x)·cos(x) + ln(x)·sin(x)

(1/x)·sin(x)

Ln(x)·cos(x)

Using the product rule with u = ln(x) and v = sin(x):

u' = 1/x

v' = cos(x)

r'(x) = u'·v + u·v' = (1/x)·sin(x) + ln(x)·cos(x)

Explanation

Using the product rule with u = ln(x) and v = sin(x):

u' = 1/x

v' = cos(x)

r'(x) = u'·v + u·v' = (1/x)·sin(x) + ln(x)·cos(x)

9) Differentiate y = x f(x) g(x).

F(x)g(x) + xf'(x)g(x) + xf(x)g'(x)

F'(x)g'(x) + x

F(x)g(x) + f'(x)g'(x)

X[f'(x)g(x) + f(x)g'(x)]

We can group terms to use the Product Rule. Let u = x and v = f(x)g(x). Then u' = 1. To find v', we use the Product Rule on f(x)g(x), so v' = f'(x)g(x) + f(x)g'(x).

Now apply the Product Rule to the whole expression: y' = u'v + uv' = 1·[f(x)g(x)] + x·[f'(x)g(x) + f(x)g'(x)]. Distributing the x gives f(x)g(x) + xf'(x)g(x) + xf(x)g'(x).

Explanation

We can group terms to use the Product Rule. Let u = x and v = f(x)g(x). Then u' = 1. To find v', we use the Product Rule on f(x)g(x), so v' = f'(x)g(x) + f(x)g'(x).

Now apply the Product Rule to the whole expression: y' = u'v + uv' = 1·[f(x)g(x)] + x·[f'(x)g(x) + f(x)g'(x)]. Distributing the x gives f(x)g(x) + xf'(x)g(x) + xf(x)g'(x).

10) If f(x)=x·g(x), what is f'(x)/f(x)?

G'(x)/g(x)

1/x + g'(x)/g(x)

1 + g'(x)

1/x

First, find f'(x) using the Product Rule: f'(x) = 1·g(x) + x·g'(x). Now divide by f(x) = x·g(x):

[g(x) + x·g'(x)] / [x·g(x)] = g(x)/[x·g(x)] + [x·g'(x)]/[x·g(x)] = 1/x + g'(x)/g(x).

Explanation

First, find f'(x) using the Product Rule: f'(x) = 1·g(x) + x·g'(x). Now divide by f(x) = x·g(x):

[g(x) + x·g'(x)] / [x·g(x)] = g(x)/[x·g(x)] + [x·g'(x)]/[x·g(x)] = 1/x + g'(x)/g(x).

11) If f(x)=x·ln(x), what is f'(e)?

1/1

1/2

E

1 + e

First find f'(x) using the product rule:

u = x, u' = 1

v = ln(x), v' = 1/x

f'(x) = 1·ln(x) + x·(1/x) = ln(x) + 1

Now evaluate at x = e:

f'(e) = ln(e) + 1 = 1 + 1 = 2

Explanation

First find f'(x) using the product rule:

u = x, u' = 1

v = ln(x), v' = 1/x

f'(x) = 1·ln(x) + x·(1/x) = ln(x) + 1

Now evaluate at x = e:

f'(e) = ln(e) + 1 = 1 + 1 = 2

12) When would the product rule give f'(x)=0?

When both functions equal zero

When u'(x)·v(x) = -u(x)·v'(x)

When both functions are constants

When u'(x)=v'(x)=0

The product rule gives f'(x) = u'(x)·v(x) + u(x)·v'(x). For this to equal zero, we need u'(x)·v(x) + u(x)·v'(x) = 0, which means u'(x)·v(x) = -u(x)·v'(x). Option d would also work if both derivatives are zero, but this is a special case. Option c is incorrect because if both functions are constants, we wouldn't use the product rule (the derivative of a constant product is just 0). Option b is the most general condition.

Explanation

The product rule gives f'(x) = u'(x)·v(x) + u(x)·v'(x). For this to equal zero, we need u'(x)·v(x) + u(x)·v'(x) = 0, which means u'(x)·v(x) = -u(x)·v'(x). Option d would also work if both derivatives are zero, but this is a special case. Option c is incorrect because if both functions are constants, we wouldn't use the product rule (the derivative of a constant product is just 0). Option b is the most general condition.

13) Let h(x)=x²f(x). Given f(2)=3 and f'(2)=-4, compute h'(2).

12/14

12/26

1/4

1/12

We treat x² as the first function u(x) and f(x) as the second function v(x). u(x) = x², so u'(x) = 2x. v(x) = f(x), so v'(x) = f'(x). The Product Rule gives h'(x) = 2xf(x) + x²f'(x). At x = 2: h'(2) = 2(2)f(2) + (2)²f'(2) = 4(3) + 4(-4) = 12 - 16 = -4.

Explanation

We treat x² as the first function u(x) and f(x) as the second function v(x). u(x) = x², so u'(x) = 2x. v(x) = f(x), so v'(x) = f'(x). The Product Rule gives h'(x) = 2xf(x) + x²f'(x). At x = 2: h'(2) = 2(2)f(2) + (2)²f'(2) = 4(3) + 4(-4) = 12 - 16 = -4.

14) The graphs of f(x) and g(x) intersect at x=2 with f(2)=g(2)=3, f'(2)=1, g'(2)=-2. Find slope of h(x)=f(x)g(x) at x=2.

12/27

12/28

-1

1/9

We are given f(2) = 3, g(2) = 3, f'(2) = 1, and g'(2) = -2. We need to find h'(2). Using the Product Rule: h'(2) = f'(2)g(2) + f(2)g'(2) = (1)(3) + (3)(-2) = 3 - 6 = -3.

Explanation

We are given f(2) = 3, g(2) = 3, f'(2) = 1, and g'(2) = -2. We need to find h'(2). Using the Product Rule: h'(2) = f'(2)g(2) + f(2)g'(2) = (1)(3) + (3)(-2) = 3 - 6 = -3.

15) Which statement about the product rule is FALSE?

The product rule can be extended to three or more functions

The product rule only applies when both functions are polynomials

The product rule requires both functions to be differentiable

The order in which you apply the product rule doesn't affect the final answer

This statement is false because the product rule applies to ANY two differentiable functions, not just polynomials. We can use it with exponential functions, trigonometric functions, logarithmic functions, or any combination. Options a, c, and d are all true statements about the product rule.

Explanation

This statement is false because the product rule applies to ANY two differentiable functions, not just polynomials. We can use it with exponential functions, trigonometric functions, logarithmic functions, or any combination. Options a, c, and d are all true statements about the product rule.