Product Rule With Trig, Exponential & Log Functions

1) What is the derivative of f(x) = (2x - 3)(4x + 1)?

2(4x + 1) + (2x - 3)·4

8x - 2

2·4

8x² - 10x - 3

Using the product rule with u = 2x - 3 and v = 4x + 1:

u' = 2

v' = 4

f'(x) = u'·v + u·v' = 2(4x + 1) + (2x - 3)·4 This simplifies to 8x + 2 + 8x - 12 = 16x - 10, but the product rule application is shown in option a.

Explanation

Using the product rule with u = 2x - 3 and v = 4x + 1:

u' = 2

v' = 4

f'(x) = u'·v + u·v' = 2(4x + 1) + (2x - 3)·4 This simplifies to 8x + 2 + 8x - 12 = 16x - 10, but the product rule application is shown in option a.

2) Find the derivative of f(x) = x²·sin(x).

2x·sin(x) + x²·cos(x)

2x·cos(x) + x²·sin(x)

2x·sin(x)

X²·cos(x)

Using the product rule with u = x² and v = sin(x):

u' = 2x

v' = cos(x)

f'(x) = u'·v + u·v' = 2x·sin(x) + x²·cos(x)

Explanation

Using the product rule with u = x² and v = sin(x):

u' = 2x

v' = cos(x)

f'(x) = u'·v + u·v' = 2x·sin(x) + x²·cos(x)

3) What is the derivative of g(x) = (3x + 1)(x² - 4)?

3(x² - 4) + (3x + 1)(2x)

3x + 2x

9x² + 2x - 12

(3x + 1)(2x)

Using the product rule with u = 3x + 1 and v = x² - 4:

u' = 3

v' = 2x

g'(x) = u'·v + u·v' = 3(x² - 4) + (3x + 1)(2x) Note: This can be simplified further to 9x² + 2x - 12, but the product rule application is shown in option a.

Explanation

Using the product rule with u = 3x + 1 and v = x² - 4:

u' = 3

v' = 2x

g'(x) = u'·v + u·v' = 3(x² - 4) + (3x + 1)(2x) Note: This can be simplified further to 9x² + 2x - 12, but the product rule application is shown in option a.

4) Find the derivative of h(x) = eˣ·x³.

Eˣ·x³ + 3x²

Eˣ·3x²

Eˣ·x³ + eˣ·3x²

Eˣ + 3x²

Using the product rule with u = eˣ and v = x³:

u' = eˣ (derivative of eˣ is itself)

v' = 3x²

h'(x) = u'·v + u·v' = eˣ·x³ + eˣ·3x²

Explanation

Using the product rule with u = eˣ and v = x³:

u' = eˣ (derivative of eˣ is itself)

v' = 3x²

h'(x) = u'·v + u·v' = eˣ·x³ + eˣ·3x²

5) What is the derivative of f(x) = (x + 5)(x - 2)?

1 + 1

(x + 5) + (x - 2)

1(x - 2) + (x + 5)·1

2x + 3

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

u' = 1

v' = 1

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

Explanation

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

u' = 1

v' = 1

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

6) If f(x) = x·cos(x), what is f'(π/2)?

-π/2

12/12

1/1

π/2

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

u = x, u' = 1

v = cos(x), v' = -sin(x)

f'(x) = 1·cos(x) + x·(-sin(x)) = cos(x) - x·sin(x)

Now evaluate at x = π/2, f'(π/2) = cos(π/2) - (π/2)·sin(π/2) = 0 - (π/2)(1) = -π/2

Explanation

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

u = x, u' = 1

v = cos(x), v' = -sin(x)

f'(x) = 1·cos(x) + x·(-sin(x)) = cos(x) - x·sin(x)

Now evaluate at x = π/2, f'(π/2) = cos(π/2) - (π/2)·sin(π/2) = 0 - (π/2)(1) = -π/2

7) Find the derivative of y = x²·ln(x).

2x·ln(x) + x²·(1/x)

2x·ln(x)

X²/x

2x·ln(x) + x

Using the product rule with u = x² and v = ln(x):

u' = 2x

v' = 1/x

y' = u'·v + u·v' = 2x·ln(x) + x²·(1/x) This can be simplified to 2x·ln(x) + x, but the product rule application is shown in option a.

Explanation

Using the product rule with u = x² and v = ln(x):

u' = 2x

v' = 1/x

y' = u'·v + u·v' = 2x·ln(x) + x²·(1/x) This can be simplified to 2x·ln(x) + x, but the product rule application is shown in option a.

8) What is the derivative of f(x) = (x² + 3x)(x³ - 2)?

(2x + 3)(x³ - 2) + (x² + 3x)(3x²)

(2x + 3)(3x²)

5x⁴ + 9x³ - 4x - 6

(x² + 3x)(3x²)

Using the product rule with u = x² + 3x and v = x³ - 2:

u' = 2x + 3

v' = 3x²

f'(x) = u'·v + u·v' = (2x + 3)(x³ - 2) + (x² + 3x)(3x²)

Explanation

Using the product rule with u = x² + 3x and v = x³ - 2:

u' = 2x + 3

v' = 3x²

f'(x) = u'·v + u·v' = (2x + 3)(x³ - 2) + (x² + 3x)(3x²)

9) The graph of u(x) has slope 2; v(x) = 4. If p(x) = u(x)v(x), what is p'(3)?

0

8

12

24

From the descriptions, u(x) = 2x (so u'(x) = 2) and v(x) = 4 (so v'(x) = 0). Using the Product Rule: p'(x) = u'(x)v(x) + u(x)v'(x) = (2)(4) + (2x)(0) = 8. Alternatively, p(x) = 8x, so p'(x) = 8. The derivative is constant everywhere, including at x = 3.

Explanation

From the descriptions, u(x) = 2x (so u'(x) = 2) and v(x) = 4 (so v'(x) = 0). Using the Product Rule: p'(x) = u'(x)v(x) + u(x)v'(x) = (2)(4) + (2x)(0) = 8. Alternatively, p(x) = 8x, so p'(x) = 8. The derivative is constant everywhere, including at x = 3.

10) Use the Product Rule to find the derivative of g(x)=x⁴·√x.

4x³·√x + x⁴·(1/2√x)

4x³·√x

X⁴/(2√x)

4x³ + 1/(2√x)

Using the product rule with u = x⁴ and v = √x = x^(½):

u' = 4x³

v' = (½)x^(-½) = 1/(2√x)

g'(x) = u'·v + u·v' = 4x³·√x + x⁴·(1/(2√x))

Explanation

Using the product rule with u = x⁴ and v = √x = x^(½):

u' = 4x³

v' = (½)x^(-½) = 1/(2√x)

g'(x) = u'·v + u·v' = 4x³·√x + x⁴·(1/(2√x))

11) If h(x) = (4x⁸ - x⁶ + 2x⁴)(3x⁵ + 2x³ - x), what is h'(x)?

(32x⁷ - 6x⁵ + 8x³)(3x⁵ + 2x³ - x) + (4x⁸ - x⁶ + 2x⁴)(15x⁴ + 6x² - 1)

(32x⁷ - 6x⁵ + 8x³)(15x⁴ + 6x² - 1) + (4x⁸ - x⁶ + 2x⁴)(3x⁵ + 2x³ - x)

96x¹² + 64x¹⁰ - 32x⁸ - 18x¹⁰ - 12x⁸ + 6x⁶ + 24x⁸ + 16x⁶ - 8x⁴

(32x⁷ - 6x⁵ + 8x³)(3x⁵ + 2x³ - x) - (4x⁸ - x⁶ + 2x⁴)(15x⁴ + 6x² - 1)

Using the Product Rule with u(x) = 4x⁸ - x⁶ + 2x⁴, u'(x) = 32x⁷ - 6x⁵ + 8x³, and v(x) = 3x⁵ + 2x³ - x, v'(x) = 15x⁴ + 6x² - 1, we have

h'(x) = u'(x)v(x) + u(x)v'(x) = (32x⁷ - 6x⁵ + 8x³)(3x⁵ + 2x³ - x) + (4x⁸ - x⁶ + 2x⁴)(15x⁴ + 6x² - 1)

Explanation

Using the Product Rule with u(x) = 4x⁸ - x⁶ + 2x⁴, u'(x) = 32x⁷ - 6x⁵ + 8x³, and v(x) = 3x⁵ + 2x³ - x, v'(x) = 15x⁴ + 6x² - 1, we have

h'(x) = u'(x)v(x) + u(x)v'(x) = (32x⁷ - 6x⁵ + 8x³)(3x⁵ + 2x³ - x) + (4x⁸ - x⁶ + 2x⁴)(15x⁴ + 6x² - 1)

12) When using the product rule to differentiate f(x)=g(x)·h(x), which statement is TRUE?

You only need to differentiate the first function

Both functions must be differentiable at the point in question

The order of multiplication doesn't matter, so g'(x)h(x)+g(x)h'(x)=h(x)g'(x)+h'(x)g(x)

You can choose to differentiate only one of the two functions

For the product rule to be applied, both g(x) and h(x) must be differentiable at the point where we're finding the derivative. If either function is not differentiable at that point, the product rule cannot be used. Option c is also mathematically true due to commutativity of multiplication, but option b is the more fundamental requirement for applying the product rule.

Explanation

For the product rule to be applied, both g(x) and h(x) must be differentiable at the point where we're finding the derivative. If either function is not differentiable at that point, the product rule cannot be used. Option c is also mathematically true due to commutativity of multiplication, but option b is the more fundamental requirement for applying the product rule.

13) What is the derivative of f(x)=sin(x)·cos(x)?

Cos(x)·cos(x) + sin(x)·(-sin(x))

Cos(x)·cos(x) - sin(x)·sin(x)

Cos²(x) - sin²(x)

All of the above

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

u' = cos(x)

v' = -sin(x)

f'(x) = cos(x)·cos(x) + sin(x)·(-sin(x)) (option a)

This equals cos(x)·cos(x) - sin(x)·sin(x) (option b)

Which can also be written as cos²(x) - sin²(x) (option c) All three expressions are equivalent, so all answers are correct.

Explanation

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

u' = cos(x)

v' = -sin(x)

f'(x) = cos(x)·cos(x) + sin(x)·(-sin(x)) (option a)

This equals cos(x)·cos(x) - sin(x)·sin(x) (option b)

Which can also be written as cos²(x) - sin²(x) (option c) All three expressions are equivalent, so all answers are correct.

14) Why might you choose the product rule instead of expanding first?

The product rule is always faster

When expanding would create a more complicated expression

You must always use the product rule

When both functions are polynomials

Sometimes expanding a product before differentiating leads to a more complicated expression, especially with functions like x²·eˣ or x·sin(x). In these cases, using the product rule directly is more efficient. For simple polynomials, expanding first might be easier, but for transcendental functions (exponential, logarithmic, trigonometric), the product rule is often the better choice.

Explanation

Sometimes expanding a product before differentiating leads to a more complicated expression, especially with functions like x²·eˣ or x·sin(x). In these cases, using the product rule directly is more efficient. For simple polynomials, expanding first might be easier, but for transcendental functions (exponential, logarithmic, trigonometric), the product rule is often the better choice.

15) Find the equation of the tangent line to f(x)=x·sin(x) at x=π.

Y = -π(x - π)

Y = π(x - π)

Y = -x + π

Y = 0

First, find the point: f(π) = π·sin(π) = π(0) = 0. The point is (π, 0). Next, find the derivative using the Product Rule: f'(x) = 1·sin(x) + x·cos(x). Evaluate at x = π: f'(π) = sin(π) + π·cos(π) = 0 + π(-1) = -π. The slope is -π. Using point-slope form: y - 0 = -π(x - π).

Explanation

First, find the point: f(π) = π·sin(π) = π(0) = 0. The point is (π, 0). Next, find the derivative using the Product Rule: f'(x) = 1·sin(x) + x·cos(x). Evaluate at x = π: f'(π) = sin(π) + π·cos(π) = 0 + π(-1) = -π. The slope is -π. Using point-slope form: y - 0 = -π(x - π).