Quotient Rule Basics

1) Find the derivative of f(x) = (sin x)/(x + 1).

(cos x (x + 1) - sin x (1))/(x + 1)²

(cos x (x + 1) + sin x (1))/(x + 1)²

(cos x)/(1)

(sin x)/(1)

To find the derivative of f(x) = (sin x)/(x + 1), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = sin x, so f'(x) = cos x. And g(x) = x + 1, so g'(x) = 1. Applying the quotient rule: (cos x (x + 1) - sin x (1))/(x + 1)².

Explanation

To find the derivative of f(x) = (sin x)/(x + 1), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = sin x, so f'(x) = cos x. And g(x) = x + 1, so g'(x) = 1. Applying the quotient rule: (cos x (x + 1) - sin x (1))/(x + 1)².

2) Find the derivative of h(x) = (2x + 5)/(3x - 2).

(2(3x - 2) - (2x + 5)(3))/(3x - 2)²

(2(3x - 2) + (2x + 5)(3))/(3x - 2)²

(2)/(3)

(2x + 5)/(3)

To find the derivative of h(x) = (2x + 5)/(3x - 2), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = 2x + 5, so f'(x) = 2. And g(x) = 3x - 2, so g'(x) = 3. Applying the quotient rule: (2(3x - 2) - (2x + 5)(3))/(3x - 2)² = (6x - 4 - 6x - 15)/(3x - 2)² = (-19)/(3x - 2)².

Explanation

To find the derivative of h(x) = (2x + 5)/(3x - 2), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = 2x + 5, so f'(x) = 2. And g(x) = 3x - 2, so g'(x) = 3. Applying the quotient rule: (2(3x - 2) - (2x + 5)(3))/(3x - 2)² = (6x - 4 - 6x - 15)/(3x - 2)² = (-19)/(3x - 2)².

3) What is the derivative of f(x) = (x³ - 2x)/(x² + 1)?

((3x² - 2)(x² + 1) - (x³ - 2x)(2x))/(x² + 1)²

((3x² - 2)(x² + 1) + (x³ - 2x)(2x))/(x² + 1)²

(3x² - 2)/(2x)

(x³ - 2x)/(2x)

To find the derivative of f(x) = (x³ - 2x)/(x² + 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x³ - 2x, so f'(x) = 3x² - 2. And g(x) = x² + 1, so g'(x) = 2x. Applying the quotient rule: ((3x² - 2)(x² + 1) - (x³ - 2x)(2x))/(x² + 1)².

Explanation

To find the derivative of f(x) = (x³ - 2x)/(x² + 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x³ - 2x, so f'(x) = 3x² - 2. And g(x) = x² + 1, so g'(x) = 2x. Applying the quotient rule: ((3x² - 2)(x² + 1) - (x³ - 2x)(2x))/(x² + 1)².

4) Find the derivative of g(x) = (ln x)/(x²).

((1/x)(x²) - (ln x)(2x))/(x²)²

((1/x)(x²) + (ln x)(2x))/(x²)²

(1/x)/(2x)

(ln x)/(2x)

To find the derivative of g(x) = (ln x)/(x²), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = ln x, so f'(x) = 1/x. And g(x) = x², so g'(x) = 2x. Applying the quotient rule: ((1/x)(x²) - (ln x)(2x))/(x²)² = (x - 2x ln x)/x⁴.

Explanation

To find the derivative of g(x) = (ln x)/(x²), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = ln x, so f'(x) = 1/x. And g(x) = x², so g'(x) = 2x. Applying the quotient rule: ((1/x)(x²) - (ln x)(2x))/(x²)² = (x - 2x ln x)/x⁴.

5) What is the derivative of h(x) = (x² + 3x - 4)/(2x - 5)?

((2x + 3)(2x - 5) - (x² + 3x - 4)(2))/(2x - 5)²

((2x + 3)(2x - 5) + (x² + 3x - 4)(2))/(2x - 5)²

(2x + 3)/(2)

(x² + 3x - 4)/(2)

To find the derivative of h(x) = (x² + 3x - 4)/(2x - 5), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x² + 3x - 4, so f'(x) = 2x + 3. And g(x) = 2x - 5, so g'(x) = 2. Applying the quotient rule: ((2x + 3)(2x - 5) - (x² + 3x - 4)(2))/(2x - 5)².

Explanation

To find the derivative of h(x) = (x² + 3x - 4)/(2x - 5), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x² + 3x - 4, so f'(x) = 2x + 3. And g(x) = 2x - 5, so g'(x) = 2. Applying the quotient rule: ((2x + 3)(2x - 5) - (x² + 3x - 4)(2))/(2x - 5)².

6) Find the derivative of f(x) = (cos x)/(sin x).

-1 /(sin x)²

-1/(cos x)²

(-sin x)/(cos x)

(cos x)/(sin x)

We use the quotient rule: (f/g)' = (f'g - fg')/g², where f(x) = cos x and g(x) = sin x. So f'(x) = -sin x and g'(x) = cos x. Applying the quotient rule: (-sin x (sin x) - cos x (cos x))/(sin x)² = (-sin² x - cos² x)/sin² x = -(sin² x + cos² x)/sin² x= -1/sin² x.

Explanation

We use the quotient rule: (f/g)' = (f'g - fg')/g², where f(x) = cos x and g(x) = sin x. So f'(x) = -sin x and g'(x) = cos x. Applying the quotient rule: (-sin x (sin x) - cos x (cos x))/(sin x)² = (-sin² x - cos² x)/sin² x = -(sin² x + cos² x)/sin² x= -1/sin² x.

7) When finding the derivative of a quotient, which of the following is the correct formula?

(f/g)' = (f'g - fg')/g²

(f/g)' = (f'g + fg')/g²

(f/g)' = (f'g - fg')/g

(f/g)' = (fg' - f'g)/g²

The quotient rule states that the derivative of a quotient f(x)/g(x) is given by (f/g)' = (f'g - fg')/g². This formula is essential for finding the derivative of functions that are expressed as one function divided by another. The numerator is the derivative of the top function times the bottom function minus the top function times the derivative of the bottom function, all divided by the bottom function squared.

Explanation

The quotient rule states that the derivative of a quotient f(x)/g(x) is given by (f/g)' = (f'g - fg')/g². This formula is essential for finding the derivative of functions that are expressed as one function divided by another. The numerator is the derivative of the top function times the bottom function minus the top function times the derivative of the bottom function, all divided by the bottom function squared.

8) If f(x) and g(x) are both differentiable functions and g(x) ≠ 0, what condition would make the derivative of f(x)/g(x) equal to zero?

F'(x)g(x) = f(x)g'(x)

F'(x) = 0 and g'(x) = 0

F(x) = 0

G(x) = 1

Using the quotient rule, the derivative of f(x)/g(x) is (f'g - fg')/g². For this derivative to equal zero, the numerator must equal zero (as long as the denominator is not zero). Therefore, f'(x)g(x) - f(x)g'(x) = 0, which simplifies to f'(x)g(x) = f(x)g'(x). This condition means that the rate of change of the numerator relative to the denominator is balanced, resulting in a zero derivative for the quotient.

Explanation

Using the quotient rule, the derivative of f(x)/g(x) is (f'g - fg')/g². For this derivative to equal zero, the numerator must equal zero (as long as the denominator is not zero). Therefore, f'(x)g(x) - f(x)g'(x) = 0, which simplifies to f'(x)g(x) = f(x)g'(x). This condition means that the rate of change of the numerator relative to the denominator is balanced, resulting in a zero derivative for the quotient.

9) When applying the quotient rule, what common error might lead to an incorrect result?

Forgetting to square the denominator

Adding instead of subtracting in the numerator

Differentiating the denominator incorrectly

All of the above

When applying the quotient rule, several common errors can occur: (1) Forgetting to square the denominator in the final result, (2) Adding instead of subtracting in the numerator (using f'g + fg' instead of f'g - fg'), and (3) Differentiating either the numerator or denominator incorrectly. All of these errors will lead to an incorrect result when applying the quotient rule.

Explanation

When applying the quotient rule, several common errors can occur: (1) Forgetting to square the denominator in the final result, (2) Adding instead of subtracting in the numerator (using f'g + fg' instead of f'g - fg'), and (3) Differentiating either the numerator or denominator incorrectly. All of these errors will lead to an incorrect result when applying the quotient rule.

10) The quotient rule can be derived from which other differentiation rules?

Product rule and chain rule

Power rule and product rule

Chain rule only

Product rule only

The quotient rule can be derived by expressing the quotient f(x)/g(x) as f(x)[g(x)]^(-1) and then applying the product rule and chain rule. Using the product rule, we get f'(x)[g(x)]^(-1) + f(x)(-1)[g(x)]^(-2)g'(x), which simplifies to (f'(x)g(x) - f(x)g'(x))/[g(x)]², the quotient rule. This derivation shows how the quotient rule is fundamentally connected to both the product rule and the chain rule.

Explanation

The quotient rule can be derived by expressing the quotient f(x)/g(x) as f(x)[g(x)]^(-1) and then applying the product rule and chain rule. Using the product rule, we get f'(x)[g(x)]^(-1) + f(x)(-1)[g(x)]^(-2)g'(x), which simplifies to (f'(x)g(x) - f(x)g'(x))/[g(x)]², the quotient rule. This derivation shows how the quotient rule is fundamentally connected to both the product rule and the chain rule.

11) What is the derivative of f(x) = (x² + 3)/(x - 1)?

(2x(x - 1) - (x² + 3))/(x - 1)²

(2x(x - 1) + (x² + 3))/(x - 1)²

(2x)/(x - 1)

(x² + 3)/(x - 1)²

To find the derivative of f(x) = (x² + 3)/(x - 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x² + 3, so f'(x) = 2x. And g(x) = x - 1, so g'(x) = 1. Applying the quotient rule: (2x(x - 1) - (x² + 3)(1))/(x - 1)² = (2x² - 2x - x² - 3)/(x - 1)² = (x² - 2x - 3)/(x - 1)².

Explanation

To find the derivative of f(x) = (x² + 3)/(x - 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x² + 3, so f'(x) = 2x. And g(x) = x - 1, so g'(x) = 1. Applying the quotient rule: (2x(x - 1) - (x² + 3)(1))/(x - 1)² = (2x² - 2x - x² - 3)/(x - 1)² = (x² - 2x - 3)/(x - 1)².

12) Find the derivative of g(x) = (3x + 2)/(x² - 4).

(3(x² - 4) - (3x + 2)(2x))/(x² - 4)²

(3(x² - 4) + (3x + 2)(2x))/(x² - 4)²

(3)/(2x)

(3x + 2)/(2x)

To find the derivative of g(x) = (3x + 2)/(x² - 4), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = 3x + 2, so f'(x) = 3. And g(x) = x² - 4, so g'(x) = 2x. Applying the quotient rule: (3(x² - 4) - (3x + 2)(2x))/(x² - 4)² = (3x² - 12 - 6x² - 4x)/(x² - 4)² = (-3x² - 4x - 12)/(x² - 4)².

Explanation

To find the derivative of g(x) = (3x + 2)/(x² - 4), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = 3x + 2, so f'(x) = 3. And g(x) = x² - 4, so g'(x) = 2x. Applying the quotient rule: (3(x² - 4) - (3x + 2)(2x))/(x² - 4)² = (3x² - 12 - 6x² - 4x)/(x² - 4)² = (-3x² - 4x - 12)/(x² - 4)².

13) What is the derivative of h(x) = (eˣ)/(x³)?

(eˣ * x³ - eˣ * 3x²)/(x³)²

(eˣ * x³ + eˣ * 3x²)/(x³)²

(eˣ)/(3x²)

(eˣ * 3x²)/(x³)

To find the derivative of h(x) = (eˣ)/(x³), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = eˣ, so f'(x) = eˣ. And g(x) = x³, so g'(x) = 3x². Applying the quotient rule: (eˣ * x³ - eˣ * 3x²)/(x³)² = (eˣ * x³ - eˣ * 3x²)/x⁶.

Explanation

To find the derivative of h(x) = (eˣ)/(x³), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = eˣ, so f'(x) = eˣ. And g(x) = x³, so g'(x) = 3x². Applying the quotient rule: (eˣ * x³ - eˣ * 3x²)/(x³)² = (eˣ * x³ - eˣ * 3x²)/x⁶.

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

((2x + 4)(x² - 1) - (x² + 4x + 4)(2x))/(x² - 1)²

((2x + 4)(x² - 1) + (x² + 4x + 4)(2x))/(x² - 1)²

(2x + 4)/(2x)

(x² + 4x + 4)/(2x)

To find the derivative of g(x) = (x² + 4x + 4)/(x² - 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x² + 4x + 4, so f'(x) = 2x + 4. And g(x) = x² - 1, so g'(x) = 2x. Applying the quotient rule: ((2x + 4)(x² - 1) - (x² + 4x + 4)(2x))/(x² - 1)².

Explanation

To find the derivative of g(x) = (x² + 4x + 4)/(x² - 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x² + 4x + 4, so f'(x) = 2x + 4. And g(x) = x² - 1, so g'(x) = 2x. Applying the quotient rule: ((2x + 4)(x² - 1) - (x² + 4x + 4)(2x))/(x² - 1)².

15) Why is the quotient rule necessary instead of just using the product rule with a negative exponent?

The quotient rule provides a direct method without rewriting the function

The quotient rule always gives a simpler result

The product rule cannot be applied to quotients

The quotient rule avoids chain rule complications

While you could technically rewrite a quotient f(x)/g(x) as f(x)[g(x)]^(-1) and then apply the product rule and chain rule, this approach often requires more steps and can lead to more complex algebraic manipulations. The quotient rule provides a direct method specifically designed for finding derivatives of quotients without having to rewrite the function first. This makes the process more efficient and less prone to errors.

Explanation

While you could technically rewrite a quotient f(x)/g(x) as f(x)[g(x)]^(-1) and then apply the product rule and chain rule, this approach often requires more steps and can lead to more complex algebraic manipulations. The quotient rule provides a direct method specifically designed for finding derivatives of quotients without having to rewrite the function first. This makes the process more efficient and less prone to errors.