Quotient Rule with Rational Functions

1) What is the derivative of f(x) = (4x² - 3)/(2x + 1)?

(8x(2x + 1) - (4x² - 3)(2))/(2x + 1)²

(8x(2x + 1) + (4x² - 3)(2))/(2x + 1)²

(8x)/(2)

(4x² - 3)/(2)

To find the derivative of f(x) = (4x² - 3)/(2x + 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = 4x² - 3, so f'(x) = 8x. And g(x) = 2x + 1, so g'(x) = 2. Applying the quotient rule: (8x(2x + 1) - (4x² - 3)(2))/(2x + 1)² = (16x² + 8x - 8x² + 6)/(2x + 1)² = (8x² + 8x + 6)/(2x + 1)².

Explanation

To find the derivative of f(x) = (4x² - 3)/(2x + 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = 4x² - 3, so f'(x) = 8x. And g(x) = 2x + 1, so g'(x) = 2. Applying the quotient rule: (8x(2x + 1) - (4x² - 3)(2))/(2x + 1)² = (16x² + 8x - 8x² + 6)/(2x + 1)² = (8x² + 8x + 6)/(2x + 1)².

2) Find the derivative of g(x) = (x³ + 2x² - x)/(x² + 3).

((3x² + 4x - 1)(x² + 3) - (x³ + 2x² - x)(2x))/(x² + 3)²

((3x² + 4x - 1)(x² + 3) + (x³ + 2x² - x)(2x))/(x² + 3)²

(3x² + 4x - 1)/(2x)

(x³ + 2x² - x)/(2x)

To find the derivative of g(x) = (x³ + 2x² - x)/(x² + 3), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x³ + 2x² - x, so f'(x) = 3x² + 4x - 1. And g(x) = x² + 3, so g'(x) = 2x. Applying the quotient rule: ((3x² + 4x - 1)(x² + 3) - (x³ + 2x² - x)(2x))/(x² + 3)².

Explanation

To find the derivative of g(x) = (x³ + 2x² - x)/(x² + 3), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x³ + 2x² - x, so f'(x) = 3x² + 4x - 1. And g(x) = x² + 3, so g'(x) = 2x. Applying the quotient rule: ((3x² + 4x - 1)(x² + 3) - (x³ + 2x² - x)(2x))/(x² + 3)².

3) What is the derivative of h(x) = (tan x)/(x² + 1)?

(sec² x (x² + 1) - tan x (2x))/(x² + 1)²

(sec² x (x² + 1) + tan x (2x))/(x² + 1)²

(sec² x)/(2x)

(tan x)/(2x)

To find the derivative of h(x) = (tan x)/(x² + 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = tan x, so f'(x) = sec² x. And g(x) = x² + 1, so g'(x) = 2x. Applying the quotient rule: (sec² x (x² + 1) - tan x (2x))/(x² + 1)².

Explanation

To find the derivative of h(x) = (tan x)/(x² + 1), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = tan x, so f'(x) = sec² x. And g(x) = x² + 1, so g'(x) = 2x. Applying the quotient rule: (sec² x (x² + 1) - tan x (2x))/(x² + 1)².

4) Find the derivative of f(x) = (eˣ + 1)/(eˣ - 1).

(eˣ(eˣ - 1) - (eˣ + 1)eˣ)/(eˣ - 1)²

(eˣ(eˣ - 1) + (eˣ + 1)eˣ)/(eˣ - 1)²

(eˣ)/(eˣ)

(eˣ + 1)/(eˣ)

To find the derivative of f(x) = (eˣ + 1)/(eˣ - 1), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = eˣ + 1, so f'(x) = eˣ. And g(x) = eˣ - 1, so g'(x) = eˣ. Applying the quotient rule: (eˣ(eˣ - 1) - (eˣ + 1)eˣ)/(eˣ - 1)² = (e²ˣ - eˣ - e²ˣ - eˣ)/(eˣ - 1)² = (-2eˣ)/(eˣ - 1)².

Explanation

To find the derivative of f(x) = (eˣ + 1)/(eˣ - 1), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = eˣ + 1, so f'(x) = eˣ. And g(x) = eˣ - 1, so g'(x) = eˣ. Applying the quotient rule: (eˣ(eˣ - 1) - (eˣ + 1)eˣ)/(eˣ - 1)² = (e²ˣ - eˣ - e²ˣ - eˣ)/(eˣ - 1)² = (-2eˣ)/(eˣ - 1)².

5) What is the derivative of g(x) = (x⁴ - 5x² + 2)/(x³ - x)?

((4x³ - 10x)(x³ - x) - (x⁴ - 5x² + 2)(3x² - 1))/(x³ - x)²

((4x³ - 10x)(x³ - x) + (x⁴ - 5x² + 2)(3x² - 1))/(x³ - x)²

(4x³ - 10x)/(3x² - 1)

(x⁴ - 5x² + 2)/(3x² - 1)

To find the derivative of g(x) = (x⁴ - 5x² + 2)/(x³ - x), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x⁴ - 5x² + 2, so f'(x) = 4x³ - 10x. And g(x) = x³ - x, so g'(x) = 3x² - 1. Applying the quotient rule: ((4x³ - 10x)(x³ - x) - (x⁴ - 5x² + 2)(3x² - 1))/(x³ - x)².

Explanation

To find the derivative of g(x) = (x⁴ - 5x² + 2)/(x³ - x), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x⁴ - 5x² + 2, so f'(x) = 4x³ - 10x. And g(x) = x³ - x, so g'(x) = 3x² - 1. Applying the quotient rule: ((4x³ - 10x)(x³ - x) - (x⁴ - 5x² + 2)(3x² - 1))/(x³ - x)².

6) Find the derivative of h(x) = (√(x))/(x + 4).

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

((1/(2√(x)))(x + 4) + √(x)(1))/(x + 4)²

(1/(2√(x)))/(1)

(√(x))/(1)

To find the derivative of h(x) = (√(x))/(x + 4), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = √(x) = x^(½), so f'(x) = (½)x^(-½) = 1/(2√(x)). And g(x) = x + 4, so g'(x) = 1. Applying the quotient rule: ((1/(2√(x)))(x + 4) - √(x)(1))/(x + 4)².

Explanation

To find the derivative of h(x) = (√(x))/(x + 4), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = √(x) = x^(½), so f'(x) = (½)x^(-½) = 1/(2√(x)). And g(x) = x + 4, so g'(x) = 1. Applying the quotient rule: ((1/(2√(x)))(x + 4) - √(x)(1))/(x + 4)².

7) What is the derivative of f(x) = (3x² + 2x - 1)/(5x - 3)?

((6x + 2)(5x - 3) - (3x² + 2x - 1)(5))/(5x - 3)²

((6x + 2)(5x - 3) + (3x² + 2x - 1)(5))/(5x - 3)²

(6x + 2)/(5)

(3x² + 2x - 1)/(5)

To find the derivative of f(x) = (3x² + 2x - 1)/(5x - 3), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = 3x² + 2x - 1, so f'(x) = 6x + 2. And g(x) = 5x - 3, so g'(x) = 5. Applying the quotient rule: ((6x + 2)(5x - 3) - (3x² + 2x - 1)(5))/(5x - 3)² = (30x² - 18x + 10x - 6 - 15x² - 10x + 5)/(5x - 3)² = (15x² - 18x - 1)/(5x - 3)².

Explanation

To find the derivative of f(x) = (3x² + 2x - 1)/(5x - 3), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = 3x² + 2x - 1, so f'(x) = 6x + 2. And g(x) = 5x - 3, so g'(x) = 5. Applying the quotient rule: ((6x + 2)(5x - 3) - (3x² + 2x - 1)(5))/(5x - 3)² = (30x² - 18x + 10x - 6 - 15x² - 10x + 5)/(5x - 3)² = (15x² - 18x - 1)/(5x - 3)².

8) Find the derivative of g(x) = (x² + sin x)/(cos x).

((2x + cos x)(cos x) - (x² + sin x)(-sin x))/(cos x)²

((2x + cos x)(cos x) + (x² + sin x)(-sin x))/(cos x)²

(2x + cos x)/(-sin x)

(x² + sin x)/(-sin x)

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

Explanation

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

9) What is the derivative of h(x) = (x³ - 8)/(x² + 4)?

((3x²)(x² + 4) - (x³ - 8)(2x))/(x² + 4)²

((3x²)(x² + 4) + (x³ - 8)(2x))/(x² + 4)²

(3x²)/(2x)

(x³ - 8)/(2x)

To find the derivative of h(x) = (x³ - 8)/(x² + 4), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x³ - 8, so f'(x) = 3x². And g(x) = x² + 4, so g'(x) = 2x. Applying the quotient rule: ((3x²)(x² + 4) - (x³ - 8)(2x))/(x² + 4)² = (3x⁴ + 12x² - 2x⁴ + 16x)/(x² + 4)² = (x⁴ + 12x² + 16x)/(x² + 4)².

Explanation

To find the derivative of h(x) = (x³ - 8)/(x² + 4), we use the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = x³ - 8, so f'(x) = 3x². And g(x) = x² + 4, so g'(x) = 2x. Applying the quotient rule: ((3x²)(x² + 4) - (x³ - 8)(2x))/(x² + 4)² = (3x⁴ + 12x² - 2x⁴ + 16x)/(x² + 4)² = (x⁴ + 12x² + 16x)/(x² + 4)².

10) Find the derivative of f(x) = (log_2 x)/(x³).

((1/(x ln 2))(x³) - (log_2 x)(3x²))/(x³)²

((1/(x ln 2))(x³) + (log_2 x)(3x²))/(x³)²

(1/(x ln 2))/(3x²)

(log_2 x)/(3x²)

To find the derivative of f(x) = (log_2 x)/(x³), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = log_2 x, so f'(x) = 1/(x ln 2). And g(x) = x³, so g'(x) = 3x². Applying the quotient rule: ((1/(x ln 2))(x³) - (log_2 x)(3x²))/(x³)² = (x²/(ln 2) - 3x² log_2 x)/x⁶.

Explanation

To find the derivative of f(x) = (log_2 x)/(x³), we apply the quotient rule: (f/g)' = (f'g - fg')/g². Here, f(x) = log_2 x, so f'(x) = 1/(x ln 2). And g(x) = x³, so g'(x) = 3x². Applying the quotient rule: ((1/(x ln 2))(x³) - (log_2 x)(3x²))/(x³)² = (x²/(ln 2) - 3x² log_2 x)/x⁶.

11) When finding the derivative of a quotient, what is the purpose of squaring the denominator in the quotient rule formula?

To ensure the denominator is always positive

To maintain the correct units of the derivative

To properly account for the rate of change of the denominator

To simplify the algebraic expression

The denominator is squared in the quotient rule formula to properly account for the rate of change of the denominator function. When deriving the quotient rule from the limit definition, the denominator g(x+h)g(x) appears, which becomes g(x)² in the limit as h approaches 0. This squaring is mathematically necessary to correctly represent how changes in both the numerator and denominator affect the overall rate of change of the quotient.

Explanation

The denominator is squared in the quotient rule formula to properly account for the rate of change of the denominator function. When deriving the quotient rule from the limit definition, the denominator g(x+h)g(x) appears, which becomes g(x)² in the limit as h approaches 0. This squaring is mathematically necessary to correctly represent how changes in both the numerator and denominator affect the overall rate of change of the quotient.

12) If the denominator of a quotient is a constant function, how does the quotient rule simplify?

It reduces to the power rule

It becomes equivalent to the constant multiple rule

It simplifies to just the derivative of the numerator divided by the constant

It becomes the product rule with a negative exponent

If the denominator g(x) is a constant function c, then g'(x) = 0. Applying the quotient rule: (f/g)' = (f'g - fg')/g² = (f'·c - f·0)/c² = (f'·c)/c² = f'/c. This means the derivative of f(x)/c is simply f'(x)/c, which is the derivative of the numerator divided by the constant denominator. This is consistent with the constant multiple rule of differentiation.

Explanation

If the denominator g(x) is a constant function c, then g'(x) = 0. Applying the quotient rule: (f/g)' = (f'g - fg')/g² = (f'·c - f·0)/c² = (f'·c)/c² = f'/c. This means the derivative of f(x)/c is simply f'(x)/c, which is the derivative of the numerator divided by the constant denominator. This is consistent with the constant multiple rule of differentiation.

13) In what scenario would the quotient rule be more efficient to use than the product rule with a negative exponent?

When the denominator is a simple monomial

When the denominator is a complex expression with multiple terms

When the numerator is a constant

When both numerator and denominator are trigonometric functions

The quotient rule is generally more efficient than using the product rule with a negative exponent when the denominator is a complex expression with multiple terms. In such cases, applying the product rule would require differentiating a negative power of a complex expression, which involves the chain rule and can lead to more complicated algebra. The quotient rule provides a direct approach that avoids these additional complications.

Explanation

The quotient rule is generally more efficient than using the product rule with a negative exponent when the denominator is a complex expression with multiple terms. In such cases, applying the product rule would require differentiating a negative power of a complex expression, which involves the chain rule and can lead to more complicated algebra. The quotient rule provides a direct approach that avoids these additional complications.

14) What is the relationship between the quotient rule and the product rule?

The quotient rule is a special case of the product rule

The product rule is a special case of the quotient rule

They are independent rules with no relationship

They can both be derived from the chain rule

The quotient rule can be derived from the product rule by expressing a quotient f(x)/g(x) as f(x)[g(x)]^(-1) and then applying the product rule and chain rule. This shows that the quotient rule is not independent of the product rule but can be obtained from it. The product rule is more fundamental in this relationship, as the quotient rule relies on it for its derivation.

Explanation

The quotient rule can be derived from the product rule by expressing a quotient f(x)/g(x) as f(x)[g(x)]^(-1) and then applying the product rule and chain rule. This shows that the quotient rule is not independent of the product rule but can be obtained from it. The product rule is more fundamental in this relationship, as the quotient rule relies on it for its derivation.

15) When applying the quotient rule to find the derivative of f(x)/g(x), what does the term f'g in the numerator represent?

The rate of change of the numerator assuming the denominator is constant

The rate of change of the denominator assuming the numerator is constant

The product of the derivatives of the numerator and denominator

The rate of change of the entire quotient function

In the quotient rule formula (f'g - fg')/g², the term f'g represents the rate of change of the numerator f(x) (which is f') multiplied by the denominator g(x). This term captures how the quotient would change if only the numerator were changing while the denominator remained constant. The subtraction of fg' then adjusts for the fact that the denominator is also changing, giving us the complete rate of change of the quotient.

Explanation

In the quotient rule formula (f'g - fg')/g², the term f'g represents the rate of change of the numerator f(x) (which is f') multiplied by the denominator g(x). This term captures how the quotient would change if only the numerator were changing while the denominator remained constant. The subtraction of fg' then adjusts for the fact that the denominator is also changing, giving us the complete rate of change of the quotient.