Quotient Rule with Rational Functions

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.