Quotient Rule Mastery & Applications

Using the Quotient Rule: If f(x) = u(x)/v(x), then f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]². Here u(x) = x² + 3 and u'(x) = 2x. v(x) = x + 1 and v'(x) = 1. Substituting gives [2x(x + 1) - (x² + 3)(1)] / (x + 1)².

For the Quotient Rule, we need u'(x)v(x) - u(x)v'(x) in the numerator. u(x) = 5x³ - 2x gives u'(x) = 15x² - 2. v(x) = x² + 4 gives v'(x) = 2x. The correct setup is [(15x² - 2)(x² + 4) - (5x³ - 2x)(2x)] divided by (x² + 4)².

u(x) = sin(x) so u'(x) = cos(x). v(x) = cos(x) so v'(x) = -sin(x). Applying the Quotient Rule formula [u'(x)v(x) - u(x)v'(x)] / v(x)² gives [cos(x)cos(x) - sin(x)(-sin(x))] / cos²(x).

u(x) = eˣ gives u'(x) = eˣ. v(x) = x³ + 2x gives v'(x) = 3x² + 2. The Quotient Rule requires subtraction in the numerator: [u'(x)v(x) - u(x)v'(x)] / v(x)². Therefore we get [eˣ(x³ + 2x) - eˣ(3x² + 2)] / (x³ + 2x)². A common error is adding instead of subtracting.

Rewrite √x as x^(½). Then u(x) = x^(½) gives u'(x) = (½)x^(-½) = 1/(2√x). v(x) = x - 5 gives v'(x) = 1. Applying the Quotient Rule: [u'(x)v(x) - u(x)v'(x)] / v(x)² = [(1/(2√x))(x - 5) - √x(1)] / (x - 5)².

First find the general derivative using Quotient Rule: u(x) = x² + 1 gives u'(x) = 2x. v(x) = x - 3 gives v'(x) = 1. The derivative is [2x(x - 3) - (x² + 1)(1)] / (x - 3)². Then substitute x = 2: [4(2 - 3) - (4 + 1)(1)] / (2 - 3)² = [4(-1) - 5] / 1 = -9.

This requires both Chain Rule and Quotient Rule. u(x) = (3x² + 2)⁴. By Chain Rule, u'(x) = 4(3x² + 2)³(6x). v(x) = x³ - 1 gives v'(x) = 3x². Applying Quotient Rule: [u'(x)v(x) - u(x)v'(x)] / v(x)² = [4(3x² + 2)³(6x)(x³ - 1) - (3x² + 2)⁴(3x²)] / (x³ - 1)².

u(x) = x² + 5 gives u'(x) = 2x. v(x) = √(x + 2) = (x + 2)^(½). By Chain Rule, v'(x) = (½)(x + 2)^(-½) = 1/(2√(x + 2)). The Quotient Rule gives [2x√(x + 2) - (x² + 5)(1/(2√(x + 2)))] / [√(x + 2)]². Since [√(x + 2)]² = x + 2, the denominator simplifies to (x + 2).

The Quotient Rule formula is (low x derivative of high - high x derivative of low) over low squared. In mathematical notation: if f(x) = u(x)/v(x), then f'(x) = [v(x)u'(x) - u(x)v'(x)] / [v(x)]². The order of subtraction matters because the denominator's derivative is subtracted from the numerator's derivative times the denominator.

The Quotient Rule requires subtraction in the numerator: [u'(x)v(x) - u(x)v'(x)] / [v(x)]². The student's solution shows addition between the two terms in the numerator. The correct derivative should be [3x²(x² - 1) - (x³ + 2)(2x)] / (x² - 1)². This is a common mistake when applying the Quotient Rule.

The Quotient Rule is specifically designed for functions that are ratios of two differentiable functions (one divided by another). Option C is explicitly written as a fraction with numerator 4x + 5 and denominator x² - 3x. Option A would use Product Rule, B would use Chain Rule, and D is a composite function requiring Chain Rule.

To find the second derivative f''(x), you must first find the first derivative f'(x). For a quotient function, this means applying the Quotient Rule (or Product Rule with negative exponents) to differentiate f(x) once. Only after obtaining f'(x) can you then differentiate it again to get f''(x). The order of operations matters - you cannot skip directly to the second derivative.

The Quotient Rule at a specific point uses the formula [p'(c)q(c) - p(c)q'(c)] / [q(c)]². Substituting c = 2, p(2) = 5, p'(2) = -1, q(2) = 3, q'(2) = 4 gives [(-1)(3) - (5)(4)] / (3)² = (-3 - 20) / 9 = -23/9. This tests understanding of the Quotient Rule's structure without requiring symbolic differentiation.

The form [g'(x)h(x) - g(x)h'(x)] / [h(x)]² is the structural template for the Quotient Rule. This pattern specifically identifies a function that is a ratio or quotient of two functions, where g(x) is the numerator function and h(x) is the denominator function. Option B, f(x) = g(x) / h(x), matches this structure exactly. The other options represent product, composition, and sum respectively.

Recognizing that x³ - 8 is a difference of cubes allows factoring: x³ - 8 = (x - 2)(x² + 2x + 4). The original function becomes f(x) = [(x - 2)(x² + 2x + 4)] / (x - 2). For x ≠ 2, the (x - 2) terms cancel, leaving f(x) = x² + 2x + 4, which is much simpler to differentiate directly using the Power Rule. This demonstrates an important conceptual skill: looking for algebraic simplifications before automatically applying differentiation rules.