The quotient rule is needed to find q'(x) for q(x) = ln x / x³ where x > 0. Identify u = ln x with u' = 1/x and v = x³ with v' = 3x². Substituting into the rule produces [(1/x) x³ - (ln x) 3x²] / (x³)², as shown in choice A. This simplifies to (x² - 3x² ln x) / $x^6$ if desired. Choice C reverses the terms in the numerator, changing the sign and thus the derivative's value. A useful strategy is to write out the quotient rule formula explicitly before substituting to ensure correct application every time.

This problem requires the quotient rule to find the derivative of R(t) = (t² + 3t)/(t - 1). The quotient rule states that if f(x) = g(x)/h(x), then f'(x) = [g'(x)·h(x) - g(x)·h'(x)]/[h(x)]². Here, g(t) = t² + 3t with g'(t) = 2t + 3, and h(t) = t - 1 with h'(t) = 1. Applying the formula: R'(t) = [(2t + 3)(t - 1) - (t² + 3t)(1)]/(t - 1)². Choice B incorrectly reverses the order in the numerator, which would give a different result. Remember: in the quotient rule, it's always "derivative of top times bottom minus top times derivative of bottom, all over bottom squared."

To find g'(x) for g(x) = $\dfrac{\ln x}{x + 2}$, we use the quotient rule with numerator $\ln x$ (derivative: $\dfrac{1}{x}$) and denominator $x + 2$ (derivative: 1). Applying the quotient rule formula: g'(x) = $\dfrac{ \left( \dfrac{1}{x} \right) (x + 2) - (\ln x)(1) }{ (x + 2)^2 }$. This simplifies to $\dfrac{ \left( \dfrac{1}{x} \right) (x + 2) - \ln x }{ (x + 2)^2 }$. Choice B incorrectly adds instead of subtracts $\ln x$ in the numerator, violating the quotient rule's subtraction requirement. When using the quotient rule, always maintain the correct order and operation: it's subtraction, not addition, between the two products in the numerator.

Finding w'(x) for w(x) = tan x/x requires the quotient rule with numerator tan x (derivative: sec² x) and denominator x (derivative: 1). Applying the quotient rule formula: w'(x) = [(sec² x)·x - (tan x)·1]/x² = (x sec² x - tan x)/x². Choice C incorrectly reverses the order of subtraction in the numerator, which would change the sign and give an incorrect result. The quotient rule's mnemonic "low d-high minus high d-low, all over low squared" helps remember the correct order, where maintaining the subtraction order is essential for the correct derivative.

Finding h'(t) for h(t) = t/(1 + t²) requires the quotient rule with numerator t (derivative: 1) and denominator 1 + t² (derivative: 2t). The quotient rule formula gives: h'(t) = [(1)(1 + t²) - t(2t)]/(1 + t²)². This simplifies to [1 + t² - 2t²]/(1 + t²)² = (1 - t²)/(1 + t²)². Choice D incorrectly omits squaring the denominator, while choice E has the same error (since (1 + t²) without exponent means (1 + t²)¹). Remember that the quotient rule always requires squaring the original denominator in the final result.

To find P'(x) for P(x) = (x³ - 4)/√x, we apply the quotient rule with numerator f(x) = x³ - 4 and denominator g(x) = √x = x^(1/2). The derivative of the numerator is f'(x) = 3x², and the derivative of the denominator is g'(x) = (1/2)x^(-1/2) = 1/(2√x). Using the quotient rule formula: P'(x) = [(3x²)(√x) - (x³ - 4)(1/(2√x))]/(√x)². Note that (√x)² = x, not √(x²) as in choice E. Choice D incorrectly omits squaring the denominator, which is essential in the quotient rule. Remember: always square the original denominator in the quotient rule result.

To find m'(x) for m(x) = (x² + 1)/cos x, we apply the quotient rule with numerator x² + 1 (derivative: 2x) and denominator cos x (derivative: -sin x). Using the quotient rule: m'(x) = [(2x)(cos x) - (x² + 1)(-sin x)]/(cos x)². Note the double negative becomes positive: (2x cos x + (x² + 1)sin x)/(cos x)². Choice C incorrectly reverses the order of subtraction, which would give a different result. The quotient rule maintains a specific order: "top' × bottom - top × bottom'" where careful attention to signs is crucial, especially with trigonometric derivatives.

To find s'(x) for s(x) = (x² - 5x)/(x² + 4), we apply the quotient rule with numerator x² - 5x (derivative: 2x - 5) and denominator x² + 4 (derivative: 2x). The quotient rule gives: s'(x) = [(2x - 5)(x² + 4) - (x² - 5x)(2x)]/(x² + 4)². This can be expanded and simplified, but the form shown in choice A is correct. Choice D incorrectly omits squaring the denominator, a common error. Remember that the quotient rule always results in the original denominator squared in the final expression, regardless of what the original denominator was.

This problem requires the quotient rule to find the derivative of the signal function S(t) = sin t / (t² + 1). The quotient rule states that if f(t) = num(t)/den(t), then f'(t) = [num'(t) den(t) - num(t) den'(t)] / [den(t)]². Here, num(t) = sin t with num'(t) = cos t, and den(t) = t² + 1 with den'(t) = 2t. Applying the rule gives [cos t (t² + 1) - sin t (2t)] / (t² + 1)², which is correct. A tempting distractor like choice B fails because it reverses the subtraction, starting with the second term instead of the first as per the rule. Remember, a transferable strategy for the quotient rule is to always write 'low d-high minus high d-low over low squared' to keep the order and signs correct.

This problem requires the quotient rule to find the derivative of the model function $H(x) = \dfrac{x^2 - 5}{\cos x}$. The quotient rule states that if $f(x) = \dfrac{\text{num}(x)}{\text{den}(x)}$, then $f'(x) = \dfrac{\text{num}'(x) \cdot \text{den}(x) - \text{num}(x) \cdot \text{den}'(x)}{[\text{den}(x)]^2}$. Here, $\text{num}(x) = x^2 - 5$ with $\text{num}'(x) = 2x$, and $\text{den}(x) = \cos x$ with $\text{den}'(x) = -\sin x$. Applying the rule gives $ \dfrac{2x \cos x - (x^2 - 5)(-\sin x)}{(\cos x)^2} $, which accounts for the negative derivative correctly. A tempting distractor like choice B fails by reversing the subtraction and mishandling the sign of the derivative of cosine. Remember, a transferable strategy for the quotient rule is to always write 'low d-high minus high d-low over low squared' to keep the order and signs correct.