The Quotient Rule - AP Calculus BC

For $g(x)=\dfrac{\ln x}{x^3}$ , what is $g'(x)$ ?
Answer: $\dfrac{\left(\dfrac{1}{x}\right)(x^3)-(\ln x)(3x^2)}{(x^3)^2}$ To find $g'(x)$ for $g(x)=\frac{\ln x}{x^3}$ , we apply the quotient rule. The quotient rule formula is $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$ . With $u(x)=\ln x$ giving $u'(x)=\frac{1}{x}$ , and $v(x)=x^3$ giving $v'(x)=3x^2$ , we substitute: $g'(x)=\frac{\left(\frac{1}{x}\right)(x^3)-(\ln x)(3x^2)}{(x^3)^2}$ . Choice E incorrectly writes the denominator as $x^6$ instead of $(x^3)^2$ —while these are equal, the form $(x^3)^2$ directly follows from the quotient rule formula. When applying the quotient rule, write the denominator as $[v(x)]^2$ first, then simplify algebraically to maintain clarity in your work.
For $R(t)=\dfrac{t^2+3t}{t-1}$ , what is $R'(t)$ ?
Answer: $\dfrac{(2t+3)(t-1)-(t^2+3t)}{(t-1)^2}$ 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."
If $v(x)=\dfrac{\sin x}{x^2+1}$ , what is $v'(x)$ ?
Answer: $\dfrac{\cos x\,(x^2+1)-\sin x\,(2x)}{(x^2+1)^2}$ This problem asks for the derivative of v(x) = sin x/(x² + 1) using the quotient rule. The quotient rule formula is [f'(x)·g(x) - f(x)·g'(x)]/[g(x)]² where f(x) is the numerator and g(x) is the denominator. Here, the numerator is sin x with derivative cos x, and the denominator is x² + 1 with derivative 2x. Substituting into the formula: v'(x) = [cos x·(x² + 1) - sin x·(2x)]/(x² + 1)². Choice B incorrectly adds instead of subtracts in the numerator, which is a common error. Always remember the quotient rule uses subtraction: "top' × bottom - top × bottom'" in the numerator.
In a lab, concentration is modeled by $C(t)=\dfrac{t^2+3t}{t+1}$ . What is $C'(t)$ ?
Answer: $\dfrac{(2t+3)(t+1)-(t^2+3t)}{(t+1)^2}$ This problem requires the quotient rule to find the derivative of the concentration function C(t) = (t² + 3t)/(t + 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) = t² + 3t with num'(t) = 2t + 3, and den(t) = t + 1 with den'(t) = 1. Applying the rule gives [(2t + 3)(t + 1) - (t² + 3t)(1)] / (t + 1)², which simplifies to the correct expression. A tempting distractor like choice B fails because it uses a plus sign instead of a minus, which would incorrectly apply the rule by adding rather than subtracting the terms in the numerator. 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.
For $q(x)=\dfrac{\sqrt{x+1}}{x}$ , what is $q'(x)$ ?
Answer: $\dfrac{\left(\dfrac{1}{2\sqrt{x+1}}\right)x-\sqrt{x+1}}{x^2}$ For q(x) = √(x + 1)/x, we use the quotient rule with numerator √(x + 1) = $(x + 1)^{1/2}$ and denominator x. The derivative of the numerator is (1/2) $(x + 1)^{-1/2}$ = 1/(2√(x + 1)), and the derivative of the denominator is 1. Applying the quotient rule: q'(x) = [(1/(2√(x + 1)))·x - √(x + 1)·1]/x². Choice C reverses the subtraction order, changing the sign of the result. Always follow the quotient rule pattern precisely: derivative of numerator times denominator minus numerator times derivative of denominator, maintaining this exact order for correct results.
If $s(x)=\dfrac{x-2}{x^2-9}$ gives a simplified score, what is $s'(x)$ ?
Answer: $\dfrac{(1)(x^2-9)-(x-2)(2x)}{(x^2-9)^2}$ The simplified score s(x) = (x - 2) / (x² - 9) is differentiated using the quotient rule. Let u = x - 2 with u' = 1 and v = x² - 9 with v' = 2x. Applying the rule results in [1 (x² - 9) - (x - 2) 2x] / (x² - 9)², matching choice A. The numerator simplifies to x² - 9 - 2x(x - 2) for further analysis if needed. Choice C incorrectly subtracts in reverse, yielding the opposite sign. Always compute the denominator squared accurately to avoid power errors in quotient rule problems.
A car's efficiency is $E(v)=\dfrac{v}{v^2+4}$ . What is $E'(v)$ ?
Answer: $\dfrac{(v^2+4)-v(2v)}{(v^2+4)^2}$ This problem requires the quotient rule to differentiate E(v) = v/(v² + 4). The quotient rule states that for f(x) = g(x)/h(x), we have f'(x) = [h(x)·g'(x) - g(x)·h'(x)]/[h(x)]². Here, g(v) = v with g'(v) = 1, and h(v) = v² + 4 with h'(v) = 2v. Applying the quotient rule: E'(v) = [(v² + 4)·1 - v·(2v)]/[(v² + 4)²] = [(v² + 4) - 2v²]/(v² + 4)² = (4 - v²)/(v² + 4)². Choice D incorrectly uses (v² + 4) instead of (v² + 4)² in the denominator, forgetting to square the original denominator. Remember the quotient rule pattern: numerator operations stay in the numerator, but the denominator gets squared.
For $m(x)=\dfrac{x^2+1}{x^3-1}$ , what is the derivative $m'(x)$ ?
Answer: $\dfrac{(x^3-1)(2x)-(x^2+1)(3x^2)}{(x^3-1)^2}$ This problem requires the quotient rule to find m'(x) for m(x) = (x² + 1)/(x³ - 1). The quotient rule formula states: if f(x) = g(x)/h(x), then f'(x) = [h(x)·g'(x) - g(x)·h'(x)]/[h(x)]². For this function, g(x) = x² + 1 with g'(x) = 2x, and h(x) = x³ - 1 with h'(x) = 3x². Applying the quotient rule: m'(x) = [(x³ - 1)·(2x) - (x² + 1)·(3x²)]/[(x³ - 1)²]. Choice C reverses the order to (x² + 1)(3x²) - (x³ - 1)(2x), which is incorrect. The quotient rule always follows the pattern 'denominator times numerator's derivative, minus numerator times denominator's derivative.'
Let $p(x)=\dfrac{\ln x}{x}$ . What is $p'(x)$ ?
Answer: $\dfrac{x\left(\dfrac{1}{x}\right)-\ln x}{x^2}$ For p(x) = ln x/x, we apply the quotient rule to find p'(x). The quotient rule states that (f/g)' = [g·f' - f·g']/g². The numerator ln x has derivative 1/x, and the denominator x has derivative 1. Substituting: p'(x) = [x·(1/x) - ln x·1]/x² = [1 - ln x]/x². Choice C incorrectly reverses the subtraction, giving [ln x - 1]/x² instead, which would make the derivative positive for x > e rather than x < e. When applying the quotient rule, maintain the exact order of operations: denominator times numerator's derivative, then subtract numerator times denominator's derivative.
Let $h(x)=\dfrac{x-1}{x^2+3x}$ . What is $h'(x)$ ?
Answer: $\dfrac{(1)(x^2+3x)-(x-1)(2x+3)}{(x^2+3x)^2}$ Finding $h'(x)$ for $h(x)=\frac{x-1}{x^2+3x}$ requires the quotient rule. The quotient rule states that for $\frac{u(x)}{v(x)}$ , the derivative is $\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}$ . Here, $u(x)=x-1$ with $u'(x)=1$ , and $v(x)=x^2+3x$ with $v'(x)=2x+3$ . Substituting: $h'(x)=\frac{(1)(x^2+3x)-(x-1)(2x+3)}{(x^2+3x)^2}$ . Choice C reverses the order of subtraction in the numerator, which would yield the negative of the correct derivative. The quotient rule mnemonic "low d-high minus high d-low, all over low squared" helps maintain the correct order.
A rate is modeled by $q(x)=\dfrac{\sqrt{x+3}}{x-2}$ . What is $q'(x)$ ?
Answer: $\dfrac{\left(\dfrac{1}{2\sqrt{x+3}}\right)(x-2)-\sqrt{x+3}(1)}{(x-2)^2}$ To find $q'(x)$ for $q(x)=\frac{\sqrt{x+3}}{x-2}$ , we apply the quotient rule. The quotient rule formula is $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$ . With $u(x)=\sqrt{x+3}$ giving $u'(x)=\frac{1}{2\sqrt{x+3}}$ , and $v(x)=x-2$ giving $v'(x)=1$ , we substitute: $q'(x)=\frac{\left(\frac{1}{2\sqrt{x+3}}\right)(x-2)-\sqrt{x+3}(1)}{(x-2)^2}$ . Choice C reverses the subtraction order, placing the second term first, which would negate the derivative. When using the quotient rule, maintain the exact order: (top derivative)(bottom) minus (top)(bottom derivative), ensuring accurate results.
If the average cost per item is $C(x)=\dfrac{5x^2+1}{x+3}$ , what is $C'(x)$ ?
Answer: $\dfrac{(10x)(x+3)-(5x^2+1)(1)}{(x+3)^2}$ This problem requires the quotient rule to find the derivative of the average cost function C(x) = (5x² + 1)/(x + 3). To apply the quotient rule, identify the numerator as u = 5x² + 1 with u' = 10x and the denominator as v = x + 3 with v' = 1. The derivative is then [u' v - u v'] / v², which substitutes to [10x (x + 3) - (5x² + 1)(1)] / (x + 3)². Simplifying the numerator gives 10x(x + 3) - 5x² - 1, confirming the expression in choice C. A tempting distractor like choice E reverses the subtraction in the numerator, which would incorrectly apply the rule by subtracting in the wrong order. Always remember to apply the quotient rule as 'lo d hi minus hi d lo over lo squared' to ensure the correct order of operations.
A bacteria culture's concentration is modeled by $C(t)=\dfrac{t^2+3t}{\sqrt{t+1}}$ . What is $C'(t)$ ?
Answer: $\dfrac{(2t+3)\sqrt{t+1}-(t^2+3t)\left(\dfrac{1}{2\sqrt{t+1}}\right)}{t+1}$ This problem requires the quotient rule to differentiate C(t) = (t² + 3t)/√(t + 1). The quotient rule states that if f(x) = g(x)/h(x), then f'(x) = [h(x)·g'(x) - g(x)·h'(x)]/[h(x)]². Here, g(t) = t² + 3t with g'(t) = 2t + 3, and h(t) = √(t + 1) = $(t + 1)^{1/2}$ with h'(t) = 1/(2√(t + 1)). Applying the quotient rule: C'(t) = [√(t + 1)·(2t + 3) - (t² + 3t)·(1/(2√(t + 1)))]/[(√(t + 1))²] = [(2t + 3)√(t + 1) - (t² + 3t)/(2√(t + 1))]/(t + 1). Choice C incorrectly omits the derivative of the denominator in the numerator, forgetting that we must subtract g(x)·h'(x). When using the quotient rule, remember: 'bottom times derivative of top, minus top times derivative of bottom, all over bottom squared.'
Let $m(x)=\dfrac{x^2+1}{x^3}$ . What is $m'(x)$ ?
Answer: $\dfrac{2x\cdot x^3-(x^2+1)\cdot 3x^2}{(x^3)^2}$ Finding m'(x) for m(x) = (x² + 1)/x³ requires the quotient rule. The quotient rule states (u/v)' = (u'v - uv')/v². The numerator (x² + 1) has derivative 2x, and the denominator x³ has derivative 3x². Applying: m'(x) = [2x·x³ - (x² + 1)·3x²]/(x³)² = [2x⁴ - 3x⁴ - 3x²]/x⁶ = [-x⁴ - 3x²]/x⁶. Choice C forgets to square the denominator, showing x³ instead of (x³)² = x⁶. When using the quotient rule, always remember to square the entire denominator expression in your final answer.
A growth index is $G(x)=\dfrac{e^x}{x^2-1}$ . Find $G'(x)$ .
Answer: $\dfrac{e^x(x^2-1)-e^x(2x)}{(x^2-1)^2}$ This problem requires the quotient rule to find the derivative of the growth index G(x) = $e^x$ / (x² - 1). The quotient rule states that if f(x) = num(x)/den(x), then f'(x) = [num'(x) den(x) - num(x) den'(x)] / [den(x)]². Here, num(x) = $e^x$ with num'(x) = $e^x$ , and den(x) = x² - 1 with den'(x) = 2x. Applying the rule gives [ $e^x$ (x² - 1) - $e^x$ (2x)] / (x² - 1)², matching the correct choice. A tempting distractor like choice D fails by not squaring the denominator, which is essential to the quotient rule formula. 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.
Suppose $m(x)=\dfrac{x^2}{x^2+1}$ . What is $m'(x)$ ?
Answer: $\dfrac{(2x)(x^2+1)-(x^2)(2x)}{(x^2+1)^2}$ Finding $m'(x)$ for $m(x)=\frac{x^2}{x^2+1}$ requires the quotient rule. The quotient rule states that $\frac{d}{dx}\left[\frac{u}{v}\right]=\frac{u'v-uv'}{v^2}$ . Here, $u(x)=x^2$ with $u'(x)=2x$ , and $v(x)=x^2+1$ with $v'(x)=2x$ . Applying the formula: $m'(x)=\frac{(2x)(x^2+1)-(x^2)(2x)}{(x^2+1)^2}$ . Choice C reverses the terms in the numerator, which would give the negative of the correct answer. The quotient rule pattern is consistent: derivative of numerator times denominator, minus numerator times derivative of denominator, all divided by denominator squared.
A projectile's height ratio is $R(t)=\dfrac{t^3}{t^2+4}$ . What is $R'(t)$ ?
Answer: $\dfrac{3t^2(t^2+4)-t^3(2t)}{(t^2+4)^2}$ This problem requires the quotient rule to find the derivative of the height ratio function $R(t) = \dfrac{t^3}{t^2 + 4}$ . The quotient rule states that if $f(t) = \dfrac{\text{num}(t)}{\text{den}(t)}$ , then $f'(t) = \dfrac{\text{num}'(t) \cdot \text{den}(t) - \text{num}(t) \cdot \text{den}'(t)}{[\text{den}(t)]^2}$ . Here, $\text{num}(t) = t^3$ with $\text{num}'(t) = 3t^2$ , and $\text{den}(t) = t^2 + 4$ with $\text{den}'(t) = 2t$ . Applying the rule gives $\dfrac{3t^2(t^2 + 4) - t^3(2t)}{(t^2 + 4)^2}$ , which is the correct expression. A tempting distractor like choice D fails because it omits the square on the denominator, violating the rule's requirement to square the bottom function. 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.
If $m(x)=\dfrac{x^2+1}{\cos x}$ , what is $m'(x)$ ?
Answer: $\dfrac{(2x)\cos x-(x^2+1)(-\sin x)}{(\cos x)^2}$ 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.
A concentration is modeled by $C(t)=\dfrac{e^t}{t+1}$ . What is $C'(t)$ ?
Answer: $\dfrac{e^t(t+1)-e^t(1)}{(t+1)^2}$ Differentiating $C(t)=\frac{e^t}{t+1}$ requires the quotient rule. The quotient rule states that $\frac{d}{dt}\left[\frac{u(t)}{v(t)}\right]=\frac{u'(t)v(t)-u(t)v'(t)}{[v(t)]^2}$ . Here, $u(t)=e^t$ with $u'(t)=e^t$ , and $v(t)=t+1$ with $v'(t)=1$ . Applying the formula: $C'(t)=\frac{e^t(t+1)-e^t(1)}{(t+1)^2}$ . Choice E incorrectly simplifies by canceling $e^t$ from numerator and denominator before differentiating, which violates the order of operations in calculus. Remember: differentiate first using the complete quotient rule formula, then simplify algebraically if possible.
A rate is defined by $h(t)=\dfrac{t}{1+t^2}$ . What is $h'(t)$ ?
Answer: $\dfrac{(1)(1+t^2)-t(2t)}{(1+t^2)^2}$ 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.

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