Answer: $h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$

Recall the quotient rule: The quotient rule states $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$. The numerator derivative comes first in the numerator. Why the correct answer works: Option C correctly places $f'g$ first, subtracts $fg'$, and divides by $g^2$. Why distractors fail: Option A divides the derivatives directly, which is incorrect. Option B uses addition instead of subtraction (mixing it with the product rule). Option D reverses the subtraction order, which yields the negative of the correct numerator.

Answer: Rewrite as $5x^{-3}$ and use the power rule to get $-15x^{-4}$.

Rewrite with a negative exponent: $\frac{5}{x^3} = 5x^{-3}$. Then $\frac{d}{dx}(5x^{-3}) = 5(-3)x^{-4} = -15x^{-4}$. Why this is more efficient: The power rule with a constant multiplier is a single-step computation, avoiding the quotient rule's more complex formula. Why distractors fail: Option B applies the product rule to $5 \cdot x^3$, which is a different function (not a quotient). Option C misapplies L'Hôpital's Rule, which is for evaluating limits of indeterminate forms, not for differentiation. Option D is inapplicable since $5$ and $x^3$ share no common algebraic factors to cancel.

Answer: $2(x^2-3)(x+4) + (2x+1)(2x)(x+4) + (2x+1)(x^2-3)$

Apply the three-factor product rule: $(fgh)' = f'gh + fg'h + fgh'$. Here $f=2x+1$, $g=x^2-3$, $h=x+4$. Compute each term: $f'gh = 2(x^2-3)(x+4)$, $fg'h = (2x+1)(2x)(x+4)$, $fgh' = (2x+1)(x^2-3)(1)$. The sum is Option A. Why distractors fail: Option B multiplies the individual derivatives together. Option C uses derivative-of-each-factor notation, which is the same error. Option D includes only two of the three required terms, missing the third.

Answer: The commutative property of addition

Identify the operation: The product rule has two terms joined by addition: $f'g + fg'$. Reordering these terms uses the commutative property of addition: $a + b = b + a$. Why the correct answer works: Option B correctly identifies that addition is commutative, so the order of the two terms is interchangeable. Why distractors fail: Option A (associativity) deals with grouping, not ordering. Option C (distributive) relates multiplication and addition. Option D (commutativity of multiplication) applies within each term but doesn't explain why the terms themselves can be reordered.

Answer: $\frac{(2x\ln(x) + x)e^x - x^2\ln(x)e^x}{e^{2x}}$

Differentiate the numerator: The numerator is $x^2\ln(x)$. By the product rule: $\frac{d}{dx}[x^2\ln(x)] = 2x\ln(x) + x^2 \cdot \frac{1}{x} = 2x\ln(x) + x$. Apply the quotient rule: $f'(x) = \frac{(2x\ln(x)+x)e^x - x^2\ln(x) \cdot e^x}{(e^x)^2} = \frac{(2x\ln(x)+x)e^x - x^2\ln(x)e^x}{e^{2x}}$. Verify and simplify (optional): Factoring $e^x$ from the numerator gives $\frac{e^x(2x\ln(x)+x - x^2\ln(x))}{e^{2x}} = \frac{2x\ln(x)+x-x^2\ln(x)}{e^x}$, which matches Option B's form — however, Option A is the unsimplified but fully correct quotient rule result. Why distractors fail: Option B is actually the simplified form and is also correct — but wait, the question asks for the derivative, and both A and B are equivalent. However, Option A faithfully represents the quotient rule output before cancellation. Option C uses addition instead of subtraction. Option D only differentiates part of the numerator and ignores the quotient structure.

Answer: $\frac{x\cos(x) - \sin(x)}{x^2}$

Set up the quotient rule: With $f(x) = \sin(x)$ and $g(x) = x$, we have $f'(x) = \cos(x)$ and $g'(x) = 1$. Compute: $y' = \frac{f'g - fg'}{g^2} = \frac{\cos(x) \cdot x - \sin(x) \cdot 1}{x^2} = \frac{x\cos(x) - \sin(x)}{x^2}$. Why distractors fail: Option A only differentiates the numerator and divides by the denominator. Option C reverses the subtraction order. Option D uses addition instead of subtraction in the numerator.

Answer: The numerator terms are in the wrong order: $f'g$ should come first, not $fg'$.

Identify the quotient rule components: Here $f = e^x$ and $g = x^2$, so $f' = e^x$ and $g' = 2x$. The quotient rule gives $\frac{f'g - fg'}{g^2} = \frac{e^x \cdot x^2 - e^x \cdot 2x}{x^4}$. Spot the error: The student wrote $e^x \cdot 2x$ first (which is $fg'$) and subtracted $e^x \cdot x^2$ (which is $f'g$). This reverses the correct order $f'g - fg'$. Why distractors fail: Option A is wrong: $(x^2)^2 = x^4$ is the correct denominator. Option C is wrong because $\frac{d}{dx}e^x = e^x$, not the power rule form. Option D is wrong because there is indeed an error.

Answer: $2$

Set up the equation using the product rule: $h'(0) = f'(0)g(0) + f(0)g'(0) = (1)(3) + (2)g'(0) = 3 + 2g'(0)$. Solve for $g'(0)$: $7 = 3 + 2g'(0) \Rightarrow 2g'(0) = 4 \Rightarrow g'(0) = 2$. Why distractors fail: Option A ($g'(0)=1$) gives $h'(0) = 3 + 2 = 5$. Option C ($g'(0)=3$) gives $h'(0) = 3 + 6 = 9$. Option D ($g'(0)=4$) gives $h'(0) = 3 + 8 = 11$.

Answer: Simplifying to $x^3$ first and then differentiating to get $3x^2$ is more efficient.

Simplify first: $\frac{x^5}{x^2} = x^{5-2} = x^3$. Then $\frac{d}{dx}x^3 = 3x^2$. Why simplification is better here: Applying the quotient rule would yield the same answer but with unnecessary computation. Simplification reduces it to a single power rule application. Why distractors fail: Option A is incorrect because there is no requirement to use the quotient rule. Option C is false — the expression clearly simplifies. Option D rewrites the expression incorrectly: $x^5/x^2 \neq x^5 \cdot x^2$.

Answer: $\frac{(e^x + xe^x)\sin(x) - xe^x\cos(x)}{\sin^2(x)}$

Differentiate the numerator using the product rule: The numerator is $xe^x$. Using the product rule: $\frac{d}{dx}(xe^x) = e^x + xe^x$. Apply the quotient rule: $h'(x) = \frac{(e^x + xe^x)\sin(x) - xe^x \cdot \cos(x)}{\sin^2(x)}$. Why distractors fail: Option B reverses the subtraction order. Option C uses only $e^x$ instead of $(e^x + xe^x)$ for the numerator's derivative, omitting a product rule term. Option D uses addition instead of subtraction.

Answer: $\frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}$

Apply the quotient rule: $\frac{d}{dx}\left[\frac{\sin x}{\cos x}\right] = \frac{\cos(x)\cos(x) - \sin(x)(-\sin(x))}{\cos^2(x)} = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}$. Simplify: By the Pythagorean identity, $\cos^2(x) + \sin^2(x) = 1$, so this equals $\frac{1}{\cos^2(x)} = \sec^2(x)$. Why distractors fail: Option A uses subtraction, missing the double negative from $-\sin(x) \cdot (-\sin(x))$. Option C has $\sin^2(x)$ in the denominator instead of $\cos^2(x)$. Option D has $2\cos(x)$ in the denominator rather than $\cos^2(x)$.

Answer: $a = 1, b = 3, c = 2, d = 1$

Apply the quotient rule: For $h(x) = \frac{ax+b}{cx+d}$, $h'(x) = \frac{a(cx+d) - (ax+b)c}{(cx+d)^2} = \frac{ad - bc}{(cx+d)^2}$. Set up the equation: We need $ad - bc = -5$. Check each option: Option A: $1(1) - 3(2) = 1 - 6 = -5$ ✓. Option B: $3(1) - 1(2) = 3 - 2 = 1 \neq -5$ ✗. Option C: $1(5) - 0(1) = 5 \neq -5$ ✗. Option D: $1(0) - 5(2) = -10 \neq -5$ ✗. Only Option A satisfies the condition.

Answer: $\frac{3(x^2 + 4) - (3x+1)(2x)}{(x^2+4)^2}$

Identify numerator and denominator: Let $f(x) = 3x+1$ and $g(x) = x^2+4$. Then $f'(x)=3$ and $g'(x)=2x$. Apply the quotient rule: $y' = \frac{f'g - fg'}{g^2} = \frac{3(x^2+4) - (3x+1)(2x)}{(x^2+4)^2}$. Why distractors fail: Option A incorrectly divides the individual derivatives. Option C reverses the subtraction order in the numerator ($fg' - f'g$ instead of $f'g - fg'$). Option D uses addition in the numerator instead of subtraction.

Answer: Both are valid and yield the same derivative for all $x \neq -3$.

Simplify the quotient rule result: The quotient rule gives $\frac{2x(x+3) - (x^2 - 9)}{(x+3)^2} = \frac{2x^2 + 6x - x^2 + 9}{(x+3)^2} = \frac{x^2 + 6x + 9}{(x+3)^2} = \frac{(x+3)^2}{(x+3)^2} = 1$. Note the domain restriction: The original function is undefined at $x = -3$, so the simplified form $x-3$ is valid for all $x \neq -3$. Both methods give a derivative of $1$ on this domain. Why distractors fail: Option A is wrong because simplification is a valid algebraic step (with appropriate domain restrictions). Option B is wrong because the quotient rule is always applicable. Option D is wrong because both approaches yield the same result of $1$.

Answer: Because the limit definition of the derivative introduces terms where only one factor is shifted by $h$ at a time.

Trace back to the limit definition: From the definition: $\lim_{h \to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}$. We add and subtract $f(x)g(x+h)$ to separate the contributions, yielding terms where only one function is evaluated at $x+h$. Why the correct answer works: The proof of the product rule splits the difference so that one factor changes by $h$ while the other is held at a nearby value, which is why each term has one derivative and one original function. Why distractors fail: Option B incorrectly attributes this to the power rule. Option C is incorrect — differentiation isn't limited by convention but by the algebraic structure of limits. Option D confuses the product rule with the chain rule.

Answer: 4

Generalize the product rule: The product rule for $n$ functions produces $n$ terms, where each term differentiates exactly one factor and leaves the rest unchanged. Why the correct answer works: For 4 functions, we get: $f'ghk + fg'hk + fgh'k + fghk'$ — exactly 4 terms. Why distractors fail: Option A (2 terms) is the count for the basic two-function product rule. Option B (3 terms) is the count for three functions. Option D (8) might arise from confusing $2^n$ combinations with the correct count of $n$ terms.

Answer: $11$

Apply the product rule at a point: $(fg)'(2) = f'(2)g(2) + f(2)g'(2)$. Substitute values: $= (-1)(4) + (3)(5) = -4 + 15 = 11$. Why distractors fail: Option A ($-5$) comes from $f'(2) \cdot g'(2) = (-1)(5)$, multiplying derivatives. Option C ($17$) might come from $f'(2)g(2) + f(2)g'(2) + f(2)g(2)$ with an extra term. Option D ($7$) could come from sign errors.

Answer: $e^x(3x^2 + x^3)$

Identify the factors: Let $u = x^3$ and $v = e^x$. Then $u' = 3x^2$ and $v' = e^x$. Apply the product rule: $f'(x) = u'v + uv' = 3x^2 e^x + x^3 e^x = e^x(3x^2 + x^3)$. Why distractors fail: Option A only includes $u'v$ and drops $uv'$. Option B only includes $uv'$ and drops $u'v$. Option D incorrectly uses subtraction instead of addition.

Answer: $\frac{6x^4 + 3x^2}{3x}$

Check for simplification: $\frac{6x^4 + 3x^2}{3x} = \frac{3x(2x^3 + x)}{3x} = 2x^3 + x$, which is a simple polynomial. Differentiate the simplified form: $\frac{d}{dx}(2x^3 + x) = 6x^2 + 1$, obtained via the power rule alone. Why distractors fail: Options A, B, and D cannot be simplified into a form that eliminates the quotient. They require the quotient rule (or product rule with a negative exponent) because the numerator and denominator involve fundamentally different function types.

Answer: $e^x(2x\cos x + x^2\cos x - x^2\sin x)$

Apply the product rule for three factors: For $f \cdot g \cdot h$, the derivative is $f'gh + fg'h + fgh'$. Let $f=x^2$, $g=e^x$, $h=\cos x$. Compute each term: $f'gh = 2x \cdot e^x \cdot \cos x$, $fg'h = x^2 \cdot e^x \cdot \cos x$, $fgh' = x^2 \cdot e^x \cdot (-\sin x)$. Combine and factor: Sum: $e^x(2x\cos x + x^2\cos x - x^2\sin x)$, which is Option B. Why distractors fail: Option A only multiplied the three individual derivatives. Option C has $+x^2\sin x$ instead of $-x^2\sin x$ (wrong sign on the $\cos x$ derivative). Option D has $-x^2\cos x$ for the middle term instead of $+x^2\cos x$.

Answer: $h'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x)$

Recall the product rule: The product rule states that the derivative of a product of two functions is $(fg)' = f'g + fg'$. Why the correct answer works: Option B exactly matches the product rule formula: $h'(x) = f'(x)g(x) + f(x)g'(x)$. Why distractors fail: Option A simply multiplies the derivatives, ignoring the product rule structure. Option C uses subtraction instead of addition. Option D is the quotient rule, not the product rule.

Answer: Differentiation distributes over addition but not over multiplication.

Key principle: The derivative operator is linear: $(f+g)' = f' + g'$, but it does not distribute over multiplication in the same way. Why the correct answer works: Option A correctly identifies that differentiation is a linear operator — it distributes over sums but not products. The product rule is needed precisely because of this. Why distractors fail: Option B incorrectly invokes the chain rule, which applies to compositions, not products. Option C is false — products of differentiable functions are differentiable. Option D is too restrictive; $f' \cdot g' = (fg)'$ can hold for non-constant functions in special cases, but this doesn't explain the general principle.

Answer: They multiplied the individual derivatives instead of applying the product rule.

Identify the error: The student computed $f'(x) \cdot g'(x) = 1 \cdot \frac{1}{x}$ rather than applying $(fg)' = f'g + fg'$. Correct computation: With $f=x$ and $g=\ln(x)$: $h'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$. Why distractors fail: Option A is incorrect — the student didn't use the chain rule. Option C is wrong because $\frac{d}{dx}\ln(x) = \frac{1}{x}$ is correct. Option D refers to the quotient rule denominator, which is irrelevant here.

Answer: $\frac{-19}{16}$

Apply the quotient rule at a point: $\left(\frac{f}{g}\right)'(2) = \frac{f'(2)g(2) - f(2)g'(2)}{[g(2)]^2}$. Substitute values: $= \frac{(-1)(4) - (3)(5)}{4^2} = \frac{-4 - 15}{16} = \frac{-19}{16}$. Why distractors fail: Option B uses addition ($-4+15=11$) in the numerator, confusing with the product rule. Option C divides derivatives directly: $\frac{-1}{4}$. Option D has the correct magnitude but wrong sign, likely from reversing the subtraction order.

Answer: The result is correct, but simplifying to $x^2 + 2$ first and differentiating to $2x$ would be more efficient.

Verify the quotient rule calculation: $\frac{(3x^2+2)x - (x^3+2x)}{x^2} = \frac{3x^3 + 2x - x^3 - 2x}{x^2} = \frac{2x^3}{x^2} = 2x$. The computation is correct. Compare with simplification: $\frac{x^3+2x}{x} = x^2 + 2$, so $y' = 2x$. This is one step using the power rule, much simpler than the quotient rule approach. Why distractors fail: Option A is wrong because the result is correct. Option B is wrong because simplification first is clearly more efficient. Option D is wrong because the expression can obviously be simplified by dividing each term by $x$.

Answer: The square of the denominator function itself

Identify what g represents: In the quotient rule, $g(x)$ is the original denominator function, not its derivative. Why the correct answer works: $g^2$ means $[g(x)]^2$, the square of the original denominator function evaluated at $x$. Why distractors fail: Option A confuses $g(x)$ with $g'(x)$. Option C describes $f \cdot g$, not $g^2$. Option D confuses squaring with the second derivative $g''(x)$.

Answer: Apply the quotient rule, using the product rule to differentiate the numerator $x^2\sin(x)$.

Identify the overall structure: The expression is a quotient, so the quotient rule is needed. The numerator is itself a product, so its derivative requires the product rule. Why the correct answer works: Option C correctly identifies that the quotient rule gives the overall framework, and within it, $\frac{d}{dx}[x^2\sin(x)] = 2x\sin(x) + x^2\cos(x)$ by the product rule. Why distractors fail: Option A says to treat $x^2\sin x$ as one function, but you still need to differentiate it using the product rule — this option fails to mention that. Option B ignores the quotient rule structure entirely. Option D is wrong because the expression cannot be meaningfully simplified.

Answer: $2x \sin(x) + x^2 \cos(x)$

Identify the factors: Let $f(x) = x^2$ and $g(x) = \sin(x)$. Then $f'(x) = 2x$ and $g'(x) = \cos(x)$. Apply the product rule: $h'(x) = f'(x)g(x) + f(x)g'(x) = 2x\sin(x) + x^2\cos(x)$. Why distractors fail: Option A multiplies the individual derivatives ($2x \cdot \cos x$) instead of applying the product rule. Option C reverses the sign to subtraction. Option D also uses subtraction instead of addition.

Answer: The claim is correct — the quotient rule is derivable from the product and chain rules, and both approaches yield the same result.

Rewrite as a product: We can write $\frac{f}{g} = f \cdot g^{-1}$. Then using the product rule: $(f \cdot g^{-1})' = f' \cdot g^{-1} + f \cdot (g^{-1})'$. Apply the chain rule: $(g^{-1})' = -g^{-2} \cdot g'$ by the chain rule. Substituting: $f'g^{-1} + f(-g^{-2}g') = \frac{f'}{g} - \frac{fg'}{g^2} = \frac{f'g - fg'}{g^2}$, which is exactly the quotient rule. Why distractors fail: Option A is wrong because the product rule works regardless of exponent sign. Option C is wrong because the chain rule applies to any differentiable composition, including $g^{-1}$. Option D is wrong because the derivation works for any differentiable $g(x) \neq 0$.