Answer: $y = x - 1$

Find f(1): $f(1) = 2(1)^2 - 3(1) + 1 = 2 - 3 + 1 = 0$. So the point is $(1, 0)$. Find f'(x) and evaluate at x = 1: $f'(x) = 4x - 3$, so $f'(1) = 4 - 3 = 1$. Write the tangent line: Point-slope: $y - 0 = 1(x - 1)$, giving $y = x - 1$. Why distractors fail: Option B ($y = x$) has the correct slope but the wrong $y$-intercept. Option C ($y = 4x - 4$) uses $f'(x) = 4x$ without subtracting $3$, or evaluates at the wrong point. Option D ($y = x + 1$) has the correct slope but shifts the line up instead of down.

Answer: $3x^{1/2} + 5x^{-2} + 4$

Differentiate each term: $\frac{d}{dx}[2x^{3/2}] = 2 \cdot \frac{3}{2}x^{1/2} = 3x^{1/2}$. $\frac{d}{dx}[-5x^{-1}] = -5 \cdot (-1)x^{-2} = 5x^{-2}$. $\frac{d}{dx}[4x] = 4$. Combine: $h'(x) = 3x^{1/2} + 5x^{-2} + 4$. Why distractors fail: Option A has $-5x^{-2}$, which neglects the double negative from $(-5)(-1) = +5$. Option C drops the derivative of $4x$, which is $4$. Option D has coefficient $1$ instead of $3$ for the first term.

Answer: $\frac{1}{2\sqrt{x}}$

Rewrite the square root: Write $\sqrt{x} = x^{1/2}$. Apply the power rule: $\frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{1/2 - 1} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. Why distractors fail: Option A ($1/\sqrt{x}$) omits the coefficient $1/2$. Option C ($2\sqrt{x}$) inverts the process (this looks like an antiderivative). Option D ($\sqrt{x}/2$) uses the wrong exponent in the result.

Answer: The student failed to recognize that the derivative of a constant is zero.

Identify the error: The student's answer includes the constant $+3$ in the derivative. The correct derivative is $f'(x) = 20x^3 + 4x - 1$. The derivative of the constant $3$ should be $0$, not $3$. Why distractors fail: Option A is incorrect because the exponents were properly reduced. Option B is incorrect because $\frac{d}{dx}[2x^2] = 4x$ was computed correctly. Option D is incorrect because there is no product involved.

Answer: $10$

Apply the power rule: $\frac{d}{dx}[x^{10}] = 10x^9$. Evaluate at x = 1: $10(1)^9 = 10$. Why distractors fail: Option A ($1$) may result from computing $1^{10} = 1$ (the function value, not the derivative). Option C ($9$) confuses the reduced exponent with the derivative value. Option D ($0$) has no basis.

Answer: The derivative of a sum equals the sum of the derivatives, and constants factor out of the derivative.

Define linearity of differentiation: An operation $L$ is linear if $L(af + bg) = aL(f) + bL(g)$ for constants $a, b$. Differentiation satisfies this: $\frac{d}{dx}[af(x) + bg(x)] = af'(x) + bg'(x)$. Why distractors fail: Option A describes a false rule (the product rule is more complex). Option C confuses 'linear operation' with 'linear function.' Option D incorrectly limits differentiation to straight lines.

Answer: $f(x) = 3x^4 - 2x^3 + 2x + 5$

Reverse the power rule (antidifferentiate): To find $f(x)$ from $f'(x) = 12x^3 - 6x^2 + 2$, reverse the power rule: $\frac{12x^4}{4} - \frac{6x^3}{3} + 2x + C = 3x^4 - 2x^3 + 2x + C$. Apply the initial condition: $f(0) = 5$ gives $C = 5$. So $f(x) = 3x^4 - 2x^3 + 2x + 5$. Verify: Check: $f'(x) = 12x^3 - 6x^2 + 2$. Correct. Why distractors fail: Option B does not divide by the new exponent (uses $12x^4$ instead of $3x^4$). Option C omits the constant $C = 5$. Option D has $-3x^3$ instead of $-2x^3$, meaning the coefficient of $x^3$ was not divided by $3$.

Answer: $a = 1$

Differentiate: $f'(x) = 3ax^2 + 2$. Set up and solve the equation: We need $f'(2) = 14$: $3a(2)^2 + 2 = 14$, so $12a + 2 = 14$, giving $12a = 12$ and $a = 1$. Why distractors fail: Option A ($a = 4$): $12(4) + 2 = 50 \neq 14$. Option B ($a = 2$): $12(2) + 2 = 26 \neq 14$. Option D ($a = 3$): $12(3) + 2 = 38 \neq 14$.

Answer: $8x^7 - 12x^3 + 2$

Differentiate term by term: Using the sum rule and power rule: $\frac{d}{dx}[x^8] = 8x^7$, $\frac{d}{dx}[-3x^4] = -12x^3$, $\frac{d}{dx}[2x] = 2$. Combine the results: So $f'(x) = 8x^7 - 12x^3 + 2$. Why distractors fail: Option B forgets to multiply the coefficient $-3$ by the exponent $4$. Option C increases the exponents (integration, not differentiation). Option D drops the derivative of $2x$.

Answer: $15x^4 - 6x^2 + 6$

Differentiate each term: $\frac{d}{dx}[3x^5] = 15x^4$, $\frac{d}{dx}[-2x^3] = -6x^2$, $\frac{d}{dx}[6x] = 6$, $\frac{d}{dx}[-9] = 0$. Combine results: $f'(x) = 15x^4 - 6x^2 + 6$. Why distractors fail: Option B includes $-9$ in the derivative (the constant does not vanish). Option C fails to multiply the coefficient $-2$ by the exponent $3$. Option D does not reduce any of the exponents.

Answer: $x = 2$ and $x = -2$

Find f'(x): $f'(x) = 3x^2 - 12$. Set derivative equal to zero: A horizontal tangent occurs when $f'(x) = 0$: $3x^2 - 12 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. Why distractors fail: Option A confuses the constant $12$ with an $x$-value. Option C solves $x^2 = 16$ instead of $x^2 = 4$. Option D finds only the positive root.

Answer: Yes, both terms are differentiated correctly.

Verify each term: $\sqrt[3]{x^2} = x^{2/3}$, so $\frac{d}{dx}[x^{2/3}] = \frac{2}{3}x^{-1/3}$. And $\frac{1}{x^5} = x^{-5}$, so $\frac{d}{dx}[x^{-5}] = -5x^{-6}$. Both are correct. Why distractors fail: Option B incorrectly claims $\sqrt[3]{x^2} = x^{3/2}$; it is actually $x^{2/3}$. Option C uses the wrong exponent rule: $-5 - 1 = -6$, not $-4$. Option D falsely restricts the sum rule — it applies to all differentiable functions.

Answer: Both approaches are valid and yield the same result, $4x^3 - 6x^2 + 2x$.

Verify both approaches: Student A: $\frac{d}{dx}[x^4] - \frac{d}{dx}[2x^3] + \frac{d}{dx}[x^2] - \frac{d}{dx}[5] = 4x^3 - 6x^2 + 2x$. Student B: $\frac{d}{dx}[x^4 + x^2] - \frac{d}{dx}[2x^3 + 5] = (4x^3 + 2x) - (6x^2) = 4x^3 - 6x^2 + 2x$. Key principle: The sum rule and constant multiple rule are linear operations, so grouping and reordering do not affect the result. Why distractors fail: Option A falsely restricts the order of differentiation. Option B falsely claims only grouping works. Option D falsely requires factoring before differentiating.

Answer: Yes, the student applied the power and sum rules correctly.

Verify the computation: $\frac{d}{dx}[x^{-2}] = -2x^{-3}$ and $\frac{d}{dx}[x^{-3}] = -3x^{-4}$. So $\frac{d}{dx}[x^{-2} + x^{-3}] = -2x^{-3} - 3x^{-4}$. The student's work is correct. Why distractors fail: Option A is false — the sum rule applies regardless of the type of exponent. Option B gives incorrect exponents (they were not reduced properly). Option D is false — negative exponents correctly produce negative coefficients when differentiated.

Answer: $8x^3 - 6x$

Simplify first: Divide each term in the numerator by $x$: $f(x) = 2x^4 - 3x^2 + 1$. Differentiate: $f'(x) = 8x^3 - 6x + 0 = 8x^3 - 6x$. Why distractors fail: Option A ($8x^3 - 6x + 1$) incorrectly includes the derivative of the constant $1$ as $1$ instead of $0$. Option C differentiates the numerator without first simplifying. Option D is the simplified function $f(x)$, not its derivative.

Answer: The student used the wrong sign for the resulting exponent; the correct derivative is $-\frac{3}{2}x^{-3/2}$.

Rewrite and differentiate correctly: Rewrite: $g(x) = 3x^{-1/2}$. Applying the power rule: $g'(x) = 3 \cdot (-\frac{1}{2})x^{-1/2 - 1} = -\frac{3}{2}x^{-3/2}$. Identify the student's mistake: The student wrote $\frac{3}{2}x^{-1/2}$, meaning they used $+\frac{1}{2}$ as the coefficient (instead of $-\frac{1}{2}$) and wrote $x^{-1/2}$ instead of $x^{-3/2}$. They did not correctly bring down the negative exponent. Why distractors fail: Option A suggests the student didn't rewrite, but the form $x^{-1/2}$ shows they did attempt it. Option C is unnecessary — the power rule works fine here. Option D states a false derivative for $\sqrt{x}$.

Answer: $\frac{2}{3}x^{-1/3} + 2x^{-2/3}$

Differentiate each term: $\frac{d}{dx}[x^{2/3}] = \frac{2}{3}x^{2/3-1} = \frac{2}{3}x^{-1/3}$. $\frac{d}{dx}[6x^{1/3}] = 6 \cdot \frac{1}{3}x^{1/3-1} = 2x^{-2/3}$. Combine: $f'(x) = \frac{2}{3}x^{-1/3} + 2x^{-2/3}$. Why distractors fail: Option B does not reduce the exponent of the first term and forgets to multiply the coefficient $6$ by $\frac{1}{3}$. Option C forgets to multiply the coefficient $6$ by $\frac{1}{3}$. Option D uses $x^{1/3}$ instead of $x^{-2/3}$ in the second term, failing to subtract $1$ from the exponent.

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

Extract conditions from the tangent line: The tangent at $x = 1$ has slope $4$, so $f'(1) = 4$. The tangent passes through $(1, f(1))$, and $y = 4(1) - 2 = 2$, so $f(1) = 2$. Set up equations: $f(x) = ax^2 + bx + c$ and $f'(x) = 2ax + b$. From $f(0) = 1$: $c = 1$. From $f(1) = 2$: $a + b + 1 = 2$, so $a + b = 1$. From $f'(1) = 4$: $2a + b = 4$. Solve the system: Subtracting: $(2a + b) - (a + b) = 4 - 1$, giving $a = 3$, then $b = -2$. So $f(x) = 3x^2 - 2x + 1$. Why distractors fail: Option B: $f'(x) = 4x$, so $f'(1) = 4$ but $f(1) = 3 \neq 2$. Option C: $f(0) = -1 \neq 1$. Option D: $f'(1) = 6 \neq 4$.

Answer: $5x^4$

Apply the power rule: The power rule states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$. Here $n = 5$, so $f'(x) = 5x^4$. Why distractors fail: Option B ($5x^5$) forgets to reduce the exponent. Option C ($4x^5$) uses the wrong coefficient and keeps the original exponent. Option D ($x^4$) reduces the exponent but drops the coefficient.

Answer: $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$

State the sum rule: The sum rule says that the derivative of a sum is the sum of the derivatives: $\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$. Why distractors fail: Option A multiplies the derivatives, which is not a standard differentiation rule. Option C is the product rule, not the sum rule. Option D divides the derivatives, which does not correspond to any standard rule.

Answer: $21x^2$

Apply the constant multiple and power rules: Using the constant multiple rule, $\frac{d}{dx}[7x^3] = 7 \cdot \frac{d}{dx}[x^3]$. Then by the power rule, $\frac{d}{dx}[x^3] = 3x^2$. So the result is $7 \cdot 3x^2 = 21x^2$. Why distractors fail: Option A ($7x^2$) reduces the exponent but forgets to multiply by the original exponent. Option B ($21x^3$) correctly computes $7 \times 3 = 21$ but forgets to reduce the exponent. Option D ($3x^2$) drops the constant multiplier $7$.

Answer: $2x^3 - 6x$

Simplify first: $\frac{x^4 - 6x^2 + 8}{2} = \frac{1}{2}x^4 - 3x^2 + 4$. Differentiate: $\frac{d}{dx}\left[\frac{1}{2}x^4 - 3x^2 + 4\right] = 2x^3 - 6x + 0 = 2x^3 - 6x$. Why distractors fail: Option A ($2x^3 - 6x + 4$) incorrectly includes the derivative of the constant $4$ as $4$ instead of $0$. Option C ($4x^3 - 12x$) differentiates the numerator without dividing by $2$ first. Option D ($2x^3 - 3x$) does not correctly multiply the coefficient $-3$ by the exponent $2$.

Answer: No, but the answer $72x^3$ happens to be incorrect; the correct derivative is $60x^4$.

Correct approach: simplify first: First simplify: $3x^2 \cdot 4x^3 = 12x^5$. Then $\frac{d}{dx}[12x^5] = 60x^4$. Identify the error: The student differentiated each factor separately and multiplied the results. This is not a valid rule — the derivative of a product is NOT the product of the derivatives. The correct result $60x^4$ does not equal the student's $72x^3$. Why distractors fail: Options A and D assert a false rule (that the derivative of a product equals the product of the derivatives). Option C claims the wrong answer is accidentally correct, but $72x^3 \neq 60x^4$.

Answer: The colleague is incorrect; this expression should be rewritten as $5x^{-4}$ and differentiated using the power rule, which is simpler.

Evaluate the claim: While the quotient rule would give the correct answer, it is unnecessarily complex. Rewriting $\frac{5}{x^4} = 5x^{-4}$ allows direct application of the power rule: $\frac{d}{dx}[5x^{-4}] = -20x^{-5}$. Why distractors fail: Option A is incorrect because the quotient rule is not required here — simpler methods exist. Option C suggests the product rule, which doesn't directly apply in this form. Option D falsely claims the power rule only works for positive integers; it works for all real exponents.

Answer: It allows you to bring the exponent down as a coefficient and reduce the exponent by one.

Understand the power rule conceptually: The power rule states $\frac{d}{dx}[x^n] = nx^{n-1}$. In plain language, you bring the exponent $n$ down to become the coefficient, and then decrease the exponent by $1$. Why distractors fail: Option A keeps the same exponent, which is incorrect — the exponent must decrease by one. Option C describes an integration-like process (increasing the exponent), not differentiation. Option D introduces logarithms, which are unrelated to the power rule.

Answer: $4x^3 + 5x^2 - 3x$

Differentiate each option: For Option A: $\frac{d}{dx}[4x^3 + 5x^2 - 3x] = 12x^2 + 10x - 3$. This matches the target. Check the other options: Option B: $\frac{d}{dx}[4x^3 + 10x^2 - 3x + 1] = 12x^2 + 20x - 3$, which has $20x$ instead of $10x$. Option C: $\frac{d}{dx}[12x^3 + 10x^2 - 3x] = 36x^2 + 20x - 3$. Option D: $\frac{d}{dx}[36x + 10] = 36$, a constant.

Answer: $y = 8x - 15$

Find f(2): $f(2) = 2^3 - 4(2) + 1 = 8 - 8 + 1 = 1$. Find f'(x) and evaluate at x = 2: $f'(x) = 3x^2 - 4$, so $f'(2) = 3(4) - 4 = 8$. Write the tangent line equation: Using point-slope form: $y - 1 = 8(x - 2)$, which simplifies to $y = 8x - 16 + 1 = 8x - 15$. Why distractors fail: Option B ($y = 8x - 17$) uses the correct slope but makes an arithmetic error in the constant. Option C ($y = 4x - 7$) uses an incorrect slope. Option D ($y = 12x - 23$) uses $f'(2) = 12$, likely from computing $3(2)^2 = 12$ without subtracting $4$.

Answer: $-\frac{12}{x^4}$

Rewrite with a negative exponent: Rewrite $\frac{4}{x^3}$ as $4x^{-3}$. Apply the power rule: $\frac{d}{dx}[4x^{-3}] = 4 \cdot (-3)x^{-3-1} = -12x^{-4} = -\frac{12}{x^4}$. Why distractors fail: Option A ($-12/x^2$) uses the wrong resulting exponent. Option B ($12/x^4$) has the correct exponent but the wrong sign. Option D ($-4/x^4$) forgets to multiply $4$ by the exponent $-3$.

Answer: $0$

Recall the constant rule: The derivative of any constant function is $0$. A constant has no rate of change, so $\frac{d}{dx}[c] = 0$ for any constant $c$. Why distractors fail: Option A ($12$) confuses the function value with its derivative. Option B ($12x$) incorrectly treats the constant as if it needs to be multiplied by $x$. Option D ($1$) may come from a misapplication of the power rule to $x^0$.