Answer: $\sin(e^{\cos(x^3)})$

Count the composition layers: Each application of the chain rule corresponds to one layer of composition beyond the outermost. Three applications means four nested layers. Analyze each option: $\sin(x^2+1)$ has two layers: one chain rule application. $e^{\cos(\ln(x))}$ has three layers ($e$, $\cos$, $\ln(x)$): two chain rule applications. $\sin(e^{\cos(x^3)})$ has four layers ($\sin$, $e$, $\cos$, $x^3$): three chain rule applications. $\ln(x) \cdot \cos(x)$ is a product, not a composition. Why the correct answer works: Differentiating $\sin(e^{\cos(x^3)})$ requires: (1) derivative of $\sin$ at $e^{\cos(x^3)}$, (2) derivative of $e$ at $\cos(x^3)$, (3) derivative of $\cos$ at $x^3$, each multiplied by the next inner derivative, plus $\frac{d}{dx}[x^3] = 3x^2$. That is exactly three chain rule applications.

Answer: $2xe^{4x} + 4x^2 e^{4x}$

Recognize the need for both rules: This is a product of $x^2$ and $e^{4x}$. The product rule gives $f'(x) = (x^2)' \cdot e^{4x} + x^2 \cdot (e^{4x})'$. The chain rule is needed for $(e^{4x})' = 4e^{4x}$. Compute the derivative: $f'(x) = 2x \cdot e^{4x} + x^2 \cdot 4e^{4x} = 2xe^{4x} + 4x^2e^{4x}$. Why distractors fail: $2xe^{4x}$ omits the second product rule term. $8xe^{4x}$ incorrectly combines the terms. $2xe^{4x} + x^2e^{4x}$ forgets the chain rule factor of $4$ on the second term.

Answer: You differentiate the outermost function first, keeping the inner function unchanged, then multiply by the derivative of the inner function.

Interpret 'outside-in': 'Outside-in' is a mnemonic for the chain rule: start by differentiating the outermost function while leaving the inner function intact, then multiply by the derivative of the inner function. Why the correct answer works: This option accurately describes the sequential process: $f'(g(x)) \cdot g'(x)$ — first $f'$ evaluated at $g(x)$, then times $g'(x)$. Why distractors fail: Evaluating the inner function first before differentiating confuses evaluation order with differentiation order. Simplifying from the outside is an algebraic strategy, not the chain rule. Mixing product and sum rules describes different derivative rules entirely.

Answer: $u^5$

Decompose the composite function: Write $y = u^5$ where $u = 2x + 1$. The outer function is the operation applied last: raising to the fifth power. Why the correct answer works: $u^5$ is the outer function because it is the final operation performed on the inner expression $u = 2x + 1$. Why distractors fail: $2x + 1$ is the inner function, not the outer. $5u^4$ is the derivative of the outer function, not the outer function itself. $2$ is the derivative of the inner function.

Answer: The student forgot to multiply by the derivative of the inner function $5x$, which is $5$.

Apply the chain rule to $\cos(5x)$: Let $u = 5x$. Then $\frac{d}{dx}[\cos(u)] = -\sin(u) \cdot \frac{du}{dx} = -\sin(5x) \cdot 5 = -5\sin(5x)$. Why the correct answer works: The student differentiated the outer function correctly ($-\sin$) but omitted the factor of $5$ from the derivative of the inner function $5x$. Why distractors fail: The derivative of cosine is indeed $-\sin$, so claiming it should be positive sine is wrong. The chain rule absolutely applies to trig functions with non-trivial arguments. The last option uses $-\cos$ instead of $-\sin$, which is the wrong trig derivative.

Answer: Yes, the rewriting is correct and the known derivative $\frac{d}{dx}[e^{kx}] = ke^{kx}$ avoids an explicit chain rule step, but this formula itself is derived from the chain rule.

Verify the algebraic rewrite: $e^{3x+1} = e^{3x} \cdot e^1 = e \cdot e^{3x}$. This is valid by exponent laws. Assess the derivative shortcut: Using $\frac{d}{dx}[e^{kx}] = ke^{kx}$, we get $\frac{d}{dx}[e \cdot e^{3x}] = e \cdot 3e^{3x} = 3e^{3x+1}$. This is correct, but the formula $\frac{d}{dx}[e^{kx}] = ke^{kx}$ is itself a consequence of the chain rule. Why distractors fail: The rewrite $e^{3x+1} = e \cdot e^{3x}$ is algebraically valid. The constant $e$ has derivative zero and is simply carried along as a constant multiple. Claiming the chain rule is never needed for exponentials is false — it underlies every exponential derivative formula with a non-trivial argument.

Answer: $2x \cdot e^{x^2} \cdot \cos(e^{x^2})$

Identify all layers: There are three layers: outermost is $\sin(\cdot)$, middle is $e^{(\cdot)}$, innermost is $x^2$. Apply the chain rule layer by layer: Start outside: $\frac{d}{du}[\sin(u)] = \cos(u)$ evaluated at $u = e^{x^2}$ gives $\cos(e^{x^2})$. Next layer: $\frac{d}{dv}[e^v] = e^v$ evaluated at $v = x^2$ gives $e^{x^2}$. Innermost: $\frac{d}{dx}[x^2] = 2x$. Multiply all: $\cos(e^{x^2}) \cdot e^{x^2} \cdot 2x$. Why distractors fail: $\cos(e^{x^2}) \cdot e^{x^2}$ omits the innermost derivative $2x$. $2x \cdot \cos(e^{x^2})$ omits the middle-layer derivative $e^{x^2}$. The fourth option incorrectly applies the power rule to $e^{x^2}$ as if it were $x^{x^2}$.

Answer: Student A is correct because the chain rule gives $e^{(x+1)^2} \cdot 2(x+1) \cdot 1$, and $\frac{d}{dx}[x+1] = 1$ keeps the factor $(x+1)$ intact.

Work through the chain rule carefully: Let $u = (x+1)^2$. Then $y = e^u$, so $\frac{dy}{du} = e^u$. Now $u = v^2$ where $v = x+1$. Then $\frac{du}{dv} = 2v = 2(x+1)$ and $\frac{dv}{dx} = 1$. Combining: $y' = e^{(x+1)^2} \cdot 2(x+1) \cdot 1 = 2(x+1)e^{(x+1)^2}$. Identify Student B's error: Student B computed $\frac{d}{dx}[(x+1)^2]$ as $2x$ instead of $2(x+1)$. The derivative of $(x+1)^2$ requires keeping the base $x+1$ and multiplying by the derivative of $x+1$, which is $1$, yielding $2(x+1)$. Why other distractors fail: Derivatives differing by a constant are NOT interchangeable — $2(x+1) = 2x + 2 \neq 2x$. There are indeed multiple chain rule layers, but Student A accounts for all of them correctly.

Answer: $\cos(x) \cdot e^{\sin(x)}$

Identify the composition layers: Outer: $e^u$. Inner: $u = \sin(x)$. The derivative of $e^u$ is $e^u$, and the derivative of $\sin(x)$ is $\cos(x)$. Apply the chain rule: $\frac{d}{dx}[e^{\sin(x)}] = e^{\sin(x)} \cdot \cos(x)$. Why distractors fail: $e^{\cos(x)}$ replaces the exponent with its derivative instead of multiplying. $\sin(x) \cdot e^{\sin(x)}$ uses the inner function instead of its derivative. $e^{\sin(x)} \cdot e^{\cos(x)}$ incorrectly factors the derivative as a product of two exponentials.

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

Apply the chain rule: Outer: $\sin(u)$, derivative $\cos(u)$. Inner: $u = x^2$, derivative $2x$. So $\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x = 2x\cos(x^2)$. Why the correct answer works: $2x\cos(x^2)$ includes both the derivative of sine (yielding cosine) and the derivative of the inner function $x^2$ (yielding $2x$). Why distractors fail: $\cos(x^2)$ forgets the inner derivative. $2x\sin(x^2)$ uses sine instead of cosine for the outer derivative. $x^2\cos(x^2)$ uses the inner function instead of its derivative.

Answer: $3e^{3x}$

Set up the chain rule: Outer: $e^u$. Inner: $u = 3x$. Since $\frac{d}{du}[e^u] = e^u$ and $\frac{du}{dx} = 3$, we get $\frac{d}{dx}[e^{3x}] = e^{3x} \cdot 3 = 3e^{3x}$. Why the correct answer works: $3e^{3x}$ correctly includes the derivative of the inner function $3x$. Why distractors fail: $e^{3x}$ omits the inner derivative. $3xe^{3x-1}$ incorrectly applies the power rule to the exponent as if it were $x^n$. $e^3$ treats the exponent as a constant.

Answer: $12(4x - 7)^2$

Identify inner and outer functions: Outer: $u^3$. Inner: $u = 4x - 7$. Apply the chain rule: $\frac{d}{dx}[u^3] = 3u^2 \cdot \frac{du}{dx} = 3(4x-7)^2 \cdot 4 = 12(4x-7)^2$. Why distractors fail: $3(4x-7)^2$ omits the inner derivative factor of $4$. $4(4x-7)^3$ uses the wrong exponent and omits the power rule coefficient. $12(4x-7)^3$ keeps the original exponent $3$ instead of reducing it to $2$.

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

Set up the chain rule: Outer: $u^4$. Inner: $u = x^3 + 2x$. Derivatives: $\frac{d}{du}[u^4] = 4u^3$, $\frac{du}{dx} = 3x^2 + 2$. Compute the result: $\frac{dy}{dx} = 4(x^3+2x)^3 \cdot (3x^2+2)$. Why distractors fail: $4(x^3+2x)^3$ is missing the inner derivative. $12x^2(x^3+2x)^3$ only uses $3x^2$ for the inner derivative, forgetting the $+2$. $(3x^2+2)^4$ replaces the original base with the inner derivative raised to the original power.

Answer: $-\tan(x)$

Apply the chain rule: Outer: $\ln(u)$, derivative $\frac{1}{u}$. Inner: $u = \cos(x)$, derivative $-\sin(x)$. So $\frac{d}{dx}[\ln(\cos(x))] = \frac{1}{\cos(x)} \cdot (-\sin(x)) = -\frac{\sin(x)}{\cos(x)} = -\tan(x)$. Why the correct answer works: $-\tan(x)$ is the simplified result of $\frac{-\sin(x)}{\cos(x)}$. Why distractors fail: $\frac{1}{\cos(x)}$ omits the inner derivative $-\sin(x)$. $\tan(x)$ has the wrong sign. $-\frac{\cos(x)}{\sin(x)} = -\cot(x)$, which has the ratio inverted.

Answer: $\frac{5}{2\sqrt{5x + 2}}$

Rewrite and identify layers: $\sqrt{5x+2} = (5x+2)^{1/2}$. Outer: $u^{1/2}$. Inner: $u = 5x + 2$. Apply the chain rule: $\frac{d}{dx}[(5x+2)^{1/2}] = \frac{1}{2}(5x+2)^{-1/2} \cdot 5 = \frac{5}{2\sqrt{5x+2}}$. Why distractors fail: $\frac{1}{2\sqrt{5x+2}}$ omits the factor of $5$ from the inner derivative. $\frac{5}{\sqrt{5x+2}}$ omits the $\frac{1}{2}$ from the power rule. $\frac{5}{2}(5x+2)^{1/2}$ uses a positive exponent instead of the required $-1/2$.

Answer: $\frac{6x[\ln(x^2+1)]^2}{x^2+1}$

Identify the layers: Outermost: $u^3$ with $u = \ln(v)$. Middle: $\ln(v)$ with $v = x^2+1$. Innermost: $x^2+1$. Apply the chain rule through all layers: $h'(x) = 3[\ln(x^2+1)]^2 \cdot \frac{1}{x^2+1} \cdot 2x = \frac{6x[\ln(x^2+1)]^2}{x^2+1}$. Why distractors fail: The first option omits the innermost derivative $2x$. The third option omits the middle-layer derivative $\frac{1}{x^2+1}$. The fourth option loses the $[\ln(x^2+1)]^2$ factor entirely.

Answer: $6x\sec^2(3x^2 + 1)$

Apply the chain rule: Outer: $\tan(u)$, derivative $\sec^2(u)$. Inner: $u = 3x^2 + 1$, derivative $6x$. Therefore $\frac{d}{dx}[\tan(3x^2+1)] = \sec^2(3x^2+1) \cdot 6x$. Why the correct answer works: $6x\sec^2(3x^2+1)$ correctly combines the outer derivative $\sec^2$ with the inner derivative $6x$. Why distractors fail: $\sec^2(3x^2+1)$ omits the inner derivative. $6x\tan(3x^2+1)$ uses $\tan$ instead of $\sec^2$ for the outer derivative. $3x^2\sec^2(3x^2+1)$ uses the inner function $3x^2$ instead of its derivative $6x$.

Answer: $\cos(x^2 + 1)$

Determine when the chain rule is needed: The chain rule applies when a function is composed with another function — i.e., there is a non-trivial 'inner function' that is not simply $x$. Why the correct answer works: $\cos(x^2+1)$ has the outer function $\cos(u)$ and the inner function $u = x^2+1$, which is not just $x$. The chain rule is necessary. Why distractors fail: $3x^5 + 2x - 7$ uses only the power and sum rules. $e^x \cdot \sin(x)$ requires the product rule but each factor has argument $x$ (no composition). $x^3/\ln(x)$ requires the quotient rule, but neither $x^3$ nor $\ln(x)$ involves a composition.

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

Simplify first (optional but helpful): $\ln(\sqrt{1+x^2}) = \frac{1}{2}\ln(1+x^2)$. Now differentiate: $\frac{1}{2} \cdot \frac{2x}{1+x^2} = \frac{x}{1+x^2}$. Alternative: direct multi-layer chain rule: Layers: $\ln(u)$, $u = \sqrt{v} = v^{1/2}$, $v = 1+x^2$. Derivatives: $\frac{1}{u}$, $\frac{1}{2}v^{-1/2}$, $2x$. Product: $\frac{1}{\sqrt{1+x^2}} \cdot \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{1+x^2}$. Why distractors fail: $\frac{1}{\sqrt{1+x^2}}$ omits derivatives of the inner layers. $\frac{2x}{\sqrt{1+x^2}}$ misses the $\frac{1}{2}$ factor. $\frac{x}{2(1+x^2)}$ keeps an extra factor of $\frac{1}{2}$ that should cancel with the $2x$.

Answer: No, because the chain rule requires an additional factor of $2$ (the derivative of $2x+3$), making the correct answer $8(2x+3)^3$.

Apply the chain rule correctly: $\frac{d}{dx}[(2x+3)^4] = 4(2x+3)^3 \cdot \frac{d}{dx}[2x+3] = 4(2x+3)^3 \cdot 2 = 8(2x+3)^3$. Why the student's claim fails: The power rule alone applies to $x^n$. When the base is a function of $x$ (here $2x+3$), the chain rule multiplies by the derivative of that function. Why other distractors fail: The power rule alone is insufficient whenever the base is not simply $x$. The quotient rule is irrelevant here. The chain rule applies to all inner functions, including linear ones — the factor of $2$ is still required.

Answer: $3x$

Identify outer vs. inner function: In a composite function $f(g(x))$, the inner function is $g(x)$ — the part that is 'inside' the outer function. Here $h(x) = \sin(3x)$, so the outer function is $\sin(u)$ and the inner function is $u = 3x$. Why the correct answer works: $3x$ is the expression being fed into the sine function, making it the inner function. Why distractors fail: $\sin(x)$ is the outer function with its argument changed, not the inner function. $\sin(3)$ is a constant, not a function of $x$. $3\sin(x)$ represents multiplication outside sine, which is a different expression entirely.

Answer: $-6\cos(3x)\sin(3x)$

Identify the layers: Outermost: $u^2$. Middle: $u = \cos(v)$. Innermost: $v = 3x$. Three layers require two applications of the chain rule. Differentiate step by step: $\frac{d}{dx}[(\cos(3x))^2] = 2\cos(3x) \cdot \frac{d}{dx}[\cos(3x)]$. Now $\frac{d}{dx}[\cos(3x)] = -\sin(3x) \cdot 3 = -3\sin(3x)$. Combining: $2\cos(3x) \cdot (-3\sin(3x)) = -6\cos(3x)\sin(3x)$. Why distractors fail: $-2\cos(3x)\sin(3x)$ omits the factor of $3$ from the innermost layer. The third option, $2\cos(3x)(-\sin(3x)) = -2\cos(3x)\sin(3x)$, also omits the factor of $3$. $-6\sin^2(3x)$ replaces $\cos$ with $\sin$ in the base.

Answer: $2e^{2x}\sin(5x) + 5e^{2x}\cos(5x)$

Apply the product rule: $f'(x) = (e^{2x})' \cdot \sin(5x) + e^{2x} \cdot (\sin(5x))'$. Use the chain rule for each factor: $(e^{2x})' = 2e^{2x}$ and $(\sin(5x))' = 5\cos(5x)$. So $f'(x) = 2e^{2x}\sin(5x) + 5e^{2x}\cos(5x)$. Why distractors fail: $10e^{2x}\cos(5x)$ only differentiates part of the expression and merges the coefficients. The third option has a minus sign before $5\cos(5x)$, but the derivative of sine is positive cosine. The fourth option multiplies terms rather than adding via the product rule.

Answer: $\frac{2\cos(2x) - 3\sin(2x)}{e^{3x}}$

Apply the quotient rule: $g'(x) = \frac{(\sin(2x))' \cdot e^{3x} - \sin(2x) \cdot (e^{3x})'}{(e^{3x})^2}$. Use the chain rule for each part: $(\sin(2x))' = 2\cos(2x)$ and $(e^{3x})' = 3e^{3x}$. So $g'(x) = \frac{2\cos(2x) \cdot e^{3x} - \sin(2x) \cdot 3e^{3x}}{e^{6x}} = \frac{2\cos(2x) - 3\sin(2x)}{e^{3x}}$. Why distractors fail: The second option has $+3\sin(2x)$ instead of $-3\sin(2x)$, a sign error in the quotient rule. The third option uses $\cos(2x)$ without the chain rule factor of $2$. The fourth option incorrectly divides the two derivatives rather than applying the quotient rule.

Answer: $f'(g(x)) \cdot g'(x)$

State the chain rule formula: The chain rule says $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$. You differentiate the outer function evaluated at the inner function, then multiply by the derivative of the inner function. Why the correct answer works: $f'(g(x)) \cdot g'(x)$ correctly keeps $g(x)$ inside $f'$ and multiplies by $g'(x)$. Why distractors fail: $f'(x) \cdot g'(x)$ incorrectly evaluates $f'$ at $x$ instead of $g(x)$. $f(g'(x))$ substitutes the derivative of $g$ into $f$ rather than into $f'$. Adding $f'(g(x)) + g'(x)$ replaces the required multiplication with addition.

Answer: The student used the power rule but forgot to keep $\sin(x)$ as the base.

Correct differentiation: $h(x) = (\sin(x))^3$. Outer: $u^3$. Inner: $u = \sin(x)$. So $h'(x) = 3(\sin(x))^2 \cdot \cos(x)$. Identify the student's error: The student wrote $3(\cos(x))^2$, replacing $\sin(x)$ with $\cos(x)$ in the base. The correct application keeps $\sin(x)$ in the base as $3\sin^2(x)$ and multiplies by $\cos(x)$. Why distractors fail: The trig derivative $\cos(x)$ is correct — the issue is where it appears. The product rule is not needed here since this is a composition, not a product. $(\sin(x))^3 \neq \sin(x^3)$; these are fundamentally different expressions.

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

Set up the chain rule: Outer: $\ln(u)$. Inner: $u = x^2 + 1$. We have $\frac{d}{du}[\ln(u)] = \frac{1}{u}$ and $\frac{du}{dx} = 2x$. Compute the derivative: $\frac{d}{dx}[\ln(x^2+1)] = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$. Why distractors fail: $\frac{1}{x^2+1}$ omits the inner derivative $2x$. $\frac{2x}{\ln(x^2+1)}$ incorrectly places $\ln$ in the denominator. $2x \cdot \ln(x^2+1)$ multiplies instead of dividing.