Answer: $\frac{\cos x - \sin x}{e^x}$

Apply the quotient rule: For $\frac{u}{v}$ with $u = \sin x$ and $v = e^x$, the quotient rule gives $\frac{u'v - uv'}{v^2}$. Compute: $u' = \cos x$, $v' = e^x$. So $\frac{\cos x \cdot e^x - \sin x \cdot e^x}{(e^x)^2} = \frac{e^x(\cos x - \sin x)}{e^{2x}} = \frac{\cos x - \sin x}{e^x}$. Why distractors fail: Option B uses a plus sign, which would come from incorrectly adding the terms instead of subtracting. Option C omits the $-\sin x$ term from the quotient rule. Option D reverses the signs entirely.

Answer: $2 \cdot 3^{2t} \cdot \ln 3$

Rewrite and apply the chain rule: We have $Q(t) = 3^{2t}$. Using $\frac{d}{dt}[a^{u}] = a^{u} \ln a \cdot u'$ with $a = 3$ and $u = 2t$, so $u' = 2$. Compute: $Q'(t) = 3^{2t} \cdot \ln 3 \cdot 2 = 2 \cdot 3^{2t} \cdot \ln 3$. Why distractors fail: Option B omits the chain rule factor of $2$. Option C misapplies the power rule. Option D rewrites $3^{2t} = 9^t$ but then incorrectly uses $\ln 6$ instead of $\ln 9$.

Answer: $\cos x$

Recall the standard derivative: The derivative of $\sin x$ with respect to $x$ is $\cos x$. This is a fundamental result that should be memorized. Why the correct answer works: By definition, $\frac{d}{dx}[\sin x] = \cos x$, which matches option B. Why distractors fail: Option A ($-\cos x$) confuses the derivative of $\sin x$ with a sign error. Option C ($\tan x$) is not the derivative of any basic trig function directly. Option D ($-\sin x$) is the derivative of $\cos x$, not $\sin x$.

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

Recall the logarithmic derivative: The derivative of the natural logarithm is $\frac{d}{dx}[\ln x] = \frac{1}{x}$ for $x > 0$. Why the correct answer works: Option A gives $\frac{1}{x}$, which is the standard result. Why distractors fail: Option B ($\ln x$) confuses the function with its derivative. Option C ($x \ln x - x$) is the antiderivative of $\ln x$, not its derivative. Option D ($e^x$) is the inverse function of $\ln x$, not its derivative.

Answer: The student forgot to apply the chain rule; the correct derivative is $e^{\sin x} \cdot \cos x$.

Identify the correct differentiation method: Since $e^{\sin x}$ is a composition $e^u$ with $u = \sin x$, we apply the chain rule: $\frac{d}{dx}[e^u] = e^u \cdot u'$. Compute correctly: $f'(x) = e^{\sin x} \cdot \cos x$. The student incorrectly replaced the exponent $\sin x$ with its derivative $\cos x$ instead of multiplying by it. Why distractors fail: Option B incorrectly suggests the product rule; this is a composition, not a product. Option C misapplies the power rule to an exponential function. Option D is wrong because the student's answer is incorrect.

Answer: $\sec x \cdot \tan x$

Recall the derivative of sec x: The derivative of $\sec x$ is $\sec x \tan x$. This can be derived by writing $\sec x = \frac{1}{\cos x}$ and applying the quotient rule. Why the correct answer works: Option C gives $\sec x \cdot \tan x$, which matches the standard formula. Why distractors fail: Option A ($\sec x \cdot \csc x$) incorrectly pairs secant with cosecant. Option B ($\sec^2 x$) is the derivative of $\tan x$, not $\sec x$. Option D has the correct form but an incorrect negative sign; the negative sign belongs to the derivative of $\csc x$.

Answer: $x^2 e^x + 2xe^x$

Identify the rule needed: The function $x^2 e^x$ is a product of two functions: $u = x^2$ and $v = e^x$. Use the product rule: $(uv)' = u'v + uv'$. Compute: $u' = 2x$ and $v' = e^x$. So $\frac{d}{dx}[x^2 e^x] = 2x \cdot e^x + x^2 \cdot e^x = e^x(x^2 + 2x)$. Why distractors fail: Option A only computes $u'v$ and omits $uv'$. Option C replaces $v' = e^x$ with just $1$ in the second term. Option D multiplies $x^2 e^x$ by $2x$, which is not any correct rule.

Answer: $-\tan x$

Apply the chain rule to ln(u): For $f(x) = \ln(u)$ where $u = \cos x$, we have $f'(x) = \frac{u'}{u} = \frac{-\sin x}{\cos x}$. Simplify: $\frac{-\sin x}{\cos x} = -\tan x$, which is option B. Why distractors fail: Option A ($1/\cos x = \sec x$) forgets to multiply by the derivative of the inner function. Option C ($\tan x$) has the wrong sign. Option D incorrectly places $x$ in the denominator instead of $\cos x$.

Answer: Because $\ln e = 1$, so the factor $\ln a$ becomes $1$.

Apply the general exponential rule: For any positive base $a$, $\frac{d}{dx}[a^x] = a^x \cdot \ln a$. When $a = e$, substitute to get $e^x \cdot \ln e$. Simplify using a key log property: Since $\ln e = 1$, we have $e^x \cdot 1 = e^x$. This is why $e^x$ is its own derivative. Why distractors fail: Option A states $e^e = 1$, which is false ($e^e \approx 15.15$). Option C incorrectly claims the formula doesn't apply; the formula is general for all $a > 0$. Option D wrongly invokes the power rule, which applies to $x^n$, not $a^x$.

Answer: $5^x \ln 5$

Apply the general exponential derivative rule: For $f(x) = a^x$ where $a$ is a positive constant, $f'(x) = a^x \ln a$. Here $a = 5$. Why the correct answer works: Substituting $a = 5$ gives $f'(x) = 5^x \ln 5$, which is option B. Why distractors fail: Option A ($x \cdot 5^{x-1}$) misapplies the power rule (which is for $x^n$, not $a^x$). Option C ($5^x \cdot 5$) replaces $\ln 5$ with $5$. Option D ($5^x / \ln 5$) inverts the $\ln 5$ factor, which would arise in integration, not differentiation.

Answer: $2e^x \cos x$

Find the first derivative: Using the product rule on $g(x) = e^x \sin x$: $g'(x) = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$. Find the second derivative: Apply the product rule again to $g'(x) = e^x(\sin x + \cos x)$: $g''(x) = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = e^x(2\cos x) = 2e^x \cos x$. Why distractors fail: Option B ($e^x(\sin x + \cos x)$) is $g'(x)$, not $g''(x)$. Option C ($e^x(\sin x - \cos x)$) has incorrect signs. Option D ($2e^x \sin x$) replaces $\cos x$ with $\sin x$.

Answer: $3\sec^2(3x)$

Apply the chain rule with the tangent derivative: We have $\frac{d}{dx}[\tan(u)] = \sec^2(u) \cdot u'$, where $u = 3x$ and $u' = 3$. Compute the result: $\frac{d}{dx}[\tan(3x)] = \sec^2(3x) \cdot 3 = 3\sec^2(3x)$. Why distractors fail: Option A ($\sec^2(3x)$) forgets the chain rule factor of $3$. Option C confuses the derivative of tangent with the derivative of secant. Option D introduces an incorrect negative sign.

Answer: $-2e^{-2x}$

Apply the chain rule: Let $u = -2x$. Then $\frac{d}{dx}[e^u] = e^u \cdot u' = e^{-2x} \cdot (-2)$. Why the correct answer works: The result is $-2e^{-2x}$, which is option B. Why distractors fail: Option A forgets the chain rule factor $(-2)$. Option C has the wrong sign on the coefficient. Option D misapplies the power rule to the exponent.

Answer: $-\csc x \cdot \cot x$

Recall the derivative formula: $\frac{d}{dx}[\csc x] = -\csc x \cot x$. This can be derived from $\csc x = 1/\sin x$ using the quotient rule. Why the correct answer works: Option C matches the standard formula with the correct negative sign. Why distractors fail: Option A omits the crucial negative sign. Option B ($-\csc^2 x$) confuses this with the derivative of $\cot x$. Option D ($\sec x \tan x$) is the derivative of $\sec x$, not $\csc x$.

Answer: $1 - \cot x$

Simplify using logarithm properties: $\ln\left(\frac{e^x}{\sin x}\right) = \ln(e^x) - \ln(\sin x) = x - \ln(\sin x)$. Differentiate term by term: $\frac{d}{dx}[x] = 1$ and $\frac{d}{dx}[\ln(\sin x)] = \frac{\cos x}{\sin x} = \cot x$. So the derivative is $1 - \cot x$. Why distractors fail: Option A ($1 + \cot x$) uses the wrong sign from the subtraction. Option C applies the quotient rule to the argument before using logarithm properties, which is valid but yields a non-simplified form that doesn't match upon simplification—it actually equals option B when simplified correctly, but as written it is the derivative of $e^x/\sin x$ divided by itself, not the derivative of the logarithm. Option D confuses $\cot x$ with $\tan x$.

Answer: $e^x$

Recall the exponential derivative rule: The function $e^x$ is unique in that it is its own derivative: $\frac{d}{dx}[e^x] = e^x$. Why the correct answer works: Option C states $e^x$, which is exactly the derivative of $e^x$. Why distractors fail: Option A ($x \cdot e^{x-1}$) incorrectly applies the power rule as if $e^x$ were $x^e$. Option B ($e^x \cdot \ln x$) confuses the rule for $a^x$ where $\frac{d}{dx}[a^x] = a^x \ln a$, but when $a = e$, $\ln e = 1$. Option D ($\frac{1}{e^x}$) is $e^{-x}$, which is not the derivative.

Answer: The derivative of $\cos x$ is $-\sin x$; the student forgot the negative sign that arises from the limit definition.

Identify the correct derivative: The standard derivative is $\frac{d}{dx}[\cos x] = -\sin x$. The negative sign is essential. Why the correct answer works: Option B correctly identifies the missing negative sign and references the limit definition where the sign naturally appears. Why distractors fail: Option A is factually wrong; tangent is not involved. Option C is misleading; the formula $\frac{d}{dx}[\cos x] = -\sin x$ holds when $x$ is in radians, and the derivative is not $\sin x$ in any unit system. Option D incorrectly claims cosine is its own derivative.

Answer: $\frac{1}{x \ln 3}$

Apply the logarithmic derivative formula: For $\log_a x$, the derivative is $\frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}$. Here $a = 3$. Why the correct answer works: Substituting gives $\frac{1}{x \ln 3}$, which is option C. Why distractors fail: Option A ($1/x$) is the derivative of $\ln x$, not $\log_3 x$. Option B ($\ln 3 / x$) places $\ln 3$ in the numerator instead of the denominator. Option D ($3/x$) replaces $\ln 3$ with $3$.