Answer: $y = 1 - \dfrac{1}{e}x$

Differentiate implicitly: $e^y \frac{dy}{dx} + y + x\frac{dy}{dx} = 0$. So $\frac{dy}{dx}(e^y + x) = -y$ and $\frac{dy}{dx} = \frac{-y}{e^y + x}$. Evaluate at (0, 1): $\frac{dy}{dx}\bigg|_{(0,1)} = \frac{-1}{e^1 + 0} = -\frac{1}{e}$. Write the tangent line: $y - 1 = -\frac{1}{e}(x - 0) \implies y = 1 - \frac{1}{e}x$. Why distractors fail: Option B has the wrong sign for the slope. Options C and D incorrectly use $\frac{1}{e+1}$ instead of $\frac{1}{e}$ in the slope, which would result from an error in the implicit differentiation step.

Answer: $\dfrac{xy - y}{x + xy}$

Differentiate both sides: $\frac{d}{dx}[\ln(xy)] = \frac{1}{xy}(y + x\frac{dy}{dx})$ and $\frac{d}{dx}[x - y] = 1 - \frac{dy}{dx}$. Set equal and solve: $\frac{y + x\frac{dy}{dx}}{xy} = 1 - \frac{dy}{dx}$. Multiply both sides by $xy$: $y + x\frac{dy}{dx} = xy - xy\frac{dy}{dx}$. Collect $\frac{dy}{dx}$: $x\frac{dy}{dx} + xy\frac{dy}{dx} = xy - y$, so $\frac{dy}{dx}(x + xy) = xy - y$ and $\frac{dy}{dx} = \frac{xy - y}{x + xy}$. Why distractors fail: Option A has $y + xy$ in the denominator instead of $x + xy$. Option B factors incorrectly. Option D, while algebraically close, does not simplify to the correct expression when fully reduced.

Answer: Both approaches are valid, but Approach 1 only captures the branch where $y > 0$, whereas Approach 2 handles both branches simultaneously

Understand the two branches: The equation $y^2 = x^3 + 2x$ defines two branches: $y = \sqrt{x^3+2x}$ (upper) and $y = -\sqrt{x^3+2x}$ (lower). Compare the approaches: Approach 1 with $y = \sqrt{x^3+2x}$ only handles the upper branch. Implicit differentiation gives $\frac{dy}{dx} = \frac{3x^2+2}{2y}$, which works for both $y > 0$ and $y < 0$ without choosing a branch. Why distractors fail: Option A is wrong — both approaches are valid. Option B is wrong — solving for $y$ is possible here. Option D ignores the branch restriction of the explicit approach.

Answer: $\dfrac{4}{5}$

Differentiate implicitly: Differentiate $y^3 + x^3 = 9xy$: $3y^2\frac{dy}{dx} + 3x^2 = 9y + 9x\frac{dy}{dx}$. Solve for dy/dx: $3y^2\frac{dy}{dx} - 9x\frac{dy}{dx} = 9y - 3x^2 \implies \frac{dy}{dx} = \frac{9y - 3x^2}{3y^2 - 9x} = \frac{3y - x^2}{y^2 - 3x}$. Evaluate at (2, 4): First verify the point lies on the curve: $4^3 + 2^3 = 64 + 8 = 72$ and $9(2)(4) = 72$. ✓ Now substitute: $\frac{dy}{dx}\bigg|_{(2,4)} = \frac{9(4) - 3(4)}{3(16) - 9(2)} = \frac{36 - 12}{48 - 18} = \frac{24}{30} = \frac{4}{5}$. Why distractors fail: Option B inverts the fraction. Option C results from arithmetic errors. Option D results from a computational mistake in the substitution.

Answer: Chain rule

Identify the underlying principle: Since $y$ is a function of $x$, differentiating any expression $g(y)$ with respect to $x$ requires $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$, which is the chain rule. Why the correct answer works: The chain rule states that when differentiating a composite function, you differentiate the outer function and multiply by the derivative of the inner function. Here $y = y(x)$ is the inner function. Why distractors fail: The product rule handles products of two functions, the quotient rule handles ratios, and the power rule handles $x^n$ for constant $n$. None of these alone accounts for the $\frac{dy}{dx}$ factor that arises from $y$ being a function of $x$.

Answer: Both the base and the exponent are variable, so neither the power rule nor exponential rule directly applies

Identify the structural challenge: In $x^{\sin x}$, the base $x$ and the exponent $\sin x$ are both functions of $x$. The standard power rule ($\frac{d}{dx}x^n = nx^{n-1}$) requires a constant exponent, and the exponential rule ($\frac{d}{dx}a^x = a^x \ln a$) requires a constant base. Why logarithmic differentiation resolves this: Taking $\ln$ of both sides gives $\ln f(x) = \sin x \cdot \ln x$, converting the variable exponent into a product that can be differentiated using the product rule. Why distractors fail: Option A is false — $x^{\sin x}$ is defined for $x > 0$. Option C is false — $\frac{d}{dx}(\sin x) = \cos x$ is well-known. Option D is false — you still must differentiate after taking the logarithm.

Answer: Both are valid differentiation procedures grounded in the chain rule; they must produce the same derivative for any differentiable function

Core principle: A function has a unique derivative (when it exists). Different differentiation techniques are alternative methods that must all yield the same result. Why the correct answer works: Both implicit and logarithmic differentiation are rigorous applications of the chain rule and other standard rules. Applied correctly to $y = x^3$, both give $\frac{dy}{dx} = 3x^2$. Why distractors fail: Option A is false — correct methods always agree. Option B is false — logarithmic differentiation can be applied to $x^3$ (it's just unnecessary). Option D is false — implicit differentiation is exact, not an approximation.

Answer: $\dfrac{\cos^2(xy) - y}{x}$

Differentiate both sides: $\sec^2(xy)\cdot(y + x\frac{dy}{dx}) = 1$. Solve for dy/dx: $y + x\frac{dy}{dx} = \cos^2(xy)$, so $x\frac{dy}{dx} = \cos^2(xy) - y$ and $\frac{dy}{dx} = \frac{\cos^2(xy) - y}{x}$. Why distractors fail: Option B incorrectly distributes the $\sec^2(xy)$ — the numerator $1 - y\sec^2(xy)$ does not simplify to $\cos^2(xy) - y$ when divided by $x$ alone (it would need $x\sec^2(xy)$ in the denominator). Option C writes $\frac{1}{x\sec^2(xy)} - y = \frac{\cos^2(xy)}{x} - y$, which equals $\frac{\cos^2(xy) - xy}{x}$, not $\frac{\cos^2(xy) - y}{x}$. Option D makes the same structural error as Option C.

Answer: $\dfrac{-y}{x + 2y}$

Differentiate each term: Differentiate $xy + y^2 = 10$ with respect to $x$. For $xy$, use the product rule: $y + x\frac{dy}{dx}$. For $y^2$: $2y\frac{dy}{dx}$. The right side gives $0$. Solve for dy/dx: $y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx}(x + 2y) = -y \implies \frac{dy}{dx} = \frac{-y}{x + 2y}$. Why distractors fail: Option B has the wrong sign in the numerator. Option C puts $-x$ in the numerator instead of $-y$. Option D omits the $x$ term from the denominator, forgetting the product rule on $xy$.

Answer: $y\left(\dfrac{2}{x+1} + \dfrac{4}{x-3} - \dfrac{6}{2x+5}\right)$

Take ln of both sides: $\ln y = 2\ln(x+1) + 4\ln(x-3) - 3\ln(2x+5)$. Note the subtraction for the denominator. Differentiate: $\frac{1}{y}\frac{dy}{dx} = \frac{2}{x+1} + \frac{4}{x-3} - \frac{3 \cdot 2}{2x+5} = \frac{2}{x+1} + \frac{4}{x-3} - \frac{6}{2x+5}$. Solve: $\frac{dy}{dx} = y\left(\frac{2}{x+1} + \frac{4}{x-3} - \frac{6}{2x+5}\right)$. Why distractors fail: Option A uses $+\frac{6}{2x+5}$ instead of $-\frac{6}{2x+5}$, forgetting the denominator produces a negative sign. Option C incorrectly subtracts the $\frac{4}{x-3}$ term. Option D writes $\frac{3}{2x+5}$ instead of $\frac{6}{2x+5}$, missing the chain rule factor of $2$ from $\frac{d}{dx}(2x+5)$.

Answer: $-\dfrac{25}{y^3}$

First derivative: $\frac{dy}{dx} = -\frac{x}{y}$. Differentiate again: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{y \cdot 1 - x \cdot \frac{dy}{dx}}{y^2} = -\frac{y - x(-x/y)}{y^2} = -\frac{y + x^2/y}{y^2}$. Simplify using the original equation: $= -\frac{(y^2 + x^2)/y}{y^2} = -\frac{x^2 + y^2}{y^3} = -\frac{25}{y^3}$. Why distractors fail: Option B omits the $x^2$ contribution. Option C has a sign error ($x^2 - y^2$ instead of $x^2 + y^2$). Option D uses $x^2$ instead of $x^2 + y^2 = 25$ in the numerator.

Answer: $-\dfrac{x}{y}$

Differentiate both sides: Differentiating $x^2 + y^2 = 25$ with respect to $x$: $2x + 2y\frac{dy}{dx} = 0$. Solve for dy/dx: $2y\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$. Why distractors fail: Option A has the wrong sign. Option C inverts the fraction. Option D is equivalent to $\frac{x}{y}$ and also has the wrong sign; the ratio $\frac{2x}{2y}$ simplifies to $\frac{x}{y}$.

Answer: $e^{xy}(y + x\frac{dy}{dx}) = 1 + \frac{dy}{dx}$

Differentiate the left side: Using the chain rule on $e^{xy}$: $\frac{d}{dx}[e^{xy}] = e^{xy} \cdot \frac{d}{dx}(xy)$. The product rule gives $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$. So the left side becomes $e^{xy}(y + x\frac{dy}{dx})$. Differentiate the right side: $\frac{d}{dx}(x + y) = 1 + \frac{dy}{dx}$. Why distractors fail: Option B omits the $x\frac{dy}{dx}$ part of the product rule on $xy$. Option C correctly differentiates the left side but writes $1$ on the right instead of $1 + \frac{dy}{dx}$, forgetting the derivative of $y$. Option D misapplies the power rule to the exponential.

Answer: No, because computing $\frac{d^2y}{dx^2}$ requires differentiating the expression for $\frac{dy}{dx}$, which must be found first

Process for implicit second derivatives: To find $\frac{d^2y}{dx^2}$, you first find $\frac{dy}{dx}$ by implicit differentiation, then differentiate that expression with respect to $x$ again (often substituting the first derivative back in). Why the claim fails: While you can differentiate both sides of the original equation a second time, the resulting expression will still involve $\frac{dy}{dx}$ terms. You must know $\frac{dy}{dx}$ to evaluate or simplify $\frac{d^2y}{dx^2}$. Why distractors fail: Option A is absurd — the second derivative is not encoded in the equation. Option B is misleading — even if you differentiate twice without formally isolating $\frac{dy}{dx}$, the first derivative still appears and must be found. Option D is false — implicit curves can have second derivatives wherever the curve is smooth.

Answer: $x^3 + y^3 = 6xy$

Identify when implicit differentiation is needed: Implicit differentiation is used when $y$ cannot be easily solved for in terms of $x$. Options A, B, and D already express $y$ explicitly as a function of $x$. Why the correct answer works: The equation $x^3 + y^3 = 6xy$ (the folium of Descartes) mixes $x$ and $y$ in a way that makes isolating $y$ algebraically impractical, so implicit differentiation is the natural approach. Why distractors fail: Options A, B, and D are all explicit functions of $x$, so their derivatives can be found directly using standard rules without implicit differentiation.

Answer: $(\sqrt[3]{2}, \sqrt[3]{4})$

Find dy/dx: Differentiating $x^3 + y^3 = 3xy$: $3x^2 + 3y^2\frac{dy}{dx} = 3y + 3x\frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{y - x^2}{y^2 - x}$. Set numerator to zero: Horizontal tangent requires $y - x^2 = 0$, i.e., $y = x^2$. Substitute into the original equation: $x^3 + x^6 = 3x \cdot x^2 = 3x^3$, so $x^6 = 2x^3$ giving $x^3 = 2$ (excluding $x=0$), hence $x = \sqrt[3]{2}$ and $y = x^2 = 2^{2/3} = \sqrt[3]{4}$. Verify: Check the point $(\sqrt[3]{2}, \sqrt[3]{4})$: $x^3 + y^3 = 2 + 4 = 6$ and $3xy = 3\sqrt[3]{2}\cdot\sqrt[3]{4} = 3\sqrt[3]{8} = 6$. ✓ Why distractors fail: Option A swaps the coordinates. Option B: at $(1,1)$, $\frac{dy}{dx} = \frac{1-1}{1-1} = \frac{0}{0}$, which is indeterminate (the origin-like self-intersection). Option C: $(3,0)$ does not satisfy $27 + 0 = 0$.

Answer: $(\sin x)^{\ln x}\left(\dfrac{\ln(\sin x)}{x} + \ln x \cdot \cot x\right)$

Take ln of both sides: $\ln y = \ln x \cdot \ln(\sin x)$. Differentiate using the product rule: $\frac{1}{y}\frac{dy}{dx} = \frac{1}{x}\ln(\sin x) + \ln x \cdot \frac{\cos x}{\sin x} = \frac{\ln(\sin x)}{x} + \ln x \cdot \cot x$. Multiply by y: $\frac{dy}{dx} = (\sin x)^{\ln x}\left(\frac{\ln(\sin x)}{x} + \ln x \cdot \cot x\right)$. Why distractors fail: Option B only accounts for part of the derivative and omits the product rule entirely. Option C incorrectly applies the power rule as if $\ln x$ were a constant exponent. Option D has a minus sign instead of plus before $\ln x \cdot \cot x$.

Answer: Take the natural logarithm of both sides to get $\ln y = 3\ln(x^2+1) + 5\ln(x-4)$

Purpose of logarithmic differentiation: Taking $\ln$ of both sides converts products into sums and powers into coefficients, making the expression much easier to differentiate. Apply logarithm properties: $\ln y = \ln[(x^2+1)^3(x-4)^5] = 3\ln(x^2+1) + 5\ln(x-4)$. Each term can now be differentiated individually. Why distractors fail: Option A is extremely tedious and error-prone. Option C misapplies the power rule — the power rule doesn't apply to products of distinct factors this way. Option D rewrites the function incorrectly; the exponent is not $(x^2+1)(x-4)$.

Answer: The approach is incorrect because $e^x$ is not constant — logarithmic differentiation is needed and produces an additional term involving $e^x \ln x$

Correct method: Let $\ln y = e^x \ln x$. Then $\frac{1}{y}\frac{dy}{dx} = e^x \cdot \frac{1}{x} + e^x \ln x = e^x\left(\frac{1}{x} + \ln x\right)$. So $\frac{dy}{dx} = x^{e^x} \cdot e^x\left(\frac{1}{x} + \ln x\right)$. Why the proposed approach fails: The power rule $\frac{d}{dx}(x^n) = nx^{n-1}$ requires $n$ to be constant. Since $e^x$ varies with $x$, treating it as constant misses the contribution from the changing exponent, which produces the $e^x \ln x$ term. Why distractors fail: Option A incorrectly states the power rule applies to variable exponents. Option C is wrong — $x^{e^x}$ is defined for $x > 0$. Option D oversimplifies the correction needed.

Answer: $(0, 2)$ and $(0, -2)$

Find dy/dx: Implicitly differentiating: $\frac{2x}{9} + \frac{2y}{4}\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{4x}{9y}$. Set dy/dx = 0: A horizontal tangent occurs when $\frac{dy}{dx} = 0$, which requires $x = 0$. Substituting into the ellipse: $\frac{y^2}{4} = 1 \implies y = \pm 2$. So the points are $(0, 2)$ and $(0, -2)$. Why distractors fail: Option A gives the points where the tangent is vertical ($y = 0$ makes $\frac{dy}{dx}$ undefined). Option C does not lie on the ellipse — check: $\frac{9}{9} + \frac{4}{4} = 2 \neq 1$. Option D is false.

Answer: The student forgot to multiply the derivative of $\sin(y)$ by $\frac{dy}{dx}$

Correct differentiation: $\frac{d}{dx}[\sin(y)] = \cos(y)\frac{dy}{dx}$, not just $\cos(y)$. The chain rule requires the $\frac{dy}{dx}$ factor since $y$ depends on $x$. Why the correct answer works: The student correctly found $\cos(y)$ as the outer derivative but omitted the essential $\frac{dy}{dx}$ factor from the chain rule. Why distractors fail: Option A suggests the wrong trig derivative — $\frac{d}{dy}[\sin(y)] = \cos(y)$ is correct, not $-\sin(y)$. Option C is incorrect because we want $\frac{dy}{dx}$, so we differentiate with respect to $x$. Option D is unnecessary — implicit differentiation works on the equation as given.

Answer: $x^x(\ln x + 1)$

Take ln of both sides: $\ln y = x \ln x$. Differentiate both sides: $\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$ (using the product rule on $x \ln x$). Solve for dy/dx: $\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$. Why distractors fail: Option A includes an extra factor of $x$. Option C incorrectly applies the power rule as if the exponent were constant. Option D omits the $+1$ term that comes from differentiating $x\ln x$.

Answer: They applied the product rule incorrectly: the derivative of $\cos x \cdot \ln x$ also has the term $\frac{\cos x}{x}$

Correct procedure: Starting from $\ln y = \cos x \cdot \ln x$, differentiate both sides: $\frac{1}{y}\frac{dy}{dx} = -\sin x \cdot \ln x + \cos x \cdot \frac{1}{x}$. Identify the error: The student only wrote $-\sin x \cdot \ln x$, which is only the first term of the product rule applied to $\cos x \cdot \ln x$. They omitted $\frac{\cos x}{x}$. Why distractors fail: Option A: They clearly did take the log (they have $\frac{1}{y}\frac{dy}{dx}$ on the left). Option C: Logarithmic differentiation is the appropriate technique for variable base and exponent. Option D: The error is in the intermediate step, not the final multiplication.

Answer: The upper and lower halves of the circle are reflections across the $x$-axis, so their tangent lines at the same $x$-value have slopes that are negatives of each other

Symmetry of the circle: The circle $x^2 + y^2 = 25$ is symmetric about the $x$-axis. Points $(3, 4)$ and $(3, -4)$ are reflections of each other across this axis. Slopes reflect the symmetry: Since $\frac{dy}{dx} = -\frac{x}{y}$, changing $y$ to $-y$ flips the sign of the slope. This matches the geometry: the tangent at $(3,4)$ slopes downward while the tangent at $(3,-4)$ slopes upward. Why distractors fail: Option A: The circle isn't a function globally, but the derivative exists at each smooth point. Option C: There's no sign error — the formula correctly accounts for negative $y$. Option D: The slopes $-3/4$ and $3/4$ are negatives, not equal, so the lines are not parallel.

Answer: Differentiate $y^2$ as $2y \dfrac{dy}{dx}$ using the chain rule

Recognize y as a function of x: In implicit differentiation, $y$ is treated as a function of $x$. Therefore, differentiating any expression involving $y$ requires the chain rule. Apply the chain rule to $y^2$: $\frac{d}{dx}(y^2) = 2y \cdot \frac{dy}{dx}$. The $\frac{dy}{dx}$ factor appears because $y$ itself depends on $x$. Why distractors fail: Option A omits the essential $\frac{dy}{dx}$ factor. Option C confuses $y$ with $x$. Option D incorrectly assumes $y$ is independent of $x$, which defeats the purpose of implicit differentiation.

Answer: $\dfrac{1}{1 + \cos y}$

Differentiate both sides: $\frac{dy}{dx} + \cos y \cdot \frac{dy}{dx} = 1$. Factor and solve: $\frac{dy}{dx}(1 + \cos y) = 1 \implies \frac{dy}{dx} = \frac{1}{1 + \cos y}$. Why distractors fail: Option A uses $1 - \cos y$ in the denominator — a sign error from incorrectly differentiating $\sin y$. Option C inverts the relationship. Option D has an extra $\cos y$ in the numerator.

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

Take ln of both sides: $\ln y = \frac{1}{2}\ln x + \frac{3}{2}\ln(x^2+1) - \frac{1}{3}\ln(x^3-1)$. Differentiate each term: $\frac{1}{y}\frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x} + \frac{3}{2} \cdot \frac{2x}{x^2+1} - \frac{1}{3} \cdot \frac{3x^2}{x^3-1} = \frac{1}{2x} + \frac{3x}{x^2+1} - \frac{x^2}{x^3-1}$. Why distractors fail: Option B has a plus sign instead of minus for the last term (the denominator produces subtraction). Option C attempts direct differentiation without using logarithmic properties. Option D omits the chain rule factors — for instance, differentiating $\ln(x^2+1)$ gives $\frac{2x}{x^2+1}$, not $\frac{1}{x^2+1}$.

Answer: $y - 4 = -\dfrac{3}{4}(x - 3)$

Find dy/dx: From $x^2 + y^2 = 25$, we get $\frac{dy}{dx} = -\frac{x}{y}$. Evaluate at the given point: At $(3, 4)$: $\frac{dy}{dx} = -\frac{3}{4}$. Write the tangent line: Using point-slope form: $y - 4 = -\frac{3}{4}(x - 3)$. Why distractors fail: Option A has the wrong sign for the slope. Option C uses the negative reciprocal (the normal line slope). Option D swaps the coordinates in the point-slope formula.

Answer: $\dfrac{2y}{x^2}$

Find dy/dx: From $xy = 1$: $y + x\frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{y}{x}$. Differentiate dy/dx: $\frac{d^2y}{dx^2} = -\frac{x\frac{dy}{dx} - y}{x^2} = -\frac{x(-y/x) - y}{x^2} = -\frac{-y - y}{x^2} = \frac{2y}{x^2}$. Why distractors fail: Option A has $x^2y$ in the denominator instead of $x^2$. Option C has the wrong sign. Option D replaces $y$ with $1/x$ prematurely and writes $\frac{2}{x^3}$; while this is numerically equivalent (since $y = 1/x$), the question asks for the answer in terms of both $x$ and $y$.

Answer: $\dfrac{-2xy - y^2}{x^2 + 2xy}$

Differentiate each term using the product rule: For $x^2y$: $\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}$. For $xy^2$: $\frac{d}{dx}(xy^2) = y^2 + 2xy\frac{dy}{dx}$. Combine and solve: $2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$. Collect $\frac{dy}{dx}$ terms: $(x^2 + 2xy)\frac{dy}{dx} = -2xy - y^2$, so $\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy}$. Why distractors fail: Option B has incorrect signs in the numerator. Option C has a sign error in the denominator. Option D omits the $-y^2$ term in the numerator, which comes from differentiating $xy^2$.

Answer: Because the tangent line is vertical when $y = 0$, and vertical lines have undefined slope, which corresponds to division by zero in $-\frac{x}{y}$

Geometric interpretation: When $y = 0$, the points are $(r, 0)$ and $(-r, 0)$, the leftmost and rightmost points of the circle. The tangent lines there are vertical. Why the formula is undefined: $\frac{dy}{dx} = -\frac{x}{y}$ has $y$ in the denominator. When $y = 0$, this expression is undefined — which is geometrically consistent because vertical tangent lines have no finite slope. Why distractors fail: Option A is the opposite — tangent lines are vertical, not horizontal, at $y=0$. Option C is false — $(r, 0)$ satisfies $r^2 + 0 = r^2$. Option D is too restrictive — implicit differentiation works for $y < 0$ as well.