Answer: $\cos(x+y)(2t + 3t^2)$

Apply the chain rule: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} = \cos(x+y)\cdot 2t + \cos(x+y)\cdot 3t^2 = \cos(x+y)(2t + 3t^2)$. Why the correct answer works: Option A correctly factors out $\cos(x+y)$ and sums $2t$ and $3t^2$. Why distractors fail: Option B multiplies the two derivative factors instead of adding them. Option C uses $-\sin$ instead of $\cos$, confusing the derivative of sine. Option D multiplies $2t$ and $3t^2$ to get $6t^3$ rather than adding.

Answer: $-\frac{\partial z}{\partial x}\,r\sin\theta + \frac{\partial z}{\partial y}\,r\cos\theta$

Compute the partial derivatives of $x$ and $y$ with respect to $\theta$: $\frac{\partial x}{\partial \theta} = -r\sin\theta$ and $\frac{\partial y}{\partial \theta} = r\cos\theta$. Assemble the chain rule: $\frac{\partial z}{\partial \theta} = \frac{\partial z}{\partial x}(-r\sin\theta) + \frac{\partial z}{\partial y}(r\cos\theta)$. Why distractors fail: Option A uses $\partial x/\partial r$ and $\partial y/\partial r$ instead of the $\theta$-derivatives. Option C is missing the negative sign on the first term. Option D is missing the factor of $r$.

Answer: $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Identify the derivative type: Since $x$ and $y$ each depend only on the single variable $t$, their derivatives with respect to $t$ are ordinary derivatives $dx/dt$ and $dy/dt$, not partial derivatives. Why the correct answer works: Option B correctly uses ordinary derivatives because $x = g(t)$ and $y = h(t)$ are functions of a single variable. Why distractors fail: Option A uses partial derivative notation, which is incorrect when $x$ and $y$ depend on only one variable. Options C and D invert the derivatives.

Answer: A derivative (ordinary or partial) of the variable at the tail with respect to the variable at the tip

Purpose of tree diagram branches: In a tree diagram, each branch connecting two variables represents the derivative of the variable at the start (tail) of the branch with respect to the variable at the end (tip). Why the correct answer works: Option B correctly identifies that each branch is a derivative — partial if there are multiple variables at that level, ordinary if there is only one. Why distractors fail: Option A confuses which variables the derivative is taken with respect to. Option C refers to the total differential, which is a sum of terms, not a single branch. Option D introduces integration, which has no role in the tree diagram.

Answer: $e^{xy}(y + x)$

Apply the chain rule with two parameters: $\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$. Now $\frac{\partial z}{\partial x} = ye^{xy}$, $\frac{\partial x}{\partial s} = 1$, $\frac{\partial z}{\partial y} = xe^{xy}$, $\frac{\partial y}{\partial s} = 1$. Simplify: $\frac{\partial z}{\partial s} = ye^{xy} + xe^{xy} = e^{xy}(x + y)$. Why distractors fail: Option A uses $x - y$, which would arise if one of the partial derivatives of $x$ or $y$ with respect to $s$ were $-1$ instead of $1$. Option C uses $y - x$, which would result from $\partial y/\partial s = -1$ (that is actually $\partial y/\partial t$). Option D incorrectly substitutes back to get $2s$.

Answer: $2(2x+y) - (x+2y)$

Compute partial derivatives of $w$: $\frac{\partial w}{\partial x} = 2x + y$ and $\frac{\partial w}{\partial y} = x + 2y$. Compute partial derivatives of $x$ and $y$ with respect to $t$: $\partial x/\partial t = 2$ and $\partial y/\partial t = -1$. Assemble the chain rule: $\frac{\partial w}{\partial t} = (2x+y)(2) + (x+2y)(-1) = 2(2x+y) - (x+2y)$. Why distractors fail: Option A uses $\partial y/\partial t = +1$ instead of $-1$. Option B uses $\partial x/\partial t = 2$ correctly but takes $\partial y/\partial t = 3$ (the $s$-derivative of $y$). Option D uses $\partial x/\partial t = 1$ and $\partial y/\partial t = 3$, which are the $s$-derivatives.

Answer: $-\frac{F_x}{F_y}$

Implicit differentiation via the chain rule: Differentiating $F(x,y) = 0$ with respect to $x$ gives $F_x + F_y \frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{F_x}{F_y}$. Why the correct answer works: Option B gives $-F_x/F_y$, which matches the derived formula. Why distractors fail: Option A is missing the negative sign. Option C has the ratio inverted ($F_y/F_x$ instead of $F_x/F_y$). Option D has both the wrong sign and the wrong ratio.

Answer: Two

Trace paths through the tree: From $s$, there are branches to $x$ and $y$ (since both depend on $s$). From $x$ and $y$, there is one branch each to $u$. Total paths: $s \to x \to u$ and $s \to y \to u$, giving two paths. Why the correct answer works: The tree diagram has exactly two paths from $s$ to $u$, corresponding to the two terms in $\frac{\partial u}{\partial s}$. Why distractors fail: Option A miscounts. Option C counts all paths in the diagram (including those through $t$), not just those through $s$. Option D is wrong because the structure is determined by the dependency relationships, not the specific formulas.

Answer: $\frac{6yz - 3x^2}{3z^2 - 6xy}$

Set up $F(x,y,z) = x^3 + y^3 + z^3 - 6xyz = 0$: $F_x = 3x^2 - 6yz$ and $F_z = 3z^2 - 6xy$. Apply the formula: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{3x^2 - 6yz}{3z^2 - 6xy} = \frac{6yz - 3x^2}{3z^2 - 6xy}$. Why distractors fail: Option B omits the negative sign. Option C flips the sign in the denominator as well, resulting in the negative of the correct answer. Option D ignores the $-6xyz$ terms entirely.

Answer: The product actually equals $-1$, not $1$.

Compute each implicit derivative: $\frac{\partial z}{\partial x} = -F_x/F_z$, $\frac{\partial x}{\partial y} = -F_y/F_x$, $\frac{\partial y}{\partial z} = -F_z/F_y$. Multiply them: $\left(-\frac{F_x}{F_z}\right)\left(-\frac{F_y}{F_x}\right)\left(-\frac{F_z}{F_y}\right) = -1$. Why distractors fail: Option A gives $+1$, which would be true in single-variable calculus but fails here due to the three negative signs. Option C is generally false. Option D is incorrect because the computation is well-defined when $F_x, F_y, F_z \neq 0$.

Answer: $36$

Apply the Two-Variable Chain Rule: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$. We have $\frac{\partial z}{\partial x} = 2xy$, $\frac{dx}{dt} = 3$, $\frac{\partial z}{\partial y} = x^2$, $\frac{dy}{dt} = 2t$. Evaluate at $t = 1$: At $t=1$: $x=3$, $y=1$. So $\frac{dz}{dt} = 2(3)(1)(3) + (9)(2) = 18 + 18 = 36$. Why distractors fail: Option A (18) only includes the first term. Option C (9) only includes the second term divided by 2. Option B (27) results from an arithmetic error on the second term, mistakenly computing $9 \cdot 1 = 9$ instead of $9 \cdot 2 = 18$.

Answer: $-\frac{x}{z}$

Apply the implicit differentiation formula: $\frac{\partial z}{\partial x} = -\frac{F_x}{F_z}$. Here $F_x = 2x$ and $F_z = 2z$. Simplify: $\frac{\partial z}{\partial x} = -\frac{2x}{2z} = -\frac{x}{z}$. Why distractors fail: Option A inverts the fraction ($z/x$ instead of $x/z$). Option C omits the negative sign. Option D adds an extra $+1$ in the denominator, suggesting an incorrect computation of $F_z$.

Answer: $-\frac{e+1}{e-1}$

Compute $F_x$ and $F_y$: $F_x = ye^{xy} + 1$ and $F_y = xe^{xy} - 1$. Evaluate at $(1,1)$ and apply the formula: $F_x(1,1) = e + 1$ and $F_y(1,1) = e - 1$. So $\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{e+1}{e-1}$. Why distractors fail: Option A omits the negative sign. Option C inverts the fraction. Option D inverts the fraction and includes a spurious negative that would cancel with the formula's negative.

Answer: Three

List the needed partial derivatives: $\frac{\partial z}{\partial x} = -F_x/F_z$ requires $F_x$ and $F_z$. $\frac{\partial z}{\partial y} = -F_y/F_z$ requires $F_y$ and $F_z$. Count distinct derivatives: The union is $\{F_x, F_y, F_z\}$, which has three elements. Why distractors fail: Options A and B undercount. Option D overcounts — there are only three first-order partials of $F(x,y,z)$.

Answer: $-\frac{z\cos(yz)}{x + y\cos(yz)}$

Compute partial derivatives of $F$: $F_y = z\cos(yz)$ and $F_z = x + y\cos(yz)$. Apply the formula: $\frac{\partial z}{\partial y} = -\frac{F_y}{F_z} = -\frac{z\cos(yz)}{x + y\cos(yz)}$. Why distractors fail: Option B is missing the negative sign. Option C swaps $y$ and $z$ in several places. Option D omits the $y\cos(yz)$ term from $F_z$.

Answer: No; each term has units of $[T]/[\text{time}]$, so both terms share the same units.

Analyze units of each term: The first term has units $\frac{[T]}{[P]}\cdot\frac{[P]}{[\text{time}]} = \frac{[T]}{[\text{time}]}$. The second term has units $\frac{[T]}{[V]}\cdot\frac{[V]}{[\text{time}]} = \frac{[T]}{[\text{time}]}$. Why the correct answer works: Both terms have identical units of $[T]/[\text{time}]$, so they can be added. The critique is not valid. Why distractors fail: Option A is incorrect because the intermediate units cancel properly. Option C adds a false requirement. Option D dismisses unit analysis, which is actually relevant and confirms the formula's validity.

Answer: The student multiplied the two path contributions instead of adding them.

Identify the structural error: The multivariable chain rule sums the contributions from each path, but the student used multiplication between the two path expressions. Why the correct answer works: The correct formula is $\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$. The student replaced the $+$ with $\cdot$. Why distractors fail: Option A is wrong because partial derivatives are appropriate here. Option C is wrong because $z$ depends on $s$ only through $x$ and $y$. Option D suggests a meaningless algebraic operation.

Answer: $\left(\frac{\partial z}{\partial r}\right)^2 + \frac{1}{r^2}\left(\frac{\partial z}{\partial \theta}\right)^2 = \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2$

Derive the chain rule expressions: $\frac{\partial z}{\partial r} = f_x\cos\theta + f_y\sin\theta$ and $\frac{\partial z}{\partial \theta} = -f_x r\sin\theta + f_y r\cos\theta$. Square and add: $(z_r)^2 = f_x^2\cos^2\theta + 2f_x f_y\cos\theta\sin\theta + f_y^2\sin^2\theta$. $\frac{1}{r^2}(z_\theta)^2 = f_x^2\sin^2\theta - 2f_x f_y\cos\theta\sin\theta + f_y^2\cos^2\theta$. Adding: $f_x^2(\cos^2\theta + \sin^2\theta) + f_y^2(\sin^2\theta + \cos^2\theta) = f_x^2 + f_y^2$. Why distractors fail: Options B, C, and D propose identities that do not follow from the chain rule and can be disproved by simple examples.

Answer: $\cos(x+y) - y\sin(xy)$

Differentiate each term with respect to $x$: $\frac{\partial}{\partial x}\sin(x+y) = \cos(x+y)$ and $\frac{\partial}{\partial x}\cos(xy) = -\sin(xy) \cdot y$. Combine: $F_x = \cos(x+y) - y\sin(xy)$. Why distractors fail: Option A forgot the chain rule on $\cos(xy)$. Option C has the wrong sign on the second term. Option D differentiated $xy$ with respect to $x$ and got $x$ instead of $y$.

Answer: Two

Identify which intermediate variables depend on $v$: $x = g(u,v)$ depends on $v$, so $\partial x/\partial v$ exists. $y = h(u)$ does not depend on $v$. $z = k(v)$ depends on $v$. Count the non-zero paths: $\frac{\partial w}{\partial v} = f_x \frac{\partial x}{\partial v} + f_z \frac{dz}{dv}$, two terms. Why distractors fail: Option A misses one path. Option C would require all three intermediate variables to depend on $v$, but $y$ does not. Option D overcounts.

Answer: Two

Determine which intermediate variables depend on $t$: $x = s$ does not depend on $t$, so $\partial x/\partial t = 0$. Both $y = s+t$ and $z = t$ depend on $t$, giving $\partial y/\partial t = 1$ and $\partial z/\partial t = 1$. Count non-zero terms: $\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x}(0) + \frac{\partial w}{\partial y}(1) + \frac{\partial w}{\partial z}(1) = \frac{\partial w}{\partial y} + \frac{\partial w}{\partial z}$, which has two terms. Why distractors fail: Option A undercounts. Option C counts all three intermediate variables, but the $x$ path contributes zero. Option D overcounts.

Answer: $17$

Identify values at $t = 0$: At $t=0$: $x = 2$, $y = 3$. So we use $f_x(2,3) = 5$ and $f_y(2,3) = -2$. Compute derivatives of the parameterization: $dx/dt = 3$ and $dy/dt = -1$. Apply the chain rule: $\frac{dz}{dt} = 5(3) + (-2)(-1) = 15 + 2 = 17$. Why distractors fail: Option A ($13$) may come from $5(3) + (-2)(1) = 13$, using $dy/dt = 1$ instead of $-1$. Option C ($11$) may use $dx/dt = 3$ and $dy/dt = 2$. Option D ($7$) may arise from $5(1) + (-2)(-1)$.

Answer: $-8$

Set up the chain rule: $\frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} + \frac{\partial T}{\partial y}\frac{dy}{dt}$. We have $\frac{dx}{dt} = -2\sin t$ and $\frac{dy}{dt} = 3\cos t$. Evaluate at $t = \pi/2$: $\frac{dx}{dt} = -2\sin(\pi/2) = -2$ and $\frac{dy}{dt} = 3\cos(\pi/2) = 0$. So $\frac{dT}{dt} = 4(-2) + (-1)(0) = -8$. Why distractors fail: Option B ($-11$) might arise from using wrong trig values such as $\sin(\pi/2)=1$ and $\cos(\pi/2)=1$, giving $4(-2)+(-1)(3) = -11$. Option C ($-3$) could result from subtracting the partials incorrectly. Option D ($8$) flips the sign of the correct answer.

Answer: It visually maps the dependency structure among variables to determine which derivatives to multiply along each path.

Purpose of the tree diagram: A tree diagram shows how the dependent variable depends on intermediate variables, which in turn depend on independent variables. Why the correct answer works: Option B accurately describes the tree diagram as a visual tool for mapping dependencies and determining which derivatives to multiply and sum. Why distractors fail: Option A confuses the tree diagram with a surface plot. Option C describes level curves, an unrelated graphical tool. Option D attributes numerical computation to a diagram that only organizes structure.

Answer: $\frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}$

Recall the Two-Variable Chain Rule: When $z = f(x,y)$ and both $x$ and $y$ depend on a single parameter $t$, the chain rule states $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}$. Why the correct answer works: Option B matches this formula exactly, summing the two paths through which $t$ influences $z$. Why distractors fail: Option A multiplies the two partial derivatives together, ignoring the chain factors. Option C multiplies all four terms as a product rather than summing two branches. Option D omits the partial derivatives of $z$ entirely.

Answer: $\frac{\partial w}{\partial x} = f_u + f_v$ and $\frac{\partial w}{\partial y} = f_u - f_v$

Apply the chain rule for $\partial w/\partial x$: $\frac{\partial w}{\partial x} = f_u \frac{\partial u}{\partial x} + f_v \frac{\partial v}{\partial x} = f_u(1) + f_v(1) = f_u + f_v$. Apply the chain rule for $\partial w/\partial y$: $\frac{\partial w}{\partial y} = f_u \frac{\partial u}{\partial y} + f_v \frac{\partial v}{\partial y} = f_u(1) + f_v(-1) = f_u - f_v$. Why distractors fail: Option A claims both expressions are the same, missing that $\partial v/\partial y = -1$. Option C swaps the signs. Option D ignores the chain rule structure entirely.

Answer: Two

Count the intermediate variables: The number of terms equals the number of intermediate variables through which $s$ can influence $z$. Here those intermediate variables are $x$ and $y$. Why the correct answer works: $\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$, which has exactly two terms. Why distractors fail: Option A (one) would only apply if $z$ depended on $s$ through a single intermediate variable. Options C and D overcount; there are only two paths from $s$ to $z$.

Answer: $F_z \neq 0$

Identify the denominator: The formula $\frac{\partial z}{\partial x} = -F_x / F_z$ has $F_z$ in the denominator, so $F_z$ must not be zero. Why the correct answer works: Option B states $F_z \neq 0$, which is necessary by both the formula and the Implicit Function Theorem. Why distractors fail: Option A concerns $F_x$, which appears in the numerator — it can be zero (giving $\partial z/\partial x = 0$). Option C involves $F_y$, which does not appear in this formula. Option D is an arbitrary condition with no basis in the theorem.

Answer: When $f$ is not given as an explicit formula

Consider when substitution is possible: Substituting requires an explicit formula for $f(x,y)$ so that $x(t)$ and $y(t)$ can be plugged in. If $f$ is only described abstractly (e.g., via measured data), substitution and direct differentiation are not feasible. Why the correct answer works: Option B identifies the key limitation: without an explicit formula for $f$, the chain rule provides the only systematic way to express $dz/dt$. Why distractors fail: Option A: polynomials are easy to substitute into. Option C: linear functions make substitution even simpler. Option D: scalar functions are the standard case; this doesn't prevent substitution.

Answer: Both methods are valid and correctly yield $dz/dt = 0$.

Verify both approaches: The chain rule gives $dz/dt = 2x(-\sin t) + 2y(\cos t) = -2\cos t \sin t + 2\sin t \cos t = 0$. Substitution gives $z = \cos^2 t + \sin^2 t = 1$, so $dz/dt = 0$. Why the correct answer works: Both methods are mathematically valid approaches to the same problem and agree. Why distractors fail: Option A incorrectly claims substitution is invalid. Option C incorrectly claims the chain rule doesn't apply. Option D gives a wrong numerical answer.

Answer: $f_x e^s\cos t + f_y e^s\sin t$

Differentiate $x$ and $y$ with respect to $s$: $\partial x/\partial s = e^s\cos t$ and $\partial y/\partial s = e^s\sin t$. Apply the chain rule: $\frac{\partial z}{\partial s} = f_x(e^s\cos t) + f_y(e^s\sin t)$. Why distractors fail: Option B uses the $t$-derivatives instead of the $s$-derivatives. Option C forgets the exponential and trig factors. Option D replaces $\cos t$ and $\sin t$ with 1.

Answer: $\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}$

Extend the chain rule to three intermediate variables: When $w$ depends on $x$, $y$, and $z$, and each depends on $t$, there are three paths from $t$ to $w$, yielding three terms. Why the correct answer works: Option C includes all three summands, one for each intermediate variable. Why distractors fail: Option A only accounts for the path through $x$. Option B omits the $z$ path. Option D treats $w$ as if it depends directly on $t$, ignoring the intermediate variables.

Answer: Yes, the computation is correct.

Verify $F_x$ and $F_z$: $F = x^2 z + yz^2 - 5$. Treating $z$ as constant when differentiating with respect to $x$: $F_x = 2xz$. Treating $x$ as constant when differentiating with respect to $z$: $F_z = x^2 + 2yz$. Both are correct. Check the formula: $\partial z/\partial x = -F_x/F_z = -2xz/(x^2+2yz)$. This is correct. Why distractors fail: Option A incorrectly claims $F_x$ should have a $z^2$ term — that term arises from $yz^2$, but $\partial(yz^2)/\partial x = 0$ since neither $y$ nor $z$ is differentiated with respect to $x$ in $F_x$. Option C adds an extra $z$ factor to the first part of $F_z$. Option D inverts the formula.

Answer: Argue by Clairaut's theorem: since $z$ is a smooth function of $s$ and $t$, the mixed partials are equal without further computation.

Recognize the applicable theorem: If $f$ has continuous second partials and $x(s,t)$, $y(s,t)$ are smooth, then $z(s,t)$ has continuous second partials, and Clairaut's theorem guarantees equality of mixed partials. Why the correct answer works: Option B is most efficient because it avoids all computation by citing a standard result. Why distractors fail: Option A works but is highly inefficient. Option C requires knowing $f$ explicitly. Option D mischaracterizes the problem — $z$ is not implicitly defined here.

Answer: $2$

Compute partial derivatives: $\frac{\partial z}{\partial x} = \frac{2x}{x^2+y}$ and $\frac{\partial z}{\partial y} = \frac{1}{x^2+y}$. Also $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = 2t$. Evaluate at $t = 0$: At $t=0$: $x = 1$, $y = 0$, so $x^2 + y = 1$. Then $\frac{dz}{dt} = \frac{2(1)}{1}\cdot e^0 + \frac{1}{1}\cdot 0 = 2$. Why distractors fail: Option A ($0$) ignores the first term. Option C ($1$) may come from forgetting the factor of $2x$ in $\partial z/\partial x$. Option D ($3$) adds an extra unit.

Answer: $1$

Write the chain rule equation for $\partial z/\partial s$: $\frac{\partial z}{\partial s} = f_x \cdot \frac{\partial x}{\partial s} + f_y \cdot \frac{\partial y}{\partial s}$. We have $\frac{\partial x}{\partial s} = 2st$ and $\frac{\partial y}{\partial s} = t^2$. Substitute known values at $s=1, t=1$: $\frac{\partial x}{\partial s} = 2(1)(1) = 2$ and $\frac{\partial y}{\partial s} = 1^2 = 1$. So $5 = f_x(1,1)\cdot 2 + f_y(1,1)\cdot 1 = 2(2) + f_y = 4 + f_y$. Solve for $f_y$: $f_y(1,1) = 5 - 4 = 1$. Why distractors fail: Option B ($3$) incorrectly solves $2 + f_y = 5$, forgetting to multiply $f_x$ by $\partial x/\partial s = 2$. Option C ($5$) confuses $\partial z/\partial s$ with $f_y$. Option D ($2$) substitutes $f_x$ for $f_y$.

Answer: Both $x$ and $y$ depend on $r$, so the chain rule requires a term for each path from $r$ to $z$, giving two terms.

Identify all paths from $r$ to $z$: $z$ depends on $x$ and $y$, and both $x = r\cos\theta$ and $y = r\sin\theta$ depend on $r$. Hence there are two paths: $r \to x \to z$ and $r \to y \to z$. Why the correct answer works: Option B correctly states that $\frac{\partial z}{\partial r} = \frac{\partial z}{\partial x}\cos\theta + \frac{\partial z}{\partial y}\sin\theta$, a sum of two terms. Why distractors fail: Option A endorses the flawed reasoning. Option C adds a nonexistent direct dependence of $z$ on $r$. Option D incorrectly suggests the chain rule does not apply in polar coordinates.