Answer: $-\sin(xy) - xy\cos(xy)$

Compute $f_y$: $f_y = -x\sin(xy)$. Compute $f_{yx}$: $f_{yx} = \frac{\partial}{\partial x}(-x\sin(xy))$. Using the product rule: $= -\sin(xy) + (-x)(y\cos(xy)) \cdot 1 = -\sin(xy) - xy\cos(xy)$. Wait, let's be careful: $\frac{\partial}{\partial x}(-x\sin(xy)) = -\sin(xy) - x \cdot \cos(xy) \cdot y = -\sin(xy) - xy\cos(xy)$. Why distractors fail: Option A swaps $\sin$ and $\cos$. Option C is $f_{yy}$ (differentiating $f_y$ with respect to $y$). Option D would arise from differentiating $f_x$ with respect to $x$.

Answer: Yes, because $e^x \cos y + (-e^x \cos y) = 0$.

Verify $f_{xx}$: $f_x = e^x \cos y$, $f_{xx} = e^x \cos y$. ✓ Verify $f_{yy}$: $f_y = -e^x \sin y$, $f_{yy} = -e^x \cos y$. ✓ Check Laplace's equation: $f_{xx} + f_{yy} = e^x \cos y - e^x \cos y = 0$. The student's conclusion is correct. Why distractors fail: Options A and B incorrectly claim the computations are wrong. Option D adds a spurious condition — Laplace's equation only involves $f_{xx}$ and $f_{yy}$.

Answer: $2$

Compute $f_x$: $f_x = 2x e^y$. Evaluate at $(1, 0)$: $f_x(1, 0) = 2(1)e^0 = 2 \cdot 1 = 2$. Why distractors fail: Option A forgets the coefficient 2. Option C confuses the evaluation, perhaps using $y=1$. Option D evaluates at $y=1$ instead of $y=0$.

Answer: $a + c = 0$

Compute $f_{xx}$: $f_x = 2ax + by$, so $f_{xx} = 2a$. Compute $f_{yy}$: $f_y = bx + 2cy$, so $f_{yy} = 2c$. Apply Laplace's equation: $f_{xx} + f_{yy} = 2a + 2c = 0 \implies a + c = 0$, or equivalently $c = -a$. Why distractors fail: Option A gives $a = c$, which would yield $f_{xx} + f_{yy} = 4a \neq 0$ in general. Option C imposes a condition on $b$ which is irrelevant since $b$ does not appear in $f_{xx}$ or $f_{yy}$. Option D is unnecessarily restrictive.

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

Set up the chain rule: Let $u = y/x$. Then $\frac{\partial u}{\partial x} = -\frac{y}{x^2}$ and $\frac{d}{du}\arctan(u) = \frac{1}{1+u^2}$. Compute the derivative: $\frac{\partial f}{\partial x} = \frac{1}{1+(y/x)^2} \cdot \left(-\frac{y}{x^2}\right) = \frac{1}{(x^2+y^2)/x^2} \cdot \left(-\frac{y}{x^2}\right) = \frac{x^2}{x^2+y^2} \cdot \left(-\frac{y}{x^2}\right) = \frac{-y}{x^2+y^2}$. Why distractors fail: Option A has the wrong sign. Option C computes $\frac{\partial f}{\partial y}$, not $\frac{\partial f}{\partial x}$. Option D drops the $y$ factor.

Answer: $x e^{xy}$

Apply the chain rule with respect to $y$: We have $f(x,y) = e^{xy}$. Treating $x$ as constant, $\frac{\partial}{\partial y}(xy) = x$. By the chain rule, $f_y = x e^{xy}$. Why the correct answer works: Option C, $x e^{xy}$, correctly applies the chain rule. Why distractors fail: Option A gives $f_x$ instead of $f_y$. Option B omits the chain rule factor entirely. Option D multiplies by $xy$ instead of just $x$.

Answer: Yes — because $g$ is a composition of smooth (infinitely differentiable) functions, all partial derivatives are continuous, so Clairaut's theorem applies.

Check smoothness of the function: $e^{x\sin y}$ is a composition of the exponential function and $x\sin y$. Both are smooth (infinitely differentiable), so their composition is also smooth. Apply Clairaut's theorem: Since all partial derivatives are continuous everywhere, the hypothesis of Clairaut's theorem is satisfied. Hence $g_{xy} = g_{yx}$. Why distractors fail: Option A is overly cautious — Clairaut's theorem exists precisely to avoid this. Option C is false — the exponential function and trigonometric functions are smooth everywhere. Option D is wrong — the argument generalizes to any composition of smooth functions.

Answer: 4

Enumerate all second-order partials: For $f(x,y)$, the second-order partials are: $f_{xx}$, $f_{xy}$, $f_{yx}$, and $f_{yy}$. Count them: There are 4 such derivatives. If Clairaut's theorem applies, $f_{xy} = f_{yx}$ reduces this to 3 independent ones, but the question asks before applying the theorem. Why distractors fail: Option A counts only the unmixed partials. Option B counts the distinct partials after applying Clairaut's theorem. Option D is too few by any count.

Answer: $\frac{2x}{y^3}$

Find $f_y$: $f = xy^{-1}$, so $f_y = x(-1)y^{-2} = -\frac{x}{y^2}$. Find $f_{yy}$: $f_{yy} = \frac{\partial}{\partial y}\left(-xy^{-2}\right) = -x(-2)y^{-3} = \frac{2x}{y^3}$. Why distractors fail: Option A gives $|f_y|$ with the wrong sign. Option B is $f_y$, not $f_{yy}$. Option D has the wrong sign — the two negatives combine to give a positive result.

Answer: $A=0,\; B=6,\; C=1,\; D=0,\; E=0$

Compute general partials: $f_x = 2Ax + By + 3Dx^2$. Then $f_{xx} = 2A + 6Dx$ and $f_{xy} = B$. At $(0,0)$: $f_{xx}(0,0) = 2A$ and $f_{xy}(0,0) = B$. Apply the constraints: We need $f_{xx}(0,0) = 2A = 0$, so $A = 0$. We need $f_{xy}(0,0) = B = 6$. The remaining coefficients $C$, $D$, $E$ are free. Option A uses $A=0, B=6$, satisfying both constraints. Why distractors fail: Option B has $A=3$, giving $f_{xx}(0,0) = 6 \neq 0$. Option C has $B=3$, giving $f_{xy}(0,0) = 3 \neq 6$. Option D has $A=6$ and $B=0$, failing both conditions.

Answer: $3$

Find $f_x$: $f_x = 2x \sin(y) + \frac{y}{x}$. Find $f_{xy}$: $f_{xy} = \frac{\partial}{\partial y}(2x \sin(y) + \frac{y}{x}) = 2x \cos(y) + \frac{1}{x}$. Evaluate at $(1, 0)$: $f_{xy}(1,0) = 2(1)\cos(0) + \frac{1}{1} = 2 + 1 = 3$. Why distractors fail: Option A omits one of the two terms. Option B forgets the $\frac{1}{x}$ term. Option D incorrectly evaluates $\cos(0) = 0$.

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

Apply the chain rule: Let $u = x^2 + y^2$. Then $\frac{\partial f}{\partial x} = \frac{1}{u} \cdot \frac{\partial u}{\partial x} = \frac{2x}{x^2 + y^2}$. Why the correct answer works: Option B correctly applies the chain rule: derivative of $\ln(u)$ is $1/u$, and $\frac{\partial u}{\partial x} = 2x$. Why distractors fail: Option A omits the chain rule factor $2x$. Option C differentiates with respect to $y$ instead of $x$. Option D forgets the factor of 2 from the power rule on $x^2$.

Answer: Because single-variable functions have no other variables to hold constant, so the partial and ordinary derivatives are identical.

Connect partial to ordinary differentiation: The partial derivative $\frac{\partial f}{\partial x}$ holds all other variables constant and differentiates with respect to $x$. When there are no other variables, there is nothing to hold constant. Why the correct answer works: Option B correctly identifies that with only one independent variable, the procedure of 'holding other variables constant' is vacuous. Why distractors fail: Option A is incorrect — both use the same limit definition. Option C is false — single-variable functions are not always linear. Option D is irrelevant — Clairaut's theorem concerns mixed second-order partials of multivariable functions.

Answer: 10

Count without symmetry: Without any symmetry, the number of third-order partials for 3 variables is $3^3 = 27$. Apply Clairaut's theorem: With continuous partials, the order of differentiation does not matter (by repeated application of Clairaut's theorem). So we count the number of ways to choose 3 differentiations from variables $\{x, y, z\}$ with repetition allowed and order irrelevant. This is $\binom{3+3-1}{3} = \binom{5}{3} = 10$. Why distractors fail: Option A is the total without the symmetry reduction. Option C counts second-order distinct partials for 3 variables. Option D does not correspond to a standard counting result.

Answer: The mixed second partial derivatives $f_{xy}$ and $f_{yx}$ must be continuous at the point in question.

Recall the hypothesis of Clairaut's theorem: Clairaut's theorem states: if $f_{xy}$ and $f_{yx}$ are continuous at a point $(a,b)$, then $f_{xy}(a,b) = f_{yx}(a,b)$. Why the correct answer works: Option B directly states the continuity condition on the mixed partials. Why distractors fail: Option A is too restrictive — the theorem applies to any sufficiently smooth function, not just polynomials. Option C is unnecessary; $f$ need only be defined on an open set containing the point. Option D confuses the hypothesis with critical point conditions.

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

Apply the chain rule with respect to $y$: Let $u = x^2 y$. Then $f = \sin(u)$ and $\frac{\partial u}{\partial y} = x^2$. So $\frac{\partial f}{\partial y} = \cos(u) \cdot x^2 = x^2 \cos(x^2 y)$. Why the correct answer works: Option A correctly applies the chain rule treating $x$ as a constant. Why distractors fail: Option B is $\frac{\partial f}{\partial x}$, not $\frac{\partial f}{\partial y}$. Option C omits the chain rule factor $x^2$. Option D incorrectly multiplies by $x^2 y$ instead of $x^2$.

Answer: The slope of the surface in the direction of the $x$-axis at a point, holding $y$ constant

Recall the definition of a partial derivative: $\frac{\partial f}{\partial x}$ is computed by differentiating $f$ with respect to $x$ while treating $y$ as a constant. Geometric meaning: This gives the slope of the curve formed by intersecting the surface $z = f(x, y)$ with a plane of constant $y$. Hence it is the slope in the $x$-direction. Why distractors fail: Option A describes $\frac{\partial f}{\partial y}$, not $\frac{\partial f}{\partial x}$. Option C describes the total differential or gradient magnitude, not a single partial. Option D confuses slope with curvature, which involves second derivatives.

Answer: As a constant

Definition of partial differentiation: When taking $\frac{\partial f}{\partial x}$, all variables other than $x$ are held fixed — that is, treated as constants. Why the correct answer works: Option C correctly states that $y$ is treated as a constant during the differentiation process. Why distractors fail: Option A would apply if using the chain rule with $y = y(x)$, which is not the standard partial derivative setting. Option B is incorrect because $y$ is held at its current value, not set to zero. Option D describes taking the total derivative or gradient, not a partial derivative.

Answer: It measures the instantaneous rate of change of $f$ along the curve where $y$ is held constant, so the surface is sliced by a plane parallel to the $xz$-plane.

Identify which variable is fixed: When computing $\frac{\partial f}{\partial x}$, the variable $y$ is held constant at some value $y_0$. This slices the surface with the plane $y = y_0$, which is parallel to the $xz$-plane. Why the correct answer works: Option B correctly identifies that $y$ is held constant and that the resulting cross-section is a curve in a plane parallel to the $xz$-plane. Why distractors fail: Option A incorrectly holds $x$ constant, which would describe $\frac{\partial f}{\partial y}$. Option C confuses instantaneous rate of change with average rate of change. Option D describes the gradient magnitude and direction, not a single partial derivative.

Answer: $x^2 y + 3z^2$

Differentiate each term with respect to $z$: $\frac{\partial}{\partial z}(x^2 yz) = x^2 y$ (since $x^2 y$ is constant with respect to $z$) and $\frac{\partial}{\partial z}(z^3) = 3z^2$. Why the correct answer works: Combining gives $x^2 y + 3z^2$, which is Option A. Why distractors fail: Option B differentiates $x^2 yz$ with respect to $x$ for the first term. Option C fails to differentiate $z^3$. Option D drops the $y$ factor from the first term.

Answer: $3x^2 y^2 + 5$

Differentiate with respect to $x$: Treating $y$ as a constant: $\frac{\partial}{\partial x}(x^3 y^2) = 3x^2 y^2$ and $\frac{\partial}{\partial x}(5x) = 5$. Combine the results: So $\frac{\partial f}{\partial x} = 3x^2 y^2 + 5$, which is Option A. Why distractors fail: Option B computes $\frac{\partial f}{\partial y}$ of only the first term. Option C fails to differentiate the $5x$ term correctly. Option D differentiates the $y^2$ factor instead of $x^3$.

Answer: At least one of $f_{xy}$ or $f_{yx}$ is discontinuous at $(0,0)$.

Apply the contrapositive of Clairaut's theorem: Clairaut's theorem states: continuous mixed partials $\Rightarrow$ $f_{xy} = f_{yx}$. The contrapositive is: $f_{xy} \neq f_{yx}$ $\Rightarrow$ at least one mixed partial is discontinuous. Why the correct answer works: Since $f_{xy}(0,0) \neq f_{yx}(0,0)$, the hypothesis of Clairaut's theorem fails, meaning at least one mixed partial must be discontinuous at that point. Why distractors fail: Option A is not necessarily true — $f$ can still be differentiable even if mixed partials are discontinuous. Option C is false — such functions do exist (classic examples are constructed using piecewise definitions). Option D is wrong — the theorem has not been 'violated'; rather, its hypothesis is not met.

Answer: $-2x \, e^{-x^2 - y^2}$

Compute $\frac{\partial T}{\partial x}$: Using the chain rule: $\frac{\partial T}{\partial x} = e^{-x^2-y^2} \cdot (-2x) = -2x \, e^{-x^2 - y^2}$. Why the correct answer works: Option B matches this computation exactly. Why distractors fail: Option A differentiates with respect to $y$ instead. Option C computes $\frac{dT}{dt}$ along some path where both $x$ and $y$ change, not a partial derivative. Option D omits the chain rule factor entirely.

Answer: $\frac{\partial^2 f}{\partial y \, \partial x}$

Understand the order convention in Leibniz notation: In Leibniz notation $\frac{\partial^2 f}{\partial y \, \partial x}$, the differentiation is applied right-to-left: first differentiate with respect to $x$, then with respect to $y$. Why the correct answer works: Option B, $\frac{\partial^2 f}{\partial y \, \partial x}$, means $\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)$, which is first with respect to $x$, then $y$. Why distractors fail: Option A ($f_{xx}$) differentiates twice with respect to $x$. Option C ($\frac{\partial^2 f}{\partial x \, \partial y}$) means first with respect to $y$, then $x$ — the reverse order. Option D ($f_{yy}$) differentiates twice with respect to $y$.

Answer: If the mixed partial derivatives are continuous at a point, then $f_{xy} = f_{yx}$ at that point

State Clairaut's theorem: Clairaut's theorem says: if $f_{xy}$ and $f_{yx}$ are both continuous at a point, then $f_{xy} = f_{yx}$ at that point. Why the correct answer works: Option B is an accurate statement of the theorem, including the essential continuity hypothesis. Why distractors fail: Option A is false — existence is not guaranteed in general. Option C confuses mixed partials with unmixed second partials $f_{xx}$ and $f_{yy}$, which need not be equal. Option D is absurd; mixed partials can take any value.

Answer: The continuity hypothesis in Clairaut's theorem is essential — without it, mixed partials can differ.

Identify the role of this example: This is the classic counterexample showing that without continuity of the mixed partials, $f_{xy}$ and $f_{yx}$ need not be equal. Why the correct answer works: Option B correctly identifies that this example demonstrates the necessity of the continuity hypothesis in Clairaut's theorem. Why distractors fail: Option A is imprecise — Clairaut's theorem is not 'false'; its hypothesis simply is not met. Option C is incorrect — mixed partials can exist for piecewise functions. Option D is a separate issue; the function actually has first partial derivatives at the origin.

Answer: $12x^2 + 6y^2$

Find $f_x$: $f_x = 4x^3 + 6xy^2$. Find $f_{xx}$: $f_{xx} = \frac{\partial}{\partial x}(4x^3 + 6xy^2) = 12x^2 + 6y^2$. Why distractors fail: Option B is $f_x$, not the second derivative. Option C replaces $y^2$ with $x^2$ erroneously. Option D differentiates $6xy^2$ incorrectly as $6x^2 y$.

Answer: The mixed partials of any polynomial are continuous everywhere, so Clairaut's theorem guarantees $f_{xy} = f_{yx}$.

Check the hypothesis: Polynomials have continuous partial derivatives of all orders on $\mathbb{R}^2$, so the hypothesis of Clairaut's theorem is satisfied. Apply Clairaut's theorem: Since the mixed partials are continuous, we conclude $f_{xy} = f_{yx}$ everywhere, including at $(1,2)$. Why distractors fail: Option A incorrectly claims Clairaut's theorem does not apply to polynomials. Option C overgeneralizes — the equality can fail for functions with discontinuous mixed partials. Option D ignores the existence of Clairaut's theorem entirely.

Answer: The student made an algebraic or differentiation error, since $f$ is a polynomial and Clairaut's theorem guarantees $f_{xy} = f_{yx}$.

Verify the hypothesis: The function $f(x,y) = x^3 y^2$ is a polynomial. All partial derivatives of polynomials are continuous everywhere. Apply Clairaut's theorem: Since the mixed partials are continuous, Clairaut's theorem guarantees $f_{xy} = f_{yx}$. Any discrepancy must be a computation error. Why distractors fail: Option A mischaracterizes the error. Option C is false — polynomials always satisfy the hypothesis. Option D is irrelevant; standard partial differentiation is the correct approach.

Answer: $6xy^2 - 4$

Find $f_x$ first: $f_x = \frac{\partial}{\partial x}(x^2 y^3 - 4xy + 7) = 2xy^3 - 4y$. Differentiate $f_x$ with respect to $y$: $f_{xy} = \frac{\partial}{\partial y}(2xy^3 - 4y) = 6xy^2 - 4$. Why distractors fail: Option A incorrectly differentiates. Option C gives $f_x$ with an error. Option D gives $f_y$, not $f_{xy}$.

Answer: Compute only $f_{xx}$, $f_{yy}$, and $f_{zz}$ and verify their sum is zero.

Identify what Laplace's equation requires: Laplace's equation $f_{xx} + f_{yy} + f_{zz} = 0$ involves only the three unmixed second-order partial derivatives (the diagonal of the Hessian). Why the correct answer works: Option B is the most efficient and correct approach: compute only the three relevant partials. Why distractors fail: Option A wastes effort computing mixed partials that are not needed. Option C incorrectly applies Clairaut's theorem — the theorem concerns mixed partials, not the relationship between unmixed ones. Option D confuses critical point conditions with Laplace's equation.