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📝 Partial Derivatives in calculus (14 MCQs)

📖 From Calculus • 14. Partial Derivatives Calculus • 14 questions available

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What is Partial Derivatives in calculus?

Definition:
Partial derivative $f_x(a,b) = \lim_{h\to 0} \frac{f(a+h,b)-f(a,b)}{h}$ measures rate of change with respect to one variable while holding others constant.

Example:
For $f(x,y) = x^2y + y^3$ , $f_x = 2xy$ treats $y$ as constant coefficient during differentiation.

Reason:
Isolating individual variable effects decomposes complex multivariable behavior into manageable single-variable components for analysis and approximation.

Easy

Medium

Hard

📝 All Partial Derivatives in calculus MCQs

Q1. A temperature field is modeled by $T(x,y) = e^{-x^2 - y^2}$ . A particle moves along the path $x(t)=t, y(t)=t^2$ . At $t=1$ , which statement correctly interprets $\frac{dT}{dt}$ versus $\nabla T \cdot \vec{v}$ ?

A.They are numerically equal because the chain rule equates the total time derivative to the dot product of gradient and velocity vector. ✅

\frac{dT}{dt}

is larger because it includes temporal changes in the field itself independent of motion.

\nabla T \cdot \vec{v}

is zero at

t=1

since the gradient is orthogonal to parabolic paths.

D.They differ by a factor of

2t

due to the parametrization speed of the curve.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This question tests conceptual understanding of the multivariable chain rule. Students must recognize that for a time-independent scalar field, the rate of change experienced by a moving observer is exactly the directional derivative scaled by velocity. Distractors exploit confusion between partial and total derivatives or geometric orthogonality assumptions that do not apply here.

Q2. In modeling heat diffusion, an engineer writes $u_t = k(u_{xx} + u_{yy})$ . During verification, they compute $u_{xy}$ and $u_{yx}$ symbolically and find unequal expressions. What is the most rigorous error analysis conclusion?

A.The function lacks continuous second-order mixed partials at some point, violating Clairaut’s theorem conditions. ✅

B.Symbolic software always introduces truncation errors in mixed derivatives for PDEs.

C.Heat equations inherently have asymmetric cross-derivatives due to irreversible thermodynamics.

D.The engineer mistakenly differentiated with respect to different variables; mixed partials are always identical.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This requires error analysis beyond computation. While smooth physical models usually satisfy equality of mixed partials, finding inequality implies either a mathematical singularity or non-smoothness in the model. Students must distinguish between computational artifacts and genuine violations of regularity conditions, recognizing that continuity of second derivatives is a sufficient but not necessary condition.

Q3. Given contour lines of $f(x,y)$ where spacing decreases as one moves rightward, and knowing $f_x > 0$ everywhere, what can be definitively concluded about $f_{xx}$ at a point where contours are closest?

f_{xx} > 0

because decreasing contour spacing indicates increasing slope magnitude in the x-direction. ✅

f_{xx} < 0

because tighter contours imply the function is concave down near maxima.

f_{xx} = 0

because contour density relates only to first derivatives, not curvature.

D.Cannot determine sign without information about

f_y

and cross-partial behavior.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Graph-based interpretation requires linking visual contour density to second derivative signs. Decreasing spacing with positive first derivative means the rate of increase accelerates, implying positive concavity. This challenges students who memorize that tight contours mean steepness but fail to connect changing steepness to curvature, distinguishing first from second-order geometric information.

Q4. A student claims that if $f_x(a,b) = 0$ and $f_y(a,b) = 0$ , then $(a,b)$ must be a local extremum. Which counterexample best refutes this while illustrating saddle point geometry?

f(x,y) = x^2 - y^2

at origin has zero gradient but increases in x and decreases in y directions. ✅

f(x,y) = x^3 + y^3

at origin has zero gradient but no extremum due to odd symmetry.

f(x,y) = |x| + |y|

at origin has undefined partials, so premise fails.

f(x,y) = x^4 + y^4

at origin satisfies conditions and is actually a minimum.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Direct recall of critical point classification misconceptions. The hyperbolic paraboloid is the canonical saddle where vanishing gradient coexists with indefinite Hessian. Other options either violate differentiability or represent actual extrema, making them invalid counterexamples. This reinforces that stationarity is necessary but insufficient for extremality in multivariable calculus.

Q5. When optimizing $f(x,y)$ subject to $g(x,y)=c$ , a student solves $\nabla f = \lambda \nabla g$ and finds three candidates. To classify them without second-derivative test, which approach correctly applies bordered Hessian logic conceptually?

A.Evaluate determinant of matrix with border row

[0, g_x, g_y]

; negative determinant indicates constrained maximum for two variables.

B.Count sign changes in eigenvalues of unconstrained Hessian projected onto tangent space of constraint.

C.Substitute constraint into objective to reduce dimension, then apply single-variable second derivative test. ✅

D.Compare Lagrange multiplier magnitudes; largest

|\lambda|

corresponds to global optimum.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Application of alternative classification methods tests deeper understanding. While bordered Hessian is valid, substitution reduces complexity and avoids matrix machinery. Option A misstates sign convention for bordered Hessian. Option B describes correct theory but is computationally equivalent to bordered Hessian. Option D reflects common misconception that multiplier magnitude correlates with optimality rather than sensitivity.

Q6. Consider $f(x,y) = xy \sin(1/(x^2+y^2))$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$ . Which statement accurately describes partial derivatives at origin?

A.Both

f_x(0,0)

and

f_y(0,0)

exist and equal zero, but

f

is not differentiable at origin. ✅

B.Partial derivatives do not exist at origin due to oscillatory singularity.

C.Mixed partials

f_{xy}(0,0)

and

f_{yx}(0,0)

exist but are unequal, demonstrating Clairaut violation.

D.Function is differentiable at origin because partials exist and are continuous in neighborhood.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Challenging problem testing nuanced differentiability concepts. Oscillation damped by $xy$ ensures partials exist via limit definition, but lack of linear approximation prevents differentiability. Students often conflate existence of partials with differentiability or assume oscillation destroys all derivatives. This distinguishes Gateaux from Frechet differentiability in pathological examples relevant to advanced analysis.

Q7. In economic production $Q(K,L)$ , marginal products are $Q_K$ and $Q_L$ . If returns to scale are constant and $Q_{KL} > 0$ , what does this imply about factor complementarity when capital increases?

A.Marginal product of labor increases with capital, indicating factors enhance each other’s productivity synergistically. ✅

B.Marginal product of labor decreases due to diminishing returns dominating interaction effects.

C.Cross-partial positivity violates Euler’s theorem for homogeneous functions of degree one.

D.Complementarity cannot be inferred without knowing elasticity of substitution value.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Scenario-based application connecting mathematics to economics. Positive cross-partial under constant returns implies supermodularity and gross complementarity. Students must integrate homogeneity properties with interaction effects, avoiding confusion between diminishing marginal returns (own-second derivative) and complementarity (cross-derivative). Realistic modeling context tests transfer of abstract calculus concepts to applied domains.

Q8. A numerical analyst approximates $f_x(x_0,y_0)$ using central difference $[f(x_0+h,y_0)-f(x_0-h,y_0)]/(2h)$ . If $f$ has bounded third derivative, what is the dominant error term’s dependence on $h$ ?

A.Error scales as

O(h^2)

because Taylor expansion cancels odd-powered terms in symmetric stencil. ✅

B.Error scales as

O(h)

due to truncation of higher-order terms in finite difference.

C.Error scales as

O(h^3)

because central difference uses three-point approximation.

D.Error is independent of

h

for smooth functions due to exact cancellation.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Conceptual understanding of numerical differentiation accuracy. Central difference achieves second-order convergence through symmetry eliminating first-order error. Students confusing forward/backward differences select linear error. Those overestimating precision pick cubic. This connects theoretical Taylor series to practical computational trade-offs, emphasizing why method choice matters in scientific computing applications.

Q9. Given $z = f(u,v)$ where $u=x+y, v=x-y$ , a student computes $z_{xx}$ as $f_{uu} + f_{vv}$ . Identify the specific error in this derivation.

A.Missing cross-term

2f_{uv}

because chain rule produces

(u_x)^2 f_{uu} + 2u_x v_x f_{uv} + (v_x)^2 f_{vv}

. ✅

B.Incorrect coefficients; should be

f_{uu} - f_{vv}

due to opposite signs in transformation.

C.Student confused partial with total derivative; should use Jacobian determinant instead.

D.No error exists; transformation diagonalizes Laplacian for this coordinate change.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Error analysis targeting chain rule application in variable transformations. Common mistake omits mixed partial term when both intermediate variables depend on same original variable. Correct expansion requires product rule on first derivatives. This tests procedural fluency beyond mechanical computation, revealing whether students understand composition structure versus memorizing formulas for specific cases like polar coordinates.

Q10. For $f(x,y) = x^3 - 3xy^2$ , level curves near origin form three-fold symmetry. Without computation, how does this geometric feature relate to partial derivatives at critical point?

A.Vanishing gradient with nonzero third-order terms creates monkey saddle; quadratic approximation fails to capture local topology. ✅

B.Three-fold symmetry implies

f_{xx}=f_{yy}=0

but

f_{xy} \neq 0

at origin.

C.Symmetry forces all partial derivatives up to third order to vanish at critical point.

D.Level curve geometry determines only first derivatives; higher-order behavior requires explicit calculation.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Mixed concepts linking algebraic structure, geometry, and calculus. Harmonic polynomial exhibits degenerate critical point where Hessian vanishes identically. Students must recognize that standard second-derivative test is inconclusive and higher-order analysis needed. Connects complex analysis (real part of $z^3$ ) to multivariable calculus, rewarding interdisciplinary insight over rote procedure.

Q11. In machine learning, gradient descent updates $\theta \leftarrow \theta - \eta \nabla L(\theta)$ . If loss surface has high condition number, why does naive gradient descent exhibit zigzagging despite correct gradient direction?

A.Gradient points toward steepest descent but not toward minimum when level sets are elongated ellipses. ✅

B.Numerical errors accumulate causing oscillations perpendicular to true gradient direction.

C.Learning rate adapts automatically to curvature, creating instability in ill-conditioned regions.

D.Zigzagging indicates incorrect gradient computation; true gradient always points directly to minimum.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Application connecting optimization theory to practical algorithm behavior. High condition number means eigenvalue disparity stretches level sets. Steepest descent direction becomes nearly orthogonal to optimal path. Students must distinguish gradient direction from convergence trajectory, understanding why preconditioning or momentum methods address this fundamental geometric limitation rather than being mere heuristics.

Q12. Two surfaces $z=f(x,y)$ and $z=g(x,y)$ intersect tangentially at $P$ . If $\nabla f(P) = \nabla g(P) \neq \vec{0}$ , what additional condition ensures second-order contact?

A.Hessians satisfy

H_f(P) - H_g(P)

has rank zero when restricted to tangent plane of intersection.

B.Second partial derivatives match exactly:

f_{xx}=g_{xx}, f_{xy}=g_{xy}, f_{yy}=g_{yy}

P

. ✅

C.Gradients are parallel rather than equal, allowing proportional scaling of normal vectors.

D.Surfaces share identical Gaussian curvature but not necessarily mean curvature at

P

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Olympiad-style problem requiring precise characterization of contact order. Tangential intersection gives first-order agreement. Second-order contact demands matching quadratic approximations, i.e., identical Hessians. Restriction to tangent plane in option A is insufficient as normal curvature components matter. Tests deep understanding of Taylor expansion geometry beyond typical textbook treatments of osculating surfaces.

Q13. A physics model uses $F(x,y,z,t)$ where $z=z(x,y)$ defines a moving surface. When computing material derivative $DF/Dt$ , which expression correctly accounts for surface constraint?

F_t + F_x x_t + F_y y_t + F_z z_t

with

z_t = z_x x_t + z_y y_t

substituted from surface kinematics.

B.Simply

F_t + \nabla F \cdot \vec{v}

treating

z

as independent coordinate.

F_t + (F_x + F_z z_x)x_t + (F_y + F_z z_y)y_t

reducing to surface-intrinsic coordinates before time differentiation. ✅

D.Material derivative undefined on constrained manifolds without introducing Lagrange multipliers.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Multi-step reasoning combining chain rule, constraints, and continuum mechanics. Must substitute surface relation before or during differentiation consistently. Option A double-counts dependencies. Option B ignores constraint. Option C correctly reduces degrees of freedom first, ensuring chain rule respects manifold structure. Tests ability to navigate composite dependencies in applied settings beyond standard textbook examples.

Q14. Student argues that since $f(x,y) \to 0$ along every straight line through origin, $\lim_{(x,y)\to(0,0)} f(x,y) = 0$ . For $f(x,y) = x^2 y / (x^4 + y^2)$ , why is this reasoning flawed despite line limits being zero?

A.Parabolic path

y=x^2

yields limit

1/2

, proving path-dependence despite radial consistency. ✅

B.Straight-line test is sufficient for continuity; function actually has limit zero but student miscomputed.

C.Limit exists but equals infinity along parabola, contradicting boundedness on lines.

D.Function is discontinuous only because denominator vanishes; numerator behavior is irrelevant.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Conceptual understanding of multivariable limits versus single-variable intuition. Classic counterexample shows radial limits insufficient for full limit existence. Students must recognize that uncountably many nonlinear paths exist beyond lines. Reinforces epsilon-delta definition necessity and warns against overgeneralizing sequential criteria. Essential foundation for rigorous partial derivative continuity discussions later.

🔗 Related Topics (MCQs)

📝 Functions of Two or More Variables MCQs 📝 Multivariable function notation and terminology MCQs 📝 Functions of two variables from tables MCQs 📝 Level curves and contour plots MCQs 📝 Graphing functions of two variables MCQs

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