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πŸ“ Higher order partial derivatives (14 MCQs)

πŸ“– From Calculus β€’ 14. Partial Derivatives Calculus β€’ 14 questions available

What is Higher order partial derivatives?

Definition:
Successive differentiation yielding fxx,fxy,fyx,fyyf_{xx}, f_{xy}, f_{yx}, f_{yy}, etc., measuring curvature and interaction effects between variables.

Example:
For f=x3y2f=x^3y^2, fxy=βˆ‚βˆ‚y(3x2y2)=6x2yf_{xy} = \frac{\partial}{\partial y}(3x^2y^2) = 6x^2y and fyx=βˆ‚βˆ‚x(2x3y)=6x2yf_{yx} = \frac{\partial}{\partial x}(2x^3y) = 6x^2y.

Reason:
Second-order derivatives classify critical points via Hessian matrix and appear in Taylor expansions, PDEs, and error estimation formulas.

2
Easy
7
Medium
5
Hard

πŸ“ All Higher order partial derivatives MCQs

Q1. A smooth surface z=f(x,y)z = f(x,y) has a critical point at the origin where fxx>0f_{xx} > 0, fyy<0f_{yy} < 0, and fxy=0f_{xy} = 0. A student claims this must be a local minimum because fxx>0f_{xx} > 0. Which statement best identifies the flaw in this reasoning?

A.The student ignored that fyy<0f_{yy} < 0 indicates concavity downward in y, making it a saddle point regardless of fxxf_{xx}. βœ…
B.The student should have checked fxy≠0f_{xy} \neq 0 first before analyzing second derivatives.
C.The sign of fxxf_{xx} alone determines minima only when fyy=0f_{yy} = 0.
D.The critical point cannot exist if mixed partials are zero.
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: This question targets error analysis by exposing the misconception that a single positive second derivative guarantees a minimum. Students must understand that classification requires the Hessian determinant D=fxxfyyβˆ’(fxy)2D = f_{xx}f_{yy} - (f_{xy})^2. Here D<0D < 0, confirming a saddle point despite fxx>0f_{xx} > 0, demonstrating multi-step reasoning about curvature in multiple directions.

Q2. Given f(x,y)=x3y2+sin⁑(xy)f(x,y) = x^3y^2 + \sin(xy), compute fxxyf_{xxy} at (0,0)(0,0). Which approach minimizes computational error while maintaining rigor?

A.Differentiate with respect to x twice first, then y, evaluating each intermediate step at (0,0).
B.Use Clairaut’s theorem to reorder as fyxxf_{yxx}, differentiate y first to simplify trigonometric terms before applying x-derivatives. βœ…
C.Apply product rule directly on original expression three times without reordering.
D.Compute fxyxf_{xyx} instead since mixed partials always commute numerically.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: This application question emphasizes strategic method selection over brute-force computation. Recognizing that differentiating sin⁑(xy)\sin(xy) with respect to y first yields xcos⁑(xy)x\cos(xy), which simplifies subsequent x-differentiation at origin, reduces algebraic complexity. This tests conceptual understanding of commutativity and practical efficiency in higher-order derivative evaluation.

Q3. A contour plot shows level curves of f(x,y)f(x,y) becoming increasingly dense near point P along the x-direction but uniformly spaced along y. What can be inferred about fxx(P)f_{xx}(P) and fyy(P)f_{yy}(P)?

A.fxx(P)f_{xx}(P) is large positive; fyy(P)β‰ˆ0f_{yy}(P) \approx 0
B.fxx(P)β‰ˆ0f_{xx}(P) \approx 0; fyy(P)f_{yy}(P) is large negative
C.Both second partials are approximately zero
D.fxx(P)f_{xx}(P) is large in magnitude; fyy(P)f_{yy}(P) is small in magnitude βœ…
πŸ’‘ Difficulty: medium | βœ… Correct: D

πŸ“– Explanation: This graph-based interpretation question links visual density of contours to second derivative magnitude. Dense spacing indicates rapid change in slope, implying large |f_xx|. Uniform spacing suggests constant slope, so f_yy β‰ˆ 0. Students must distinguish between first and second derivative information from contour geometry, avoiding confusion with gradient direction or absolute function values.

Q4. For f(x,y)=ex+y+x2yf(x,y) = e^{x+y} + x^2y, which equality must hold if f is CΒ³ smooth?

A.fxxy=fxyx=fyxxf_{xxy} = f_{xyx} = f_{yxx} βœ…
B.fxxy=fyyxf_{xxy} = f_{yyx}
C.fxxx=fyyyf_{xxx} = f_{yyy}
D.fxy=fyxf_{xy} = f_{yx} only at critical points
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: This direct recall question verifies foundational knowledge of Clairaut’s theorem for third-order mixed partials. Under sufficient smoothness (CΒ³), all permutations of differentiation order yield identical results. The distractors test common misconceptions: confusing orders, assuming symmetry without smoothness, or restricting equality to special points. Mastery here enables confident manipulation in complex multivariable problems.

Q5. A physical model defines temperature T(x,y,t)T(x,y,t) where TxytT_{xyt} represents rate of change of horizontal thermal gradient with time. If experimental data shows Txyt≠TxtyT_{xyt} \neq T_{xty} at some point, what is the most plausible explanation?

A.Measurement error or insufficient smoothness in the underlying physical process βœ…
B.Thermal diffusion violates Clairaut’s theorem universally
C.Time derivatives never commute with spatial ones
D.The coordinate system is non-Cartesian
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: This scenario-based error analysis question connects mathematical theory to real-world modeling. While Clairaut’s theorem guarantees equality under smoothness, physical measurements may reflect discontinuities, noise, or non-differentiable phenomena. Students must distinguish theoretical assumptions from empirical limitations, recognizing that observed inequality suggests either data issues or breakdown of model regularity rather than mathematical falsehood.

Q6. Consider f(x,y)=x4+y4βˆ’4x2y2f(x,y) = x^4 + y^4 - 4x^2y^2. At (0,0), all second partials vanish. How should one proceed to classify this critical point?

A.Conclude it is degenerate and use higher-order derivatives or directional analysis βœ…
B.Assume it is a saddle point due to the minus sign
C.Declare it a global minimum since fourth powers dominate
D.Apply second derivative test with modified formula
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: This challenging problem exposes limitations of second-order tests when Hessian is singular. Students must recognize degeneracy and transition to alternative methods like examining f along paths (e.g., y=x gives negative values, y=0 gives positive), proving saddle behavior. This integrates mixed concepts of critical point classification, polynomial analysis, and higher-order reasoning beyond standard calculus procedures.

Q7. If fxx=6xf_{xx} = 6x, fyy=2f_{yy} = 2, and fxy=0f_{xy} = 0 everywhere, reconstruct f(x,y) up to constants. Which term is necessarily absent?

A.x3x^3
B.y2y^2
C.xyxy
D.x2yx^2y βœ…
πŸ’‘ Difficulty: medium | βœ… Correct: D

πŸ“– Explanation: This reverse-engineering application tests integration of partial derivatives and understanding of cross-term generation. Integrating f_xx gives xΒ³ + A(y)x + B(y); f_xy=0 forces A’(y)=0 so A constant; f_yy=2 implies B''(y)=2 so B=yΒ²+Cy+D. No xΒ²y term arises because f_xy=0 eliminates mixed dependence. Distractors probe confusion between integration constants and functional forms.

Q8. A student computes fxyf_{xy} for f=x2ln⁑(y)f = x^2 \ln(y) as 2x/y2x/y but gets fyx=2x/y+x2/y2f_{yx} = 2x/y + x^2/y^2. They conclude Clairaut’s theorem fails. What is the actual error?

A.Incorrect differentiation of ln(y) with respect to x in f_yx calculation
B.Misapplication of product rule in f_xy
C.Failure to recognize domain restriction y>0
D.Arithmetic mistake in f_yx; correct value is also 2x/y βœ…
πŸ’‘ Difficulty: medium | βœ… Correct: D

πŸ“– Explanation: This error analysis question diagnoses computational mistakes disguised as theoretical contradictions. Both mixed partials should equal 2x/y. The erroneous extra term suggests incorrect differentiation of xΒ² w.r.t. y (treating it as variable). Students must verify calculations before questioning fundamental theorems, reinforcing precision and self-checking habits in multivariable calculus workflows.

Q9. On a topographic map, ridge lines correspond to where fxx<0f_{xx} < 0 and fyy>0f_{yy} > 0 along principal axes. If a path follows constant elevation through such a ridge, what does fxyf_{xy} typically indicate?

A.Rate of rotation of principal curvature axes relative to coordinate grid βœ…
B.Magnitude of steepest ascent perpendicular to path
C.Zero, since elevation is constant
D.Curvature along the path direction
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: This advanced graph-concept integration links terrain features to mixed partials. Along a ridge at constant height, f_x=f_y=0, but f_xy measures how principal curvatures rotate relative to x,y axes. Nonzero f_xy indicates ridge isn’t aligned with coordinates. This transcends basic derivative computation, requiring synthesis of differential geometry intuition with partial derivative meaning in applied contexts.

Q10. Given u(x,y)=g(x+y)+h(xβˆ’y)u(x,y) = g(x+y) + h(x-y) for arbitrary CΒ² functions g,h, which identity always holds?

A.uxxβˆ’uyy=0u_{xx} - u_{yy} = 0 βœ…
B.uxx+uyy=0u_{xx} + u_{yy} = 0
C.uxy=0u_{xy} = 0
D.uxx=uyy=uxyu_{xx} = u_{yy} = u_{xy}
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: This conceptual understanding question reveals wave equation structure hidden in functional form. Computing derivatives: u_x = g’+h’, u_xx=g’’+h’’; u_y=g’-h’, u_yy=g’’+h’’; thus u_xx-u_yy=0. Students must generalize beyond specific functions, recognizing d’Alembert solution form. Distractors test confusion with Laplace equation or incorrect chain rule application, emphasizing structural insight over computation.

Q11. In optimizing profit P(K,L)P(K,L), a firm finds PKL>0P_{KL} > 0 at current input levels. Economically, this implies:

A.Capital and labor are complementary; increasing K raises marginal product of L βœ…
B.Capital and labor are substitutes; more K reduces need for L
C.Diminishing returns to scale are present
D.Profit function is concave ensuring global maximum
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: This application question translates mathematical signs into economic meaning. Positive cross-partial means marginal productivity of one input increases with the other, defining complementarity. Students must connect abstract derivatives to real-world decision-making, avoiding confusion with own-second derivatives (diminishing returns) or global properties. This bridges pure math and interdisciplinary modeling skills essential for applied mathematics.

Q12. A function satisfies fxxx+fyyy=0f_{xxx} + f_{yyy} = 0 everywhere. Which statement about its third-order behavior is necessarily true?

A.Sum of pure third derivatives vanishes, but mixed partials remain unconstrained βœ…
B.All third-order partials must be zero
C.Function is harmonic in three dimensions
D.Mixed partials satisfy fxxy=βˆ’fxyyf_{xxy} = -f_{xyy}
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: This Olympiad-style question probes deep understanding of PDE constraints. The given condition only restricts pure third derivatives; mixed partials like f_xxy aren’t determined by it. Harmonicity involves second derivatives. Students must avoid overgeneralizing and recognize independence of derivative types, testing precise logical deduction versus pattern-matching instincts common in lower-level problems.

Q13. When approximating f(a+h,b+k)f(a+h,b+k) via Taylor series, the coefficient of hkhk term is fxy(a,b)f_{xy}(a,b). If this coefficient is negative, what geometric feature dominates near (a,b)?

A.Surface twists such that increase in x decreases slope in y-direction βœ…
B.Local maximum in both variables
C.Saddle oriented along y=x line
D.Concave down paraboloid
πŸ’‘ Difficulty: medium | βœ… Correct: A

πŸ“– Explanation: This mixed-concept question ties Taylor coefficients to surface geometry. Negative f_xy means cross-curvature opposes diagonal growth, creating twist where x-increase reduces y-slope. Students must visualize beyond simple maxima/minima, interpreting bilinear term’s role in shaping local topology. This integrates series expansion, derivative meaning, and spatial reasoning crucial for advanced multivariable analysis.

Q14. Suppose f(x,y)f(x,y) has continuous third partials and fxxy(1,2)=5f_{xxy}(1,2) = 5. Without computing, what is fyxx(1,2)f_{yxx}(1,2)?

A.5 βœ…
B.Cannot determine without explicit formula
C.βˆ’5
D.Depends on path taken to (1,2)
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: This direct recall reinforces Clairaut’s theorem applicability under continuity. Since third partials are continuous, all six permutations of x,x,y differentiation are equal. The question deliberately avoids computation to isolate conceptual mastery of symmetry conditions. Distractors target doubts about path-dependence or sign changes, ensuring students internalize sufficient conditions for interchangeability rather than memorizing formulas mechanically.

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