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πŸ“ Partial derivatives of three variable functions (14 MCQs)

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

What is Partial derivatives of three variable functions?

Definition:
Computing fx,fy,fzf_x, f_y, f_z for w=f(x,y,z)w=f(x,y,z) by differentiating with respect to one variable while fixing the other two as constants.

Example:
For f(x,y,z)=xyz+ezf(x,y,z) = xyz + e^z, fz=xy+ezf_z = xy + e^z treats x,yx,y as constants.

Reason:
Three-variable partials extend gradient concepts to R3\mathbb{R}^3, enabling analysis of volumetric fields, thermodynamic potentials, and spatial dynamics.

4
Easy
7
Medium
3
Hard

πŸ“ All Partial derivatives of three variable functions MCQs

Q1. A thermodynamic system is modeled by U(S,V,N)U(S, V, N). If entropy SS and volume VV are held constant while particle number NN changes, which partial derivative represents the chemical potential, and why is holding other variables fixed physically necessary?

A.βˆ‚U/βˆ‚N\partial U / \partial N because energy change per particle defines chemical potential regardless of constraints.
B.βˆ‚U/βˆ‚N\partial U / \partial N at constant S,VS, V because chemical potential is defined under adiabatic, rigid conditions to isolate particle exchange effects. βœ…
C.βˆ‚U/βˆ‚S\partial U / \partial S at constant V,NV, N since entropy drives particle flow in isolated systems.
D.βˆ‚U/βˆ‚V\partial U / \partial V at constant S,NS, N because volume change correlates with particle addition in compressible media.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The chemical potential is rigorously defined as the partial derivative of internal energy with respect to particle number while keeping entropy and volume constant. This constraint ensures that the measured energy change arises solely from adding particles, not from heat transfer or mechanical work, reflecting true thermodynamic conjugacy.

Q2. In a four-variable production function Q(K,L,M,T)Q(K, L, M, T), a student computes βˆ‚Q/βˆ‚K\partial Q / \partial K but forgets that technology TT depends implicitly on capital via T=g(K)T = g(K). What error does this introduce in marginal productivity analysis?

A.No error; partial derivatives by definition ignore indirect dependencies.
B.Underestimates true marginal product because it omits the positive feedback of capital-induced technological improvement. βœ…
C.Overestimates marginal product by double-counting capital’s direct and indirect effects.
D.Misattributes all output change to labor since technology is treated as exogenous.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: When computing partial derivatives in applied models, one must distinguish between explicit and implicit dependencies. If T=g(K)T = g(K), then the total derivative dQ/dKdQ/dK includes both βˆ‚Q/βˆ‚K\partial Q/\partial K and (βˆ‚Q/βˆ‚T)(dT/dK)(\partial Q/\partial T)(dT/dK). Ignoring the latter yields an incomplete marginal productivity measure, leading to flawed investment decisions in economic modeling.

Q3. Given f(x,y,z,w)=x2y+yzwβˆ’w3f(x,y,z,w) = x^2 y + y z w - w^3, a learner claims fxyz=0f_{xyz} = 0 because no term contains all three variables multiplied together. Is this reasoning valid, and what is the correct third-order mixed partial?

A.Valid; mixed partials vanish unless all variables appear in a single monomial.
B.Invalid; fxyz=1f_{xyz} = 1 because differentiating yzwyzw first by xx gives 0, but order matters and fyxz≠fxyzf_{yxz} \neq f_{xyz}.
C.Invalid; fxyz=1f_{xyz} = 1 since βˆ‚/βˆ‚x(yzw)=0\partial/\partial x (yzw) = 0, but βˆ‚/βˆ‚y(x2y)=x2\partial/\partial y (x^2 y) = x^2, then βˆ‚/βˆ‚z(x2)=0\partial/\partial z (x^2) = 0; actually fxyz=0f_{xyz} = 0, but the reasoning about monomials is flawed. βœ…
D.Invalid; fxyz=1f_{xyz} = 1 because βˆ‚3f/βˆ‚xβˆ‚yβˆ‚z\partial^3 f / \partial x \partial y \partial z applied to yzwyzw yields ww, not zero.
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: The student’s conclusion that fxyz=0f_{xyz} = 0 is numerically correct, but the justification is misleading. Mixed partials depend on sequential differentiation, not just monomial structure. Here, fy=x2+zwf_y = x^2 + zw, then fyx=2xf_{yx} = 2x, and fyxz=0f_{yxz} = 0. By Clairaut’s theorem, all orders yield zero, but the reasoning must invoke differentiation rules, not syntactic presence of variables.

Q4. A contour plot shows level surfaces of h(a,b,c,d)h(a,b,c,d) projected onto the abab-plane for fixed c=2,d=5c=2, d=5. Near point PP, contours are densely packed along aa but sparse along bb. What can be inferred about partial derivatives at PP?

A.βˆ£βˆ‚h/βˆ‚a∣β‰ͺβˆ£βˆ‚h/βˆ‚b∣|\partial h/\partial a| \ll |\partial h/\partial b| because dense contours indicate slow change.
B.βˆ£βˆ‚h/βˆ‚aβˆ£β‰«βˆ£βˆ‚h/βˆ‚b∣|\partial h/\partial a| \gg |\partial h/\partial b| since contour density reflects rate of change in that direction. βœ…
C.Both partials are zero because the projection loses information about cc and dd.
D.Cannot determine without seeing full 4D graph; 2D projections are insufficient for partial derivative inference.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Contour line spacing inversely correlates with gradient magnitude in the projected plane. Dense contours along aa imply rapid change in hh with respect to aa, hence large βˆ£βˆ‚h/βˆ‚a∣|\partial h/\partial a|. Sparse contours along bb indicate gentle variation, so smaller βˆ£βˆ‚h/βˆ‚b∣|\partial h/\partial b|. This interpretation holds even in higher dimensions when other variables are fixed, making 2D slices valid for local partial analysis.

Q5. In climate modeling, temperature T(x,y,z,t)T(x,y,z,t) depends on spatial coordinates and time. A researcher uses βˆ‚T/βˆ‚t\partial T/\partial t to assess local warming but ignores advection terms. Under what condition is this partial derivative sufficient for predicting actual temperature change at a weather station?

A.Always sufficient because stations are fixed in space.
B.Only if wind velocity is zero everywhere, eliminating advective transport. βœ…
C.Never sufficient; total derivative must always include spatial gradients dotted with velocity.
D.Sufficient only during nighttime when convection ceases.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: The partial derivative βˆ‚T/βˆ‚t\partial T/\partial t measures temperature change at a fixed location, ignoring movement of air masses. Actual observed change follows the material derivative DT/Dt=βˆ‚T/βˆ‚t+vβƒ—β‹…βˆ‡TDT/Dt = \partial T/\partial t + \vec{v} \cdot \nabla T. Thus, βˆ‚T/βˆ‚t\partial T/\partial t alone predicts real change only when advection vanishes (vβƒ—=0\vec{v} = 0). In most atmospheric contexts, neglecting advection leads to significant forecast errors, highlighting the distinction between Eulerian and Lagrangian perspectives.

Q6. Consider F(u,v,w,x)=uvwxF(u,v,w,x) = u v w x. A student argues that since FF is symmetric in all variables, all second-order mixed partials like FuvF_{uv} and FwxF_{wx} must be equal. Evaluate this claim using properties of multivariable functions.

A.Correct; symmetry implies all mixed partials are identical regardless of variable pair.
B.Incorrect; Fuv=wxF_{uv} = w x and Fwx=uvF_{wx} = u v, which are equal only if ux=vwux = vw, not universally. βœ…
C.Correct; Clairaut’s theorem guarantees equality of all mixed partials for smooth symmetric functions.
D.Incorrect; mixed partials are never equal unless the function is separable.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Symmetry of the function does not imply equality of arbitrary mixed partials. While Fuv=FvuF_{uv} = F_{vu} by Clairaut’s theorem, Fuv=wxF_{uv} = wx and Fwx=uvF_{wx} = uv involve different variable pairs and are generally unequal. Symmetry means F(u,v,w,x)=F(Οƒ(u,v,w,x))F(u,v,w,x) = F(\sigma(u,v,w,x)) for permutations Οƒ\sigma, but mixed partials transform accordingly. Equality occurs only under specific value constraints, not as a general property, revealing a common misconception about symmetry and derivatives.

Q7. An engineer models stress Οƒ(Ο΅,T,Ο΅Λ™,H)\sigma(\epsilon, T, \dot{\epsilon}, H) in a viscoelastic material, where strain Ο΅\epsilon, temperature TT, strain rate Ο΅Λ™\dot{\epsilon}, and humidity HH interact. During testing, TT and HH drift unintentionally. How should partial derivatives be interpreted in regression analysis of experimental data?

A.As true physical sensitivities since regression isolates each variable’s effect.
B.As apparent sensitivities confounded by uncontrolled covariates; true partials require controlled experiments or multivariate correction. βœ…
C.As total derivatives because real-world variables never vary independently.
D.As irrelevant; only total differentials matter in empirical settings.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: In observational or imperfectly controlled experiments, estimated coefficients approximate partial derivatives only if other variables are truly held constant. Drift in TT and HH introduces omitted variable bias, making regression estimates reflect combined effects rather than pure βˆ‚Οƒ/βˆ‚Ο΅\partial \sigma / \partial \epsilon. Valid partial derivative estimation requires either experimental control or statistical adjustment (e.g., multiple regression with all relevant predictors), emphasizing the gap between mathematical definition and empirical practice in multivariable systems.

Q8. Let G(p,q,r,s)=pq2r3s4G(p,q,r,s) = p q^2 r^3 s^4. Without computing directly, determine the value of GpqrsG_{pqr s} at (1,1,1,1)(1,1,1,1) using combinatorial reasoning about differentiation orders.

A.0, because fourth-order mixed partial of a degree-10 monomial exceeds its total degree.
B.24, since each differentiation reduces exponent by 1 and multiplies by original exponent, yielding 1β‹…2β‹…3β‹…41 \cdot 2 \cdot 3 \cdot 4. βœ…
C.1, because all exponents become 1 after differentiation and evaluation at unity.
D.Undefined, as mixed partials of order exceeding variable count do not exist.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: For a monomial paqbrcsdp^a q^b r^c s^d, the mixed partial βˆ‚a+b+c+dG/βˆ‚pβˆ‚qβˆ‚rβˆ‚s\partial^{a+b+c+d} G / \partial p \partial q \partial r \partial s equals a!b!c!d!a! b! c! d! only if differentiating each variable exactly its exponent times. Here, differentiating once per variable gives (1)(2q)(3r2)(4s3)(1)(2q)(3r^2)(4s^3) evaluated at 1, yielding 1β‹…2β‹…3β‹…4=241 \cdot 2 \cdot 3 \cdot 4 = 24. The key insight is recognizing that single differentiation per variable preserves non-zero result, and factorial logic applies only for repeated differentiation of same variable.

Q9. A student computes βˆ‚2f/βˆ‚xβˆ‚y\partial^2 f / \partial x \partial y for f(x,y,z)=xyzsin⁑(z)f(x,y,z) = x y z \sin(z) and obtains zsin⁑(z)z \sin(z). Another claims it should be zcos⁑(z)z \cos(z) due to product rule on sin⁑(z)\sin(z). Who is correct and why?

A.First student; sin⁑(z)\sin(z) is constant w.r.t. xx and yy, so derivative is zsin⁑(z)z \sin(z). βœ…
B.Second student; product rule must apply to zsin⁑(z)z \sin(z) even when differentiating w.r.t. other variables.
C.Neither; the correct answer is sin⁑(z)+zcos⁑(z)\sin(z) + z \cos(z) from full product rule.
D.First student; but only because zz is independent of x,yx,y, making βˆ‚(zsin⁑z)/βˆ‚y=0\partial (z \sin z)/\partial y = 0 before multiplying by xx.
πŸ’‘ Difficulty: easy | βœ… Correct: A

πŸ“– Explanation: When taking βˆ‚2f/βˆ‚xβˆ‚y\partial^2 f / \partial x \partial y, treat zz as constant during both differentiations since it is an independent variable. First, βˆ‚f/βˆ‚y=xzsin⁑(z)\partial f / \partial y = x z \sin(z). Then βˆ‚/βˆ‚x\partial / \partial x of that is zsin⁑(z)z \sin(z). The product rule involving sin⁑(z)\sin(z) and zz is irrelevant here because neither depends on xx or yy. This tests understanding of variable independence in partial differentiation versus ordinary calculus contexts.

Q10. In optimizing f(a,b,c,d)f(a,b,c,d) subject to two constraints, Lagrange multipliers yield a critical point. To classify it, one examines the bordered Hessian. Why can’t the standard Hessian test for unconstrained extrema be applied directly?

A.Because constraints reduce dimensionality, altering definiteness criteria; bordered Hessian incorporates constraint gradients to project curvature onto feasible subspace. βœ…
B.Standard Hessian still works but requires more computation; bordered version is merely convenient.
C.Constraints make the function non-differentiable, invalidating Hessian tests entirely.
D.Bordered Hessian accounts for numerical instability in high dimensions, not theoretical necessity.
πŸ’‘ Difficulty: hard | βœ… Correct: A

πŸ“– Explanation: Constrained optimization restricts movement to a manifold defined by constraints. The standard Hessian measures curvature in full space, including infeasible directions. The bordered Hessian augments the matrix with constraint gradients to evaluate curvature only along tangent directions of the feasible set. Its signature determines constrained extremum type via modified Sylvester’s criterion. Applying unconstrained tests ignores geometric restrictions, potentially misclassifying saddle points as minima or vice versa, underscoring the need for specialized tools in multivariable constrained analysis.

Q11. A neural network activation depends on weights w1,w2,w3,w4w_1, w_2, w_3, w_4 via A=tanh⁑(w1w2+w3w4)A = \tanh(w_1 w_2 + w_3 w_4). During backpropagation, a developer mistakenly treats w1w2w_1 w_2 as a single variable when computing βˆ‚A/βˆ‚w1\partial A / \partial w_1. What is the consequence?

A.No consequence; chain rule automatically handles composite terms.
B.Underestimates gradient magnitude by omitting multiplication by w2w_2, slowing convergence. βœ…
C.Overestimates gradient by including spurious cross-terms from w3w4w_3 w_4.
D.Causes division by zero when w2=0w_2 = 0.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Correct application of chain rule gives βˆ‚A/βˆ‚w1=sech2(w1w2+w3w4)β‹…w2\partial A / \partial w_1 = \text{sech}^2(w_1 w_2 + w_3 w_4) \cdot w_2. Treating w1w2w_1 w_2 as atomic ignores the inner derivative βˆ‚(w1w2)/βˆ‚w1=w2\partial (w_1 w_2)/\partial w_1 = w_2, yielding only sech2(β‹…)\text{sech}^2(\cdot). This omission scales the true gradient by 1/w21/w_2 (when w2β‰ 0w_2 \neq 0), distorting update steps. In deep learning, such errors propagate through layers, causing training instability or stagnation, illustrating precise partial derivative computation’s role in algorithmic correctness.

Q12. Suppose P(x,y,z,t)P(x,y,z,t) satisfies βˆ‚P/βˆ‚t=kβˆ‡2P\partial P/\partial t = k \nabla^2 P in 3D space. At a point where βˆ‚P/βˆ‚x=βˆ‚P/βˆ‚y=βˆ‚P/βˆ‚z=0\partial P/\partial x = \partial P/\partial y = \partial P/\partial z = 0 but βˆ‚2P/βˆ‚x2>0\partial^2 P/\partial x^2 > 0, what can be said about temporal evolution?

A.PP increases over time because positive Laplacian implies local minimum smoothing.
B.PP decreases because diffusion acts to flatten peaks, and positive second derivative indicates concave-up profile needing reduction.
C.Temporal change cannot be determined without knowing kk’s sign and other second derivatives. βœ…
D.PP remains constant since first spatial derivatives vanish.
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: The heat equation links temporal change to the full Laplacian βˆ‡2P=Pxx+Pyy+Pzz\nabla^2 P = P_{xx} + P_{yy} + P_{zz}. Knowing only Pxx>0P_{xx} > 0 is insufficient; PyyP_{yy} and PzzP_{zz} could dominate negatively. Even if net Laplacian were positive, kk’s sign determines increase/decrease. Vanishing first derivatives indicate a critical point, but curvature in all directions governs diffusion. This highlights that partial information about spatial derivatives cannot predict dynamics without complete second-order context in PDEs.

Q13. Compare computing βˆ‚f/βˆ‚x\partial f / \partial x for f(x,y,z)=exyzf(x,y,z) = e^{xyz} via direct differentiation versus logarithmic differentiation. In what scenario would logarithmic method offer computational advantage despite added steps?

A.Never; exponential functions differentiate cleanly without logs.
B.When ff is a product of many exponentials, converting sum simplifies repeated partials. βœ…
C.Only for functions with negative exponents to avoid sign errors.
D.When variables are correlated, logs linearize dependencies for easier partials.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Logarithmic differentiation transforms products into sums: ln⁑f=xyz\ln f = xyz for single exponential, offering no gain. But if f=∏i=1negi(x,y,z)f = \prod_{i=1}^n e^{g_i(x,y,z)}, then ln⁑f=βˆ‘gi\ln f = \sum g_i, and βˆ‚f/βˆ‚x=fβ‹…βˆ‘βˆ‚gi/βˆ‚x\partial f / \partial x = f \cdot \sum \partial g_i / \partial x. This avoids applying product rule across nn terms, reducing algebraic complexity. The advantage emerges in multiplicative models common in probability or economics, where log-space computation streamlines partial derivatives despite initial transformation cost.

Q14. A physicist asserts that for any smooth F(a,b,c,d)F(a,b,c,d), Fabcd=FdcbaF_{abcd} = F_{dcba} always holds. A mathematician counters that this requires continuous fourth-order partials. Who is correct in practical scientific computing?

A.Physicist; natural phenomena are infinitely differentiable, so equality is guaranteed.
B.Mathematician; numerical approximations may violate equality if discretization lacks sufficient smoothness or resolution. βœ…
C.Both are wrong; mixed partials commute only up to third order in four variables.
D.Physicist; computational floating-point errors are negligible compared to modeling assumptions.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Clairaut’s theorem guarantees equality of mixed partials only when those partials are continuous. In theory, physical fields are smooth, but numerical methods use discrete grids where continuity isn’t assured. Finite-difference schemes may produce asymmetric results if step sizes differ or if underlying data has noise. Thus, while mathematically true under ideal conditions, practical computation demands verification of smoothness or use of symmetric stencils. This bridges abstract analysis and applied numerics, showing why theoretical assumptions matter in implementation.

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