What is Partial derivatives of three variable functions?
Definition: Computing fxβ,fyβ,fzβ for 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+ez, fzβ=xy+ez treats x,y as constants.
Reason: Three-variable partials extend gradient concepts to R3, enabling analysis of volumetric fields, thermodynamic potentials, and spatial dynamics.
4
Easy
7
Medium
3
Hard
π All Partial derivatives of three variable functions MCQs
Q1. A thermodynamic system is modeled by U(S,V,N). If entropy S and volume V are held constant while particle number N changes, which partial derivative represents the chemical potential, and why is holding other variables fixed physically necessary?
A.βU/βN because energy change per particle defines chemical potential regardless of constraints.
B.βU/βN at constant S,V because chemical potential is defined under adiabatic, rigid conditions to isolate particle exchange effects. β
C.βU/βS at constant V,N since entropy drives particle flow in isolated systems.
D.βU/βV at constant S,N because volume change correlates with particle addition in compressible media.
π‘ Difficulty: medium | β Correct: B
π 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), a student computes βQ/βK but forgets that technology T depends implicitly on capital via 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. β
C.Overestimates marginal product by double-counting capitalβs direct and indirect effects.
D.Misattributes all output change to labor since technology is treated as exogenous.
π‘ Difficulty: hard | β Correct: B
π Explanation: When computing partial derivatives in applied models, one must distinguish between explicit and implicit dependencies. If T=g(K), then the total derivative dQ/dK includes both βQ/βK and (βQ/βT)(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+yzwβw3, a learner claims fxyzβ=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; fxyzβ=1 because differentiating yzw first by x gives 0, but order matters and fyxzβξ =fxyzβ.
C.Invalid; fxyzβ=1 since β/βx(yzw)=0, but β/βy(x2y)=x2, then β/βz(x2)=0; actually fxyzβ=0, but the reasoning about monomials is flawed. β
D.Invalid; fxyzβ=1 because β3f/βxβyβz applied to yzw yields w, not zero.
π‘ Difficulty: medium | β Correct: C
π Explanation: The studentβs conclusion that fxyzβ=0 is numerically correct, but the justification is misleading. Mixed partials depend on sequential differentiation, not just monomial structure. Here, fyβ=x2+zw, then fyxβ=2x, and fyxzβ=0. By Clairautβs 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) projected onto the ab-plane for fixed c=2,d=5. Near point P, contours are densely packed along a but sparse along b. What can be inferred about partial derivatives at P?
A.β£βh/βaβ£βͺβ£βh/βbβ£ because dense contours indicate slow change.
B.β£βh/βaβ£β«β£βh/βbβ£ since contour density reflects rate of change in that direction. β
C.Both partials are zero because the projection loses information about c and d.
D.Cannot determine without seeing full 4D graph; 2D projections are insufficient for partial derivative inference.
π‘ Difficulty: easy | β Correct: B
π Explanation: Contour line spacing inversely correlates with gradient magnitude in the projected plane. Dense contours along a imply rapid change in h with respect to a, hence large β£βh/βaβ£. Sparse contours along b indicate gentle variation, so smaller β£βh/βbβ£. 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) depends on spatial coordinates and time. A researcher uses βT/βt 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. β
C.Never sufficient; total derivative must always include spatial gradients dotted with velocity.
D.Sufficient only during nighttime when convection ceases.
π‘ Difficulty: medium | β Correct: B
π Explanation: The partial derivative βT/βt measures temperature change at a fixed location, ignoring movement of air masses. Actual observed change follows the material derivative DT/Dt=βT/βt+vβ βT. Thus, βT/βt alone predicts real change only when advection vanishes (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)=uvwx. A student argues that since F is symmetric in all variables, all second-order mixed partials like Fuvβ and Fwxβ 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; Fuvβ=wx and Fwxβ=uv, which are equal only if ux=vw, not universally. β
C.Correct; Clairautβs theorem guarantees equality of all mixed partials for smooth symmetric functions.
D.Incorrect; mixed partials are never equal unless the function is separable.
π‘ Difficulty: hard | β Correct: B
π Explanation: Symmetry of the function does not imply equality of arbitrary mixed partials. While Fuvβ=Fvuβ by Clairautβs theorem, Fuvβ=wx and Fwxβ=uv involve different variable pairs and are generally unequal. Symmetry means F(u,v,w,x)=F(Ο(u,v,w,x)) for permutations Ο, 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 Ο(Ο΅,T,Ο΅Λ,H) in a viscoelastic material, where strain Ο΅, temperature T, strain rate Ο΅Λ, and humidity H interact. During testing, T and H drift unintentionally. How should partial derivatives be interpreted in regression analysis of experimental data?
A.As true physical sensitivities since regression isolates each variableβs effect.
B.As apparent sensitivities confounded by uncontrolled covariates; true partials require controlled experiments or multivariate correction. β
C.As total derivatives because real-world variables never vary independently.
D.As irrelevant; only total differentials matter in empirical settings.
π‘ Difficulty: medium | β Correct: B
π Explanation: In observational or imperfectly controlled experiments, estimated coefficients approximate partial derivatives only if other variables are truly held constant. Drift in T and H introduces omitted variable bias, making regression estimates reflect combined effects rather than pure βΟ/βΟ΅. 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)=pq2r3s4. Without computing directly, determine the value of Gpqrsβ at (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 4. β
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.
π‘ Difficulty: easy | β Correct: B
π Explanation: For a monomial paqbrcsd, the mixed partial βa+b+c+dG/βpβqβrβs equals a!b!c!d! only if differentiating each variable exactly its exponent times. Here, differentiating once per variable gives (1)(2q)(3r2)(4s3) evaluated at 1, yielding 1β 2β 3β 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 β2f/βxβy for f(x,y,z)=xyzsin(z) and obtains zsin(z). Another claims it should be zcos(z) due to product rule on sin(z). Who is correct and why?
A.First student; sin(z) is constant w.r.t. x and y, so derivative is zsin(z). β
B.Second student; product rule must apply to zsin(z) even when differentiating w.r.t. other variables.
C.Neither; the correct answer is sin(z)+zcos(z) from full product rule.
D.First student; but only because z is independent of x,y, making β(zsinz)/βy=0 before multiplying by x.
π‘ Difficulty: easy | β Correct: A
π Explanation: When taking β2f/βxβy, treat z as constant during both differentiations since it is an independent variable. First, βf/βy=xzsin(z). Then β/βx of that is zsin(z). The product rule involving sin(z) and z is irrelevant here because neither depends on x or y. This tests understanding of variable independence in partial differentiation versus ordinary calculus contexts.
Q10. In optimizing f(a,b,c,d) subject to two constraints, Lagrange multipliers yield a critical point. To classify it, one examines the bordered Hessian. Why canβt the standard Hessian test for unconstrained extrema be applied directly?
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.
π‘ Difficulty: hard | β Correct: A
π 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 Sylvesterβs 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 w1β,w2β,w3β,w4β via A=tanh(w1βw2β+w3βw4β). During backpropagation, a developer mistakenly treats w1βw2β as a single variable when computing βA/βw1β. What is the consequence?
B.Underestimates gradient magnitude by omitting multiplication by w2β, slowing convergence. β
C.Overestimates gradient by including spurious cross-terms from w3βw4β.
D.Causes division by zero when w2β=0.
π‘ Difficulty: medium | β Correct: B
π Explanation: Correct application of chain rule gives βA/βw1β=sech2(w1βw2β+w3βw4β)β w2β. Treating w1βw2β as atomic ignores the inner derivative β(w1βw2β)/βw1β=w2β, yielding only sech2(β ). This omission scales the true gradient by 1/w2β (when w2βξ =0), distorting update steps. In deep learning, such errors propagate through layers, causing training instability or stagnation, illustrating precise partial derivative computationβs role in algorithmic correctness.
Q12. Suppose P(x,y,z,t) satisfies βP/βt=kβ2P in 3D space. At a point where βP/βx=βP/βy=βP/βz=0 but β2P/βx2>0, what can be said about temporal evolution?
A.P increases over time because positive Laplacian implies local minimum smoothing.
B.P 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 kβs sign and other second derivatives. β
D.P remains constant since first spatial derivatives vanish.
π‘ Difficulty: medium | β Correct: C
π Explanation: The heat equation links temporal change to the full Laplacian β2P=Pxxβ+Pyyβ+Pzzβ. Knowing only Pxxβ>0 is insufficient; Pyyβ and Pzzβ could dominate negatively. Even if net Laplacian were positive, kβs 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 βf/βx for f(x,y,z)=exyz 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 f is a product of many exponentials, converting sum simplifies repeated partials. β
C.Only for functions with negative exponents to avoid sign errors.
D.When variables are correlated, logs linearize dependencies for easier partials.
π‘ Difficulty: easy | β Correct: B
π Explanation: Logarithmic differentiation transforms products into sums: lnf=xyz for single exponential, offering no gain. But if f=βi=1nβegiβ(x,y,z), then lnf=βgiβ, and βf/βx=fβ ββgiβ/βx. This avoids applying product rule across n 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), Fabcdβ=Fdcbaβ 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. β
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.
π‘ Difficulty: medium | β Correct: B
π Explanation: Clairautβs 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 isnβt 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.