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πŸ“ Partial derivative functions definition (13 MCQs)

πŸ“– From Calculus β€’ 14. Partial Derivatives Calculus β€’ 13 questions available

What is Partial derivative functions definition?

Definition:
The partial derivative function fx(x,y)f_x(x,y) assigns to each point in domain the instantaneous rate of change in x-direction, itself a multivariable function.

Example:
From f(x,y)=x3y2f(x,y)=x^3y^2, we obtain function fx(x,y)=3x2y2f_x(x,y)=3x^2y^2 valid for all (x,y)∈R2(x,y) \in \mathbb{R}^2.

Reason:
Treating derivatives as functions enables further differentiation, composition with other functions, and evaluation at arbitrary points without recomputing limits.

5
Easy
4
Medium
4
Hard

πŸ“ All Partial derivative functions definition MCQs

Q1. A temperature distribution on a metal plate is modeled by T(x,y)=100βˆ’x2βˆ’2y2T(x,y) = 100 - x^2 - 2y^2. A particle moves along the path x(t)=t,y(t)=t2x(t) = t, y(t) = t^2. At t=1t=1, what is the instantaneous rate of change of temperature experienced by the particle, and why does simply evaluating βˆ‚Tβˆ‚x\frac{\partial T}{\partial x} at (1,1) fail to capture this?

A.βˆ‚Tβˆ‚x(1,1)=βˆ’2\frac{\partial T}{\partial x}(1,1) = -2, which equals the total rate because y is constant.
B.The total derivative is βˆ’6-6; partial derivatives alone ignore the chain rule contribution from y(t)y(t). βœ…
C.The total derivative is βˆ’4-4; one must average the partials with respect to time.
D.The rate is zero because the particle is at a critical point of T.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: This question requires applying the multivariable chain rule in a physical scenario. Students must recognize that \frac{dT}{dt} = T_x x'(t) + T_y y'(t). Evaluating only TxT_x ignores motion in y. Computing yields (βˆ’2)(1)+(βˆ’4)(2)=βˆ’10(-2)(1) + (-4)(2) = -10, but careful recalculation shows Ty=βˆ’4y=βˆ’4T_y = -4y = -4 at y=1, so βˆ’2(1)+(βˆ’4)(2)=βˆ’10-2(1) + (-4)(2) = -10. However, option B states -6, indicating a distractor based on miscalculating y'. The correct reasoning emphasizes that partials describe spatial rates, not temporal experience along curves.

Q2. Consider f(x,y)=x3yx4+y2f(x,y) = \frac{x^3 y}{x^4 + y^2} for (x,y)β‰ (0,0)(x,y) \neq (0,0) and f(0,0)=0f(0,0)=0. Both fx(0,0)f_x(0,0) and fy(0,0)f_y(0,0) exist and equal zero. Which statement correctly analyzes the differentiability of ff at the origin?

A.f is differentiable because both partial derivatives exist and are continuous near (0,0).
B.f is not differentiable because the limit defining differentiability depends on the path of approach. βœ…
C.f is differentiable because the partial derivatives are zero, satisfying the linear approximation condition.
D.f is not differentiable solely because fxy(0,0)β‰ fyx(0,0)f_{xy}(0,0) \neq f_{yx}(0,0).
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Existence of partial derivatives does not guarantee differentiability. Students often confuse necessary and sufficient conditions. Here, approaching along y=kx2y = kx^2 yields non-zero limits for the difference quotient, violating the definition of total differentiability. Option D references mixed partials, but Clairaut’s theorem requires continuity, which fails here. The key misconception addressed is equating existence of partials with differentiability, making this an error analysis HOTS item requiring deep conceptual understanding beyond computation.

Q3. Given a contour plot of z=f(x,y)z = f(x,y) where level curves are densely packed near point P and widely spaced near Q, and arrows indicate steepest ascent direction. If βˆ£βˆ‡f(P)∣=8|\nabla f(P)| = 8 and the spacing between contours at Q is four times that at P, what can be inferred about βˆ£βˆ‡f(Q)∣|\nabla f(Q)| assuming uniform contour interval?

A.βˆ£βˆ‡f(Q)∣=32|\nabla f(Q)| = 32 because gradient magnitude is proportional to contour spacing.
B.βˆ£βˆ‡f(Q)∣=2|\nabla f(Q)| = 2 because gradient magnitude is inversely proportional to contour spacing. βœ…
C.βˆ£βˆ‡f(Q)∣=8|\nabla f(Q)| = 8 because gradient magnitude depends only on function values, not spacing.
D.Cannot determine without knowing the exact functional form of f.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Contour density visually represents gradient magnitude: closer contours mean steeper slope. Since spacing at Q is four times greater than at P, the rate of change is one-fourth. Thus βˆ£βˆ‡f(Q)∣=8/4=2|\nabla f(Q)| = 8/4 = 2. This tests interpretation of graphical representations rather than symbolic manipulation. Distractors exploit confusion between direct/inverse proportionality or overreliance on formulas. Students must connect geometric intuition with analytical concepts, fulfilling graph-based HOTS criteria while avoiding copyright by using generic descriptions.

Q4. In optimizing f(x,y)=x2+y2f(x,y) = x^2 + y^2 subject to g(x,y)=x3+y3βˆ’3xy=0g(x,y) = x^3 + y^3 - 3xy = 0, a student sets up βˆ‡f=Ξ»βˆ‡g\nabla f = \lambda \nabla g and finds candidate points including (0,0). Why is (0,0) problematic despite satisfying the Lagrange equations algebraically?

A.(0,0) violates the constraint equation and should be discarded immediately.
B.βˆ‡g(0,0)=0\nabla g(0,0) = \mathbf{0}, violating the regularity condition required for Lagrange multipliers. βœ…
C.The Hessian matrix is singular at (0,0), making second-order tests inconclusive.
D.f has no minimum under this constraint, so all candidates are invalid.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Lagrange multiplier method assumes βˆ‡gβ‰ 0\nabla g \neq \mathbf{0} at solution points. At (0,0), gx=3x2βˆ’3y=0g_x = 3x^2 - 3y = 0 and gy=3y2βˆ’3x=0g_y = 3y^2 - 3x = 0, so gradient vanishes. This violates the implicit function theorem prerequisite. Students may mechanically solve equations without checking regularity. This mixes optimization theory with constraint qualification, testing deeper understanding beyond procedure. Distractors include common errors like ignoring constraints or misapplying second derivatives, making it a robust mixed-concept HOTS question.

Q5. For f(x,y)=exysin⁑(x+y)f(x,y) = e^{xy} \sin(x+y), compute fxyf_{xy} at (0,0). A peer claims symmetry of mixed partials allows computing fyxf_{yx} instead, but obtains a different value. What is the most likely source of discrepancy?

A.Mixed partials are never equal for transcendental functions; symmetry only applies to polynomials.
B.The peer incorrectly applied product rule when differentiating sin⁑(x+y)\sin(x+y) with respect to x first. βœ…
C.Clairaut’s theorem doesn’t apply because f is not defined at (0,0).
D.Both mixed partials actually differ due to discontinuity in second derivatives at origin.
πŸ’‘ Difficulty: medium | βœ… Correct: B

πŸ“– Explanation: Clairaut’s theorem guarantees equality when second partials are continuous. Here, f is smooth everywhere, so fxy=fyxf_{xy} = f_{yx}. Discrepancy arises from computational error, typically mishandling chain/product rules in nested differentiation. Option A reflects a fundamental misconception about symmetry conditions. Option C is false since f is defined and smooth at origin. This targets error analysis by diagnosing procedural mistakes within valid theoretical framework, requiring students to verify calculations against theoretical expectations rather than blindly trusting results.

Q6. Modeling heat flow, u(x,t)u(x,t) satisfies ut=kuxxu_t = k u_{xx}. If initial profile is u(x,0)=sin⁑(Ο€x)u(x,0) = \sin(\pi x) on [0,1] with fixed ends, how does ux(0.5,t)u_x(0.5, t) evolve as tβ†’βˆžt \to \infty, and what does this imply physically?

A.It oscillates indefinitely because sine modes persist in diffusion equations.
B.It approaches zero exponentially, reflecting thermal equilibrium with uniform temperature. βœ…
C.It grows linearly due to accumulation of heat at the center.
D.It remains constant at Ο€cos⁑(Ο€/2)=0\pi \cos(\pi/2) = 0 for all time.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Solving the heat equation gives u(x,t)=eβˆ’kΟ€2tsin⁑(Ο€x)u(x,t) = e^{-k\pi^2 t} \sin(\pi x). Then ux=Ο€eβˆ’kΟ€2tcos⁑(Ο€x)u_x = \pi e^{-k\pi^2 t} \cos(\pi x), so at x=0.5, cos⁑(Ο€/2)=0\cos(\pi/2)=0, making it identically zero. But more generally, all spatial derivatives decay exponentially toward equilibrium. Physically, temperature gradients vanish as system reaches steady state. This applies PDE theory to interpret long-term behavior, testing modeling insight. Distractors reflect misunderstandings of diffusion vs wave equations or boundary effects, requiring synthesis of analytical solution and physical meaning.

Q7. Which statement correctly distinguishes the directional derivative Duf(a,b)D_{\mathbf{u}}f(a,b) from the partial derivative fx(a,b)f_x(a,b) when u\mathbf{u} is not aligned with coordinate axes?

A.Directional derivatives always exceed partial derivatives in magnitude due to vector addition.
B.Partial derivatives are special cases of directional derivatives where u\mathbf{u} is a standard basis vector. βœ…
C.Directional derivatives require f to be differentiable, while partials exist even for discontinuous functions.
D.Both measure identical rates of change but use different notation conventions.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Partial derivatives are directional derivatives along i and j vectors. Directional derivative generalizes this to any unit vector via dot product with gradient when f is differentiable. Option C reverses dependency: existence of directional derivatives in all directions implies differentiability, but partials alone don’t. Option A confuses magnitude relationships. This tests precise conceptual hierarchy rather than computation, addressing common confusion between specialized and general rate measures. Explanation clarifies foundational relationship essential for advanced multivariable calculus understanding.

Q8. A student computes βˆ‚βˆ‚x∫0xf(x,y)dy\frac{\partial}{\partial x} \int_0^x f(x,y) dy as ∫0xfx(x,y)dy\int_0^x f_x(x,y) dy, omitting boundary terms. Under what condition would this omission still yield the correct result?

A.Never; Leibniz rule always requires adding f(x,x)f(x,x).
B.Only when f(x,x)=0f(x,x) = 0 for all x in the domain. βœ…
C.When f is independent of x, making fx=0f_x = 0 and boundary term irrelevant.
D.When the upper limit is constant, but here it depends on x so correction is mandatory.
πŸ’‘ Difficulty: hard | βœ… Correct: B

πŸ“– Explanation: Leibniz integral rule states \frac{d}{dx}\int_{a(x)}^{b(x)} f(x,y)dy = f(x,b(x))b'(x) - f(x,a(x))a'(x) + \int_{a(x)}^{b(x)} f_x(x,y)dy. Here a=0, b=x, so extra term is f(x,x). Omission is valid iff f(x,x)=0. This challenges students to recall precise conditions for rule simplification, going beyond standard textbook statements. Requires synthesizing integration and differentiation concepts at advanced level. Distractors test awareness of edge cases versus blanket rules, fitting Olympiad-style depth while remaining within partial derivative scope.

Q9. If f(x,y)=x2y+y3f(x,y) = x^2 y + y^3 and we define g(r,ΞΈ)=f(rcos⁑θ,rsin⁑θ)g(r,\theta) = f(r\cos\theta, r\sin\theta), which expression correctly represents βˆ‚gβˆ‚ΞΈ\frac{\partial g}{\partial \theta} using chain rule, and why might direct substitution followed by differentiation be preferable?

A.gΞΈ=fx(βˆ’rsin⁑θ)+fy(rcos⁑θ)g_\theta = f_x (-r\sin\theta) + f_y (r\cos\theta); direct method avoids chain rule complexity.
B.gθ=fx(rcos⁑θ)+fy(rsin⁑θ)g_\theta = f_x (r\cos\theta) + f_y (r\sin\theta); polar conversion simplifies trigonometric identities.
C.gΞΈ=βˆ’fxrsin⁑θ+fyrcos⁑θg_\theta = -f_x r\sin\theta + f_y r\cos\theta; both methods equivalent but chain rule reveals geometric meaning. βœ…
D.gΞΈ=frβ‹…0+fΞΈβ‹…1g_\theta = f_r \cdot 0 + f_\theta \cdot 1; polar coordinates decouple variables automatically.
πŸ’‘ Difficulty: medium | βœ… Correct: C

πŸ“– Explanation: Chain rule gives gΞΈ=fxxΞΈ+fyyΞΈ=fx(βˆ’rsin⁑θ)+fy(rcos⁑θ)g_\theta = f_x x_\theta + f_y y_\theta = f_x(-r\sin\theta) + f_y(r\cos\theta). Direct substitution yields same after differentiation. Option C correctly states formula and acknowledges equivalence while highlighting geometric insight (angular rate relates to tangential components). Option A misses sign, B swaps derivatives, D incorrectly uses polar partials without transformation. Tests ability to navigate multiple representations and evaluate methodological trade-offs, requiring sequential application of coordinate transforms and chain rule with critical comparison.

Q10. For f(x,y)=∣x∣+y2f(x,y) = |x| + y^2, which statement accurately describes partial derivatives at (0,0)?

A.fx(0,0)=0f_x(0,0) = 0 and fy(0,0)=0f_y(0,0) = 0 because absolute value has derivative zero at origin.
B.fx(0,0)f_x(0,0) does not exist, but fy(0,0)=0f_y(0,0) = 0 since yΒ² is smooth. βœ…
C.Both partials exist and equal zero due to symmetry of absolute value function.
D.fx(0,0)=1f_x(0,0) = 1 from right derivative, ignoring left derivative convention.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Recall that ∣x∣|x| lacks derivative at x=0, so fx(0,0)f_x(0,0) undefined. However, fy=2yf_y = 2y, so fy(0,0)=0f_y(0,0)=0. This combines basic recall with nuance about piecewise functions. While categorized as direct recall per distribution, the absolute value trap makes it higher-order by testing precise definition application. Distractors exploit common errors like assuming symmetry implies differentiability or misremembering one-sided derivatives. Ensures foundational knowledge supports advanced topics, maintaining balance in cognitive demand across question set.

Q11. Suppose f(x,y)f(x,y) has continuous second partials and fxx(a,b)>0f_{xx}(a,b) > 0, fyy(a,b)<0f_{yy}(a,b) < 0. Without computing fxyf_{xy}, what can definitively be concluded about the critical point at (a,b)?

A.It must be a saddle point because pure second derivatives have opposite signs.
B.It could be a local minimum if fxyf_{xy} is sufficiently large positive.
C.Classification is impossible without discriminant; opposite pure second derivatives suggest but don’t prove saddle.
D.It is definitely not a local maximum due to positive fxxf_{xx}. βœ…
πŸ’‘ Difficulty: hard | βœ… Correct: D

πŸ“– Explanation: While opposite signs in pure second derivatives often indicate saddle points, definitive classification requires discriminant D=fxxfyyβˆ’(fxy)2D = f_{xx}f_{yy} - (f_{xy})^2. However, fxx>0f_{xx} > 0 alone precludes local maximum (which requires fxx<0f_{xx} < 0). Option A overstates certainty, B misunderstands discriminant role, C understates what can be concluded. This tests nuanced understanding of second derivative test limitations versus definite exclusions. Students must distinguish between sufficient conditions for saddle points and necessary conditions for extrema types, refining conceptual precision beyond algorithmic application.

Q12. In approximating f(1.02,0.97)f(1.02, 0.97) for f(x,y)=x3y2f(x,y)=x^3 y^2 using linearization at (1,1), a student gets 1.08 but actual value is ~1.06. What primary factor explains this overestimation?

A.Linear approximation always overestimates for increasing functions.
B.The function has negative curvature in some direction near (1,1), making tangent plane lie above surface. βœ…
C.Calculation error in partial derivatives led to incorrect slope estimates.
D.Higher-order terms in Taylor series are negligible, so discrepancy indicates arithmetic mistake.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Linearization error depends on second-order behavior. Compute Hessian: fxx=6xy2>0f_{xx}=6xy^2>0, fyy=2x3>0f_{yy}=2x^3>0, but cross term and combined effect may create downward curvature along certain paths. Actual value being lower suggests tangent plane overestimates, implying concave-down behavior in displacement direction. This applies approximation theory diagnostically rather than computationally. Distractors blame calculation or make universal claims about monotonicity. Requires connecting numerical discrepancy to geometric properties of surfaces, testing applied understanding of linearization limitations in realistic scenarios.

Q13. Compare computing βˆ‚2fβˆ‚xβˆ‚y\frac{\partial^2 f}{\partial x \partial y} for f(x,y)=x2arctan⁑(y/x)f(x,y) = x^2 \arctan(y/x) (xβ‰ 0) via direct differentiation versus converting to polar coordinates first. Which approach better reveals structural properties and why?

A.Direct differentiation is superior because it avoids coordinate singularity issues at x=0.
B.Polar conversion simplifies to r2cos⁑2ΞΈβ‹…ΞΈr^2 \cos^2\theta \cdot \theta, making angular dependence explicit and reducing algebraic complexity. βœ…
C.Both methods are equally efficient; choice depends solely on personal preference.
D.Direct method exposes hidden symmetries through product rule cancellations that polar obscures.
πŸ’‘ Difficulty: easy | βœ… Correct: B

πŸ“– Explanation: Converting to polar: x=rcos⁑θ,y=rsin⁑θ⇒y/x=tan⁑θx=r\cos\theta, y=r\sin\theta \Rightarrow y/x=\tan\theta, so f=r2cos⁑2ΞΈβ‹…ΞΈf=r^2\cos^2\theta \cdot \theta. Mixed partial becomes cleaner in separating radial/angular behaviors. Direct method involves messy quotient/product rules with arctan derivatives. Polar reveals that f depends on ΞΈ nonlinearly but r quadratically, clarifying scaling properties. This compares methodologies structurally rather than computationally, testing metacognitive evaluation of representation choices. Addresses when coordinate transforms illuminate versus obscure, integrating conceptual understanding with practical strategy selection for complex expressions.

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