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📝 Differentiability implies continuity (13 MCQs)

📖 From Calculus • 14. Partial Derivatives Calculus • 13 questions available

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What is Differentiability implies continuity?

Definition:
If $f$ is differentiable at $(a,b)$ , then $f$ is necessarily continuous at $(a,b)$ ; converse is false.

Example:
$f(x,y) = \sqrt{x^2+y^2}$ is continuous at origin but not differentiable there; differentiable $f=x^2+y^2$ is automatically continuous.

Reason:
This implication provides a sufficient condition for continuity and reinforces that differentiability is a stricter smoothness requirement than mere continuity.

Easy

Medium

Hard

📝 All Differentiability implies continuity MCQs

Q1. A function $f(x,y)$ has partial derivatives $f_x(0,0) = 2$ and $f_y(0,0) = -1$ . However, the limit of $f(x,y)$ as $(x,y) \to (0,0)$ along the path $y=x^2$ differs from $f(0,0)$ . Which statement best analyzes this scenario?

A.The function is differentiable at the origin because partial derivatives exist.

B.The function is continuous but not differentiable since directional derivatives vary.

C.The existence of partial derivatives guarantees continuity, so the path limit must equal

f(0,0)

D.The function is neither continuous nor differentiable at the origin despite existing partials. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: This question targets error analysis regarding the common misconception that existing partial derivatives imply continuity or differentiability. Students must recognize that partial derivatives only measure behavior along coordinate axes. If the function fails to be continuous along a parabolic path, it cannot be differentiable, as differentiability requires the linear approximation to hold uniformly in all directions, which presupposes continuity.

Q2. Consider a surface modeling temperature distribution where contour lines near point $P$ are densely packed ellipses, but at point $Q$ , contours form a sharp cusp intersection. Based solely on this graphical evidence, how do the differentiability properties compare?

A.Both points are differentiable because contour maps always represent smooth functions.

B.Point

P

is likely differentiable while

Q

is not, as cusp intersections indicate undefined tangent planes. ✅

C.Point

Q

is differentiable due to symmetry, while

P

lacks sufficient gradient information.

D.Neither point is differentiable because contour lines only indicate level sets, not derivatives.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Interpreting contour graphs requires connecting visual density and geometry to analytical concepts. Dense smooth ellipses suggest a well-defined gradient and tangent plane at $P$ . Conversely, intersecting or cusped contours at $Q$ signal a singularity where no unique tangent plane exists. This tests graph-based reasoning, distinguishing between mere continuity of level sets and the stricter requirement of local linearity needed for differentiability in multivariable calculus.

Q3. An engineer models stress using $f(x,y)$ . They compute $f_x$ and $f_y$ everywhere and find them bounded. Yet, physical testing shows material failure at the origin, suggesting non-differentiability. What is the most rigorous mathematical explanation for this discrepancy?

A.Bounded partial derivatives are sufficient for differentiability; the physical model is flawed.

B.The partial derivatives exist and are bounded, but they are not continuous at the origin. ✅

C.Differentiability requires second-order partial derivatives to exist, which were not computed.

D.The function is discontinuous, making partial derivative calculations invalid at that point.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This application question addresses the subtle distinction between boundedness and continuity of partial derivatives. While continuous partials guarantee differentiability, merely bounded partials do not. The engineer’s error lies in assuming boundedness suffices. Students must apply the theorem correctly: without continuity of $f_x$ and $f_y$ at the critical point, the total differential may fail to approximate the function, explaining the physical anomaly despite valid first-order computations.

Q4. Given $f(x,y) = \frac{x^3}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$ , a student claims it is differentiable at the origin because $f_x(0,0)=1$ and $f_y(0,0)=0$ . Identify the flaw in this reasoning.

A.The student calculated partial derivatives incorrectly; both should be zero.

B.Partial derivatives existing is necessary but not sufficient; the limit definition of differentiability must also be satisfied. ✅

C.The function is not continuous at the origin, so it cannot be differentiable.

D.Differentiability requires polar coordinates to confirm, which the student omitted.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This error analysis question exposes the frequent mistake of equating partial derivative existence with total differentiability. Although $f_x$ and $f_y$ exist at the origin, students must verify if $\lim_{(h,k)\to(0,0)} \frac{f(h,k)-f(0,0)-f_x h - f_y k}{\sqrt{h^2+k^2}} = 0$ . In this case, the limit does not vanish, proving non-differentiability. The explanation reinforces that partials provide only axial information, while differentiability demands uniform linear approximation in all directions.

Q5. Two researchers analyze $g(x,y)$ . Researcher A verifies continuity and existence of $g_x, g_y$ at $(a,b)$ . Researcher B additionally confirms $g_x, g_y$ are continuous in a neighborhood of $(a,b)$ . Whose conclusion about differentiability is guaranteed correct?

A.Researcher A, because continuity plus partial existence is the definition of differentiability.

B.Researcher B, because continuous partial derivatives in a neighborhood constitute a sufficient condition for differentiability. ✅

C.Both are correct, as the conditions are logically equivalent.

D.Neither is sufficient without verifying the limit definition directly.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: This conceptual understanding question distinguishes between necessary and sufficient conditions. Researcher A’s criteria are necessary but not sufficient for differentiability. Researcher B applies the standard sufficient condition theorem: if partial derivatives exist and are continuous near a point, the function is differentiable there. Recognizing this hierarchy prevents over-reliance on weaker conditions and clarifies why textbooks emphasize $C^1$ smoothness as a practical test for differentiability in applied contexts.

Q6. In optimizing a cost function $C(x,y)$ , an algorithm uses gradient descent starting at a point where $C_x$ and $C_y$ exist but $C$ is not differentiable. What is the most likely consequence during iteration?

A.The algorithm will converge faster due to simpler derivative calculations.

B.Convergence is guaranteed because partial derivatives define a valid search direction.

C.The algorithm may oscillate or diverge because the gradient does not represent true steepest descent. ✅

D.Gradient descent automatically detects non-differentiability and switches methods.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: This application scenario links theoretical differentiability to numerical method reliability. Gradient descent assumes the gradient vector points in the direction of maximal decrease, which holds only when the function is differentiable. At non-differentiable points, partial derivatives may exist but fail to capture directional behavior accurately, leading to misleading update steps. Understanding this limitation explains why subgradient methods or smoothing techniques are required for non-smooth optimization, bridging abstract analysis with computational practice.

Q7. A student argues that since $|f(x,y)| \leq x^2 + y^2$ near the origin, $f$ must be differentiable at $(0,0)$ with zero derivative. Evaluate this claim.

A.The claim is false; the inequality only guarantees continuity, not differentiability.

B.The claim is true; the bound implies the difference quotient tends to zero, satisfying the differentiability definition. ✅

C.The claim is false unless

f(0,0) = 0

is explicitly stated.

D.The claim is true only if

f

is also known to be continuous.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This challenging problem tests deep understanding of the squeeze theorem applied to differentiability. The inequality $|f(x,y)| \leq x^2 + y^2$ implies $f(0,0)=0$ by continuity. Substituting into the differentiability limit yields $\left| \frac{f(h,k) - 0 - 0\cdot h - 0\cdot k}{\sqrt{h^2+k^2}} \right| \leq \sqrt{h^2+k^2}$ , which vanishes as $(h,k) \to (0,0)$ . Thus, the bound actually suffices to prove differentiability with zero gradient, correcting the intuitive but incorrect assumption that quadratic bounds only ensure continuity.

Q8. Which combination of properties for $f(x,y)$ at a point definitively rules out differentiability, even if partial derivatives exist?

A.The function is continuous and partial derivatives are bounded.

B.Directional derivatives exist in all directions but are not linear combinations of partials. ✅

C.The mixed partial derivatives

f_{xy}

and

f_{yx}

are unequal.

D.The function satisfies

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

symmetrically.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This mixed-concepts question probes the relationship between directional derivatives and total differentiability. Differentiability implies that every directional derivative equals the dot product of the gradient and the direction vector. If directional derivatives exist but violate this linearity, the function cannot be differentiable, regardless of partial derivative existence. Options A and D are compatible with differentiability, while C relates to second-order smoothness, not first-order differentiability. Identifying this specific failure mode strengthens conceptual precision beyond computational checks.

Q9. When analyzing $h(x,y) = \sqrt{|xy|}$ at the origin, a student computes $h_x(0,0) = 0$ and $h_y(0,0) = 0$ and concludes differentiability. What critical step was omitted?

A.Verifying that

h

is continuous at the origin.

B.Checking whether the limit defining differentiability approaches zero along all paths. ✅

C.Computing second-order partial derivatives to confirm smoothness.

D.Ensuring the domain includes negative values for absolute value handling.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This error analysis highlights incomplete verification protocols. While partials exist and the function is continuous, differentiability requires the remainder term $\frac{h(h,k) - 0 - 0\cdot h - 0\cdot k}{\sqrt{h^2+k^2}} = \frac{\sqrt{|hk|}}{\sqrt{h^2+k^2}}$ to tend to zero. Along $h=k$ , this ratio equals $\frac{|h|}{\sqrt{2}|h|} = \frac{1}{\sqrt{2}} \neq 0$ , disproving differentiability. The omission of path testing in the limit definition is a pervasive mistake; emphasizing multi-directional validation prevents false positives in differentiability assessment.

Q10. A weather model defines pressure $P(x,y)$ as continuous everywhere with existing partials, yet forecasters observe abrupt wind shifts at a location. Assuming the model is mathematically consistent, what does this imply about $P$ at that point?

P

is differentiable, and wind shifts arise from external factors.

P

is not differentiable, as wind velocity depends on the gradient of

P

. ✅

P

has discontinuous second derivatives, causing turbulent flow.

D.The partial derivatives of

P

do not exist at that point.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This scenario-based question integrates physics with mathematical analysis. Wind velocity is proportional to the negative gradient of pressure. Abrupt wind shifts indicate discontinuous or undefined gradients, implying $P$ is not differentiable at that location. Even if $P$ is continuous and partials exist pointwise, non-differentiability means the gradient vector isn’t well-defined or varies discontinuously, disrupting fluid dynamics predictions. This connects abstract differentiability to tangible meteorological phenomena, requiring synthesis of modeling context and analytical definitions.

Q11. Compare two functions at the origin: $f(x,y) = (x^2+y^2)\sin(\frac{1}{x^2+y^2})$ and $g(x,y) = x\sin(\frac{1}{x}) + y\sin(\frac{1}{y})$ . Both have partials at $(0,0)$ . Which is differentiable?

A.Only

f

is differentiable because its amplitude decays quadratically. ✅

B.Only

g

is differentiable due to separable structure.

C.Both are differentiable since oscillations are bounded.

D.Neither is differentiable due to essential discontinuities in derivatives.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: This Olympiad-style comparison demands nuanced limit analysis. For $f$ , the quadratic envelope ensures $\frac{|f(h,k)|}{\sqrt{h^2+k^2}} \leq \sqrt{h^2+k^2} \to 0$ , proving differentiability despite wild oscillation. For $g$ , although partials exist, the cross-term behavior in the differentiability limit fails; approaching along $x=y$ yields non-vanishing ratios due to unsynchronized oscillations. The key insight is that radial decay rate dominates angular pathology in $f$ , while $g$ ’s separability doesn’t compensate for lack of joint control, illustrating subtle interplay between dimensionality and regularity.

Q12. In machine learning, activation functions like ReLU are non-differentiable at zero. Why is this acceptable in backpropagation despite relying on gradients?

A.Backpropagation only requires continuity, not differentiability.

B.Subgradients or generalized derivatives extend gradient concepts to non-differentiable points. ✅

C.Neural networks avoid zero inputs through weight initialization.

D.Differentiability is only needed for output layers, not hidden units.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This application question bridges pure math and modern AI practice. Classical backpropagation assumes differentiability, but ReLU’s non-differentiability at isolated points is handled via subdifferential calculus, where any value in [0,1] serves as a valid subgradient at zero. Since the set of non-differentiable points has measure zero and optimization trajectories rarely land exactly on them, convergence remains robust. Understanding this extension explains why theoretical differentiability requirements can be relaxed in practice, highlighting the evolution of calculus concepts for real-world engineering constraints.

Q13. A student proves $f$ is continuous at $(a,b)$ and that all directional derivatives exist and equal $\nabla f(a,b) \cdot \mathbf{u}$ . They conclude $f$ is differentiable. Is this valid?

A.Yes, because matching directional derivatives to the gradient is equivalent to differentiability.

B.No, directional derivatives could match the gradient linearly yet fail the uniform limit condition. ✅

C.Yes, provided

f

is also known to be continuously differentiable elsewhere.

D.No, unless mixed partial derivatives are also shown to be equal.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This challenging conceptual trap tests precise knowledge of differentiability characterizations. While differentiability implies directional derivatives equal the gradient dot direction, the converse isn’t true without additional hypotheses like continuity of partials or Lipschitz conditions. There exist pathological functions where directional derivatives behave linearly yet the function isn’t differentiable due to non-uniform convergence in the increment ratio. The student’s reasoning overlooks this subtlety, demonstrating that even seemingly strong directional evidence doesn’t substitute for the full epsilon-delta definition of total differentiability in multivariable analysis.

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