Question bank — Clairaut's theorem — mixed partials are equal (under conditions)

False. They agree only when the mixed partials are continuous near the point; the origin counterexample $f=xy(x^2-y^2)/(x^2+y^2)$ gives $f_{xy}(0,0)=-1\ne f_{yx}(0,0)=1$.

False. Mere existence is not enough — both exist in the counterexample yet differ. What forces equality is continuity of the mixed partials, not their existence.

True. Continuity is exactly Clairaut's hypothesis, and it holds at every point of the disk, so the conclusion holds pointwise throughout.

False. It is an analysis theorem whose whole content is the continuity condition; without it the equality can genuinely fail, so the hypothesis is not decorative.

True. $f_x$ is the slope in the $x$-direction; differentiating it with respect to $y$ measures how that slope twists as you shift in $y$ — this is the geometric "warping" (Picture 2) Clairaut compares in both orders.

False. Equality of mixed partials at a point does not force full two-variable continuity of $f$: partial derivatives (even equal ones) probe only along coordinate lines, while $f$ can still jump when approached from a diagonal direction.

True. Polynomials have continuous partials of every order, so Clairaut applies at every point and every mixed partial is order-independent.

True (Schwarz's refined version). It suffices that $f_x,f_y$ exist near the point and one mixed partial is continuous there; the other then exists and equals it.

True. The Hessian's off-diagonal entries are $f_{xy}$ and $f_{yx}$; under Clairaut's continuity condition they are equal, making the Hessian Matrix symmetric — which is why its eigenvalues are real.

Error. In Leibniz notation the operator nearest $f$ acts first, so this is $y$-then-$x$. The subscript form $f_{yx}$ reads $y$ first left-to-right; the two notations run opposite directions.

Error. The point $(c_1,d_1)$ is unknown and generally not $(a,b)$; only after $h,k\to 0$ does it squeeze toward $(a,b)$, and only continuity lets its function value approach the value at $(a,b)$.

Error. Existence at the origin doesn't grant swapping; the mixed partials there are discontinuous (Picture 3), so the hypothesis fails and $f_{xy}(0,0)\ne f_{yx}(0,0)$.

Error. Each order peels $\Delta$ apart with two applications of the MVT — one in $x$, one in $y$ — producing $hk{f}_{xy}(c_1,d_1)$ one way and $hk{f}_{yx}(c_2,d_2)$ the other; four interior points total.

Error. Under continuity Clairaut lets you reorder the whole string; only the count of each variable matters, so $f_{xxy}=f_{xyx}=f_{yxx}$.

Error. Continuity of $f$ is far weaker than continuity of its second mixed partials. The counterexample function is itself continuous everywhere yet fails Clairaut at the origin.

Because $\Delta$ is visibly unchanged when the roles of $x$ and $y$ swap (see the symmetric corner-signs in Picture 1), so a single quantity can be shown to limit to $f_{xy}$ one way and $f_{yx}$ the other — the shared symmetry forces both limits to agree.

The Mean Value Theorem (MVT) hands us derivative values at unknown interior points $c_i,d_i$; continuity is the only thing that converts "value at a squeezing point" into "value at $(a,b)$," so removing it removes the conclusion.

"How the $x$-slope changes as you move in $y$" and "how the $y$-slope changes as you move in $x$" both quantify the same local warping of the surface, so a smooth surface must report the same twist either way.

The rational formula $xy(x^2-y^2)/(x^2+y^2)$ is $0/0$ at the origin, so a separate value is set there; this seam (the pinch in Picture 3) is exactly where the second mixed partials lose continuity and the two orders diverge.

Because either order gives the same result, you may choose the algebraically cleaner path — e.g. differentiate the simpler variable first — and trust the answer without redoing the hard order.

Symmetry $f_{xy}=f_{yx}$ is earned from Clairaut's continuity hypothesis; for a non-smooth function the Hessian's off-diagonal entries can differ, so symmetry is a theorem, not a built-in property.

No. $f_{xy}$ requires $f_x$ to exist in a neighbourhood before differentiating it in $y$; if $f_x$ is undefined nearby, the mixed partial isn't defined, so the equality question is void.

No, the order of the double limit matters. The proof needs $\Delta$ divided by $hk$ with both steps genuinely nonzero; setting $k=0$ collapses $\Delta$ to $0$ and destroys the second difference before any information is extracted.

On the crease the hypothesis fails, so equality may break there; away from the crease, where the partials are continuous, Clairaut still guarantees $f_{xy}=f_{yx}$ locally.

No. First-partial continuity is too weak; Clairaut demands continuity of the second mixed partials at that point, which is precisely what's missing.

Yes, under the same continuity condition. Any pair of adjacent variables can be swapped when the relevant mixed partials are continuous, so any permutation of $x,y,z$ gives the same third-order partial.

Trivially true. All partials are $0$ and continuous, so $f_{xy}=f_{yx}=0$; the theorem holds but carries no information — a useful sanity check that the hypotheses are satisfiable.