Visual walkthrough — Clairaut's theorem — mixed partials are equal (under conditions)

Step 0 — The picture everything sits on

Look at the figure. The surface is the pastel sheet. At the marked point, the coral arrow is the direction "walk East" and its steepness is $f_x$; the lavender arrow is "walk North" and its steepness is $f_y$. A slope is just rise over run for a tiny step — nothing more mysterious than the tilt of a ramp.

Now the star of the show:

The whole theorem is the claim: twist measured East-then-North = twist measured North-then-East. We now prove it by never trusting the algebra until the picture forces it.

Step 1 — Draw a tiny rectangle, read four corner heights

WHAT IT LOOKS LIKE: the four coloured dots on the base plane are the corners; the vertical sticks rising to the surface are the four heights. Name them:

$\underset{\text{far corner: East \& North}}{\underbrace{f(a+h,b+k)}}\underset{\text{East only}}{\underbrace{f(a+h,b)}}\underset{\text{North only}}{\underbrace{f(a,b+k)}}\underset{\text{base}}{\underbrace{f(a,b)}}$

Step 2 — Combine the corners into one symmetric number $\Delta$

WHAT IT LOOKS LIKE: the green $(+)$ corners and coral $(-)$ corners in the figure. Notice the magic property: swap the roles of East and North (swap $h\leftrightarrow k$, swap the axes) and the four terms just reorder — $\Delta$ is unchanged. It doesn't know or care which direction we call "first." That hidden symmetry is the entire engine of the proof.

Step 3 — Peel it as an East-difference of a North-thing

WHAT IT LOOKS LIKE: two lavender vertical bars in the figure are $g(a)$ (left) and $g(a+h)$ (right). $\Delta$ is the difference of their two lengths.

Step 4 — First Mean Value Theorem (in the East direction)

WHAT IT LOOKS LIKE: the figure marks $c_1$ somewhere inside the East interval (we don't know where — shown as a floating tick with a "?" ). At that column, we now hold $x=c_1$ and look at how the East-slope $f_x$ changed between North-height $b$ and $b+k$. That bracket is a difference of $f_x$ in $y$ — begging for a second MVT.

Step 5 — Second Mean Value Theorem (in the North direction) → $f_{xy}$

WHAT IT LOOKS LIKE: the mystery point $(c_1,d_1)$ sits somewhere inside the pastel rectangle — pinned down only to "inside," shown by the dashed box and a floating dot. Remember this: $\Delta =hk\cdot (\text{East-then-North twist there})$.

Step 6 — Peel the SAME $\Delta$ the other way → $f_{yx}$

WHAT IT LOOKS LIKE: the same rectangle, now sliced the other way (horizontal-gap-first). A second mystery point $(c_2,d_2)$ — generally a different interior point than $(c_1,d_1)$, but also trapped inside the box.

Step 7 — Set them equal and shrink the rectangle

WHAT IT LOOKS LIKE: a sequence of nested pastel rectangles shrinking to the single point $(a,b)$, both mystery dots funnelling inward. Continuity is what guarantees the twist-values ride smoothly along to the centre instead of leaping.

Step 8 — The degenerate & edge cases (never skip a scenario)

• Negative steps ($h<0$ or $k<0$): the rectangle just extends West or South instead. MVT works on $[a+h,a]$ equally well; the factor $hk$ carries the correct sign and cancels identically. Result unchanged. Picture: the box flips to the other side of $(a,b)$ — same story.
• $h=0$ or $k=0$ (a degenerate rectangle): then $\Delta =0$ trivially (two pairs of corners coincide) — no information, so we never divide by zero because we take the limit $h,k\to 0$ through nonzero values, never at zero.
• Continuity FAILS — the crease case: if $f_{xy}$ or $f_{yx}$ is not continuous at $(a,b)$, Step 7 breaks. The two hidden points can approach $(a,b)$ along different paths and land on different limiting twist-values. This is not hypothetical: $f(x,y)=\{\begin{array}{ll}\frac{xy(x^2-y^2)}{x^2+y^2},& (x,y)\ne (0,0)\\ 0,& (x,y)=(0,0)\end{array}\Rightarrow f_{xy}(0,0)=-1\ne f_{yx}(0,0)=1.$

WHAT IT LOOKS LIKE: two panels. Left — the smooth surface: both mystery dots slide to the same twist value (curves meet). Right — the creased counterexample surface: the two dots approach along different ridges, twist-values split to $-1$ and $+1$. The crease is the culprit; smoothness is what forbids it.

The one-picture summary

Everything on one canvas: the tiny rectangle with its four signed corners feeds the single symmetric number $\Delta$. Peel it East→North and out comes $hk{f}_{xy}$; peel it North→East and out comes $hk{f}_{yx}$. Since it's the same $\Delta$, divide by $hk$, shrink the box, and — provided the twist is continuous — both funnel to the identical value at $(a,b)$.

Active recall