Visual walkthrough — Differentiability in multiple variables

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Step 1 — What is a graph of a two-variable function?

WHAT. A function $f(x,y)$ takes two numbers in and gives one number out. Feed it every point $(x,y)$ of a flat floor and stack the output $z=f(x,y)$ upward as a height. The result is a surface — a landscape floating above the floor.

WHY. Before we can ask "does this landscape have a flat tilted board touching it?", we must agree on what the landscape is. Everything below lives on this surface.

PICTURE. The floor is the $xy$-plane. Above the point $(a,b)$ (marked in red) the surface rises to a single height $f(a,b)$. That red vertical stick is the value we will approximate.

See Partial Derivatives for how we probe this surface one axis at a time.

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Step 2 — What would a "tangent plane" even mean?

WHAT. A plane is a perfectly flat, infinitely wide tilted board. We want the one board that touches the surface at $\mathbf{a}$ and lies as flat against it as possible — the 2D cousin of a tangent line.

WHY. In one variable, "differentiable" meant "has a tangent line — zoom in and the curve looks straight." In two variables the honest generalisation is "zoom in and the surface looks flat." A flat thing is a plane. So differentiability = this plane exists and truly hugs the surface.

PICTURE. The black surface curves. The red plane passes through the point $(a,b,f(a,b))$ and tilts to match. Near $\mathbf{a}$ the gap between them is tiny; far away it grows.

A full treatment of the plane's equation lives in Tangent Planes and Linear Approximation.

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Step 3 — Write the candidate plane algebraically

WHAT. Any plane through the point $(\mathbf{a},f(\mathbf{a}))$ can be written $L(\mathbf{x})=\underset{\text{height at }\mathbf{a}}{\underbrace{f(\mathbf{a})}}+\underset{\text{tilt vector}}{\underbrace{\mathbf{m}}}\cdot \underset{\text{step across floor}}{\underbrace{(\mathbf{x}-\mathbf{a})}}.$

WHY. We do not yet know the correct tilt, so we call it $\mathbf{m}=(m_1,m_2)$ and solve for it later. Starting from the point $f(\mathbf{a})$ guarantees the plane touches the surface there. The dot product $\mathbf{m}\cdot (\mathbf{x}-\mathbf{a})$ is how a plane converts a floor-step into a rise.

PICTURE. Take a step $\mathbf{h}=\mathbf{x}-\mathbf{a}$ along the floor (red arrow). The plane's rise is $m_1$ times the East-part of the step plus $m_2$ times the North-part. Steeper $\mathbf{m}$ ⇒ bigger rise per step.

Here $\mathbf{h}=(h_1,h_2)$ is the step vector and $\Vert \mathbf{h}\Vert =\sqrt{h_1^2+h_2^2}$ is its length (straight-line distance stepped). See Gradient and Directional Derivatives.

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Step 4 — Measure the error, then demand it be small in the right way

WHAT. The surface says the true height at $\mathbf{a}+\mathbf{h}$ is $f(\mathbf{a}+\mathbf{h})$. The plane predicts $L(\mathbf{a}+\mathbf{h})=f(\mathbf{a})+\mathbf{m}\cdot \mathbf{h}$. Their difference is the error: $E(\mathbf{h})=\underset{\text{true height}}{\underbrace{f(\mathbf{a}+\mathbf{h})}}-\underset{\text{plane’s guess}}{\underbrace{[f(\mathbf{a})+\mathbf{m}\cdot \mathbf{h}]}}.$

WHY. A plane that merely touches ($E\to 0$ as $\mathbf{h}\to 0$) is not special — infinitely many tilted boards touch a surface at a point. To pin down the tangent plane we need the error to vanish faster than the step itself.

PICTURE. At floor-step $\mathbf{h}$, the black surface and red plane are two heights. The vertical red segment between them is $E(\mathbf{h})$. As you slide $\mathbf{h}$ back toward $0$, that gap must not merely close — it must close faster than the horizontal distance shrinks.

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Step 5 — Why divide by $\Vert \mathbf{h}\Vert$? The ratio picture

WHAT. We impose the condition ${\lim }_{\mathbf{h}\to 0}\frac{E(\mathbf{h})}{\Vert \mathbf{h}\Vert }=0.$

WHY. Dividing the vertical gap $E$ by the horizontal distance $\Vert \mathbf{h}\Vert$ turns the gap into a slope of the error. If even this slope goes to zero, the surface and plane become tangent, not merely intersecting. Concretely: near $\mathbf{a}$ the error behaves like $\Vert \mathbf{h}{\Vert }^2$ — a bad "touching" plane leaves error $\sim \Vert \mathbf{h}\Vert$, and dividing exposes the difference.

PICTURE. Two curves of "error vs step": a touching plane's error (upper) shrinks like a straight line to $0$ — but divided by $\Vert \mathbf{h}\Vert$ it flattens to a nonzero number. The tangent plane's error (red, lower) curls down like $\Vert \mathbf{h}{\Vert }^2$ — divided by $\Vert \mathbf{h}\Vert$ it still dives to $0$.

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Step 6 — Solve for the tilt $\mathbf{m}$ by walking the axes

WHAT. The condition must hold approaching from every direction. Choose the easy ones first. Step purely East, $\mathbf{h}=(h,0)$, so $\Vert \mathbf{h}\Vert =|h|$: $\frac{f(a+h,b)-f(a,b)-m_{1}h}{|h|}\underset{h\to 0}{\to }0⟹m_1={\lim }_{h\to 0}\frac{f(a+h,b)-f(a,b)}{h}=f_x(a,b).$ Step purely North, $\mathbf{h}=(0,h)$: gives $m_2=f_y(a,b)$.

WHY. We had an unknown tilt $\mathbf{m}$. Approaching along an axis kills the other component, leaving a one-variable limit — which is exactly the definition of the partial derivative. So if a tangent plane exists at all, its tilt is forced to be $(f_x,f_y)$; there is no freedom.

PICTURE. Two red slices through the surface: the East–West slice (its slope at $\mathbf{a}$ is $f_x$) and the North–South slice (slope $f_y$). The tangent plane must contain both slice-tangent-lines.

This is the gradient of Gradient and Directional Derivatives; packaged as a matrix it is the Total Derivative / Jacobian.

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Step 7 — Assemble the master formula

WHAT. Plugging the forced tilt back in, differentiability says the surface splits into plane + tiny leftover: $\boxed{f(\mathbf{a}+\mathbf{h})=\underset{\text{base height}}{\underbrace{f(\mathbf{a})}}+\underset{\text{plane’s rise}}{\underbrace{\nabla f(\mathbf{a})\cdot \mathbf{h}}}+\underset{\text{negligible error}}{\underbrace{o(\Vert \mathbf{h}\Vert )}}}$

WHY. This is the whole point in one line: a differentiable surface is its tangent plane plus an error too small to matter. It also instantly proves differentiable ⇒ continuous — as $\mathbf{h}\to 0$ the plane-rise $\to 0$ and the error $\to 0$, so $f(\mathbf{a}+\mathbf{h})\to f(\mathbf{a})$. See Continuity in Multiple Variables.

PICTURE. Three stacked bars at a fixed small step: total height (black), the part explained by the plane (red, almost all of it), and the leftover sliver of error (shrinking to nothing).

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Step 8 — The degenerate case: partials exist but NO plane hugs

WHAT. Take $f(x,y)=\{\begin{array}{ll}\frac{xy}{x^2+y^2}& (x,y)\ne (0,0)\\ 0& (0,0).\end{array}$ On each axis $f\equiv 0$, so $f_x(0,0)=f_y(0,0)=0$ — both partials exist. Yet along the diagonal $y=x$, $f(x,x)=\frac{x\cdot x}{x^2+x^2}=\frac{1}{2}\ne 0.$

WHY. The axis-walk of Step 6 only proposed a tilt; it never checked the diagonal directions of Step 5. Here the surface is a flat cross along the axes but a raised ridge on the diagonal — no single plane can match both. Since $f$ isn't even continuous at $0$ (it hits $\frac{1}{2}$ on approach), by Step 7 it cannot be differentiable.

PICTURE. Looking down from above: along the black axes the height is $0$; along the red diagonal the height jumps to $\frac{1}{2}$. A plane through the origin with zero East/North slope would be the flat floor — but the red ridge pokes through it. No hug possible.

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Step 9 — The escape hatch: $C^1$ ⇒ differentiable

WHAT. Verifying the Step 5 limit in every direction is exhausting. Shortcut: if $f_x$ and $f_y$ exist and are continuous near $\mathbf{a}$, then $f$ is automatically differentiable there. Such $f$ is called $C^1$.

WHY. Continuity of the partials means the East and North slopes don't jump around as you move — so gluing them into a plane can't leave a diagonal ridge behind. The pathological Example (Step 8) fails precisely because its partials are not continuous at the origin (they oscillate). Ruling that out rules out the trap.

PICTURE. A flowchart of the hierarchy: $C^1$ sits at the top and each arrow is a one-way implication, never reversible.

See C1 and Smooth Functions and Chain Rule (Multivariable) for where $C^1$ pays off.

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The one-picture summary

Everything at once: black surface, red tangent plane touching at $\mathbf{a}$, the step $\mathbf{h}$ on the floor, and the shrinking vertical error $E$ that must die faster than $\Vert \mathbf{h}\Vert$.

Recall Feynman retelling — the whole walk in plain words

Picture a hilly landscape (Step 1). Stand on one spot and try to slide a perfectly flat tilted board so it kisses the ground right there (Step 2). You write the board's rule: start at your height, then tilt by some amount for each step you take (Step 3). Now walk a little and compare the ground to your board — the vertical gap is the error (Step 4). Touching isn't good enough: you demand the gap shrink faster than the distance you walked, so you divide gap by distance and insist that goes to zero (Step 5). To find the right tilt, first walk exactly East and exactly North — those slopes are the partials, and they force the board's tilt to be the gradient (Step 6). Put it together: ground = board + a sliver too small to matter, which also proves a differentiable hill has no cliffs (Step 7). But beware — a hill can be perfectly flat along East and North yet hide a ridge on the diagonal; then no board fits, even though the partials exist (Step 8). The safe check: if the East/North slopes vary smoothly (the hill is $C^1$), a fitting board is guaranteed — and that covers nearly every function you'll ever meet (Step 9).

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Recall

Why must we divide the error by $\Vert \mathbf{h}\Vert$? ::: So the error is higher-order ($o(\Vert \mathbf{h}\Vert )$) — it selects the unique tangent plane, not any plane that merely touches. If $f$ is differentiable, what must its tilt $\mathbf{m}$ equal, and why? ::: The gradient $(f_x,f_y)$, forced by approaching along the two axes (each gives a partial derivative). In Step 8, why is $f$ not differentiable despite $f_x=f_y=0$? ::: It isn't continuous at the origin ($\to \frac{1}{2}$ along $y=x$); differentiable ⇒ continuous fails. What does $C^1$ buy you? ::: Continuous partials ⇒ differentiable automatically — no limit check needed.

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Connections

• 4.4.06 Differentiability in multiple variables (Hinglish)
• Partial Derivatives
• Gradient and Directional Derivatives
• Tangent Planes and Linear Approximation
• Continuity in Multiple Variables
• Total Derivative / Jacobian
• C1 and Smooth Functions
• Chain Rule (Multivariable)