Differentiability in multiple variables: Worked examples

Before we start, one reminder in plain words. A function $f (x, y)$ takes two numbers and returns one — think of it as the height of a landscape above the point $(x, y)$ on the floor. "Differentiable at a point" means: if you zoom in far enough, the landscape looks like a flat tilted board (a tangent plane). The symbol $\nabla f = (f_{x}, f_{y})$ (read "grad f") is just the pair of slopes: how fast the height climbs walking East ( $f_{x}$ ) and North ( $f_{y}$ ). See Partial Derivatives and Gradient and Directional Derivatives if either feels shaky.

The scenario matrix

Every problem in this topic falls into one of these cells. The right column names the example that clears it.

#	Case class	What makes it tricky	Cleared by
A	Nice polynomial, all signs of partials	nothing — the $C^{1}$ shortcut	Ex 1
B	Partials exist, but discontinuous	diagonal betrays the origin	Ex 2
C	Continuous, partials exist, still NOT differentiable	the sneakiest failure	Ex 3
D	Degenerate / boundary input (on a curve where formula dies)	limit that survives, one that doesn't	Ex 4
E	Directional derivatives all exist, yet not differentiable	strictly-stronger trap	Ex 5
F	Linearization estimate (real-world word problem, signed steps)	mixed-sign $h$ , units	Ex 6
G	Limiting behaviour / squeeze proves differentiability	proving the raw $o (∥ h ∥)$ limit	Ex 7
H	Exam twist: find the value of a parameter that makes $f$ differentiable	reverse-engineer the definition	Ex 8

Two symbols we will lean on:

$∥ h ∥ = h_{1}^{2} + h_{2}^{2}$ — the length of the step $h = (h_{1}, h_{2})$ , i.e. straight-line distance from the base point. It is never negative.
$o (∥ h ∥)$ ("little-o of $∥ h ∥$ ") — an error that shrinks strictly faster than the distance: divide it by $∥ h ∥$ and it still goes to $0$ .

Cell A — the easy, nice case

Worked example Example 1 (Cell A) — polynomial, both slopes nonzero

Show $f (x, y) = x^{3} - 2 x y^{2} + y$ is differentiable at $(2, - 1)$ and write its tangent plane.

Forecast: guess now — will we need the scary limit, or is there a shortcut? What sign do you expect $f_{x}$ to have (the $x^{3}$ term is climbing)?

Compute the partials. $f_{x} = 3 x^{2} - 2 y^{2}$ , $f_{y} = - 4 x y + 1$ . Why this step? The partial $f_{x}$ freezes $y$ and differentiates in $x$ only — it is the East-slope. See Partial Derivatives.
Note both partials are polynomials. Polynomials are continuous everywhere. Why this step? The $C^{1}$ theorem says: partials exist and continuous near the point $\Rightarrow$ differentiable. No limit needed. This is $C^1 \Rightarrow$ differentiable.
Evaluate at $(2, - 1)$ . $f_{x} (2, - 1) = 3 (4) - 2 (1) = 10$ , $f_{y} (2, - 1) = - 4 (2) (- 1) + 1 = 9$ . And $f (2, - 1) = 8 - 2 (2) (1) + (- 1) = 8 - 4 - 1 = 3$ . Why this step? The gradient at the point gives the exact tilt of the tangent plane.
Write the plane (Tangent Planes and Linear Approximation): $z = 3 + 10 (x - 2) + 9 (y + 1)$ .

Verify: plug the base point $(2, - 1)$ into the plane: $z = 3 + 10 (0) + 9 (0) = 3 = f (2, - 1)$ . ✓ The plane passes through the surface point, as any tangent plane must.

Cell B — partials exist but the function is broken

Worked example Example 2 (Cell B) — the classic saddle-ratio

$g (x, y) = ⎩ ⎨ ⎧ \frac{x ^{2} y}{x ^{4} + y ^{2}} 0 (x, y) \neq = (0, 0) (0, 0)$ Show both partials exist at the origin, yet $g$ is not continuous there — hence not differentiable.

Forecast: the partials only walk along the axes. What do you expect to see along the axes — is $g$ zero there? And what sneaky curve might reveal trouble?

Partial $g_{x} (0, 0)$ : on the $x$ -axis $y = 0$ , so $g (x, 0) = 0$ for all $x$ . Its slope is $0$ : $g_{x} (0, 0) = 0$ . Why this step? The partial only probes the East line. On that line $g$ is flat, so it looks perfectly tame.
Partial $g_{y} (0, 0)$ : on the $y$ -axis $x = 0$ , $g (0, y) = 0$ , so $g_{y} (0, 0) = 0$ .
Approach along the parabola $y = x^{2}$ : $g (x, x^{2}) = \frac{x ^{2} \cdot x ^{2}}{x ^{4} + x ^{4}} = \frac{x ^{4}}{2 x ^{4}} = \frac{1}{2}$ . Why this step? We hunt a path that doesn't go to $0$ . The value clings to $\frac{1}{2}$ no matter how close $x$ gets — so the limit at the origin does not exist. See Continuity in Multiple Variables.
Conclude: $g$ is not continuous $\Rightarrow$ not differentiable (differentiable $\Rightarrow$ continuous, contrapositive).

Verify: along the axes value $= 0$ ; along $y = x^{2}$ value $= \frac{1}{2} \neq = 0$ . Two different limits confirm discontinuity. ✓

Cell C — continuous, partials exist, STILL not differentiable

Worked example Example 3 (Cell C) — the sneakiest failure

$h (x, y) = ⎩ ⎨ ⎧ \frac{x ^{3}}{x ^{2} + y ^{2}} 0 (x, y) \neq = (0, 0) (0, 0)$ Show $h$ is continuous at $(0, 0)$ , its partials exist, yet $h$ is not differentiable there.

Forecast: this one passes the continuity test and the partials test. Where can it still trip? Watch the $o (∥ h ∥)$ limit.

Continuity: $∣ h ∣ = \frac{∣ x ∣ ^{3}}{x ^{2} + y ^{2}} \leq \frac{∣ x ∣ ( x ^{2} + y ^{2} )}{x ^{2} + y ^{2}} = ∣ x ∣ \to 0$ . Why this step? Since $x^{2} \leq x^{2} + y^{2}$ , we bound $∣ x ∣^{3} = ∣ x ∣ \cdot x^{2} \leq ∣ x ∣ (x^{2} + y^{2})$ ; the squeeze forces $h \to 0 = h (0, 0)$ . Continuous. ✓
Partials at origin: on $y = 0$ , $h (x, 0) = \frac{x ^{3}}{x ^{2}} = x$ , slope $h_{x} (0, 0) = 1$ . On $x = 0$ , $h (0, y) = 0$ , so $h_{y} (0, 0) = 0$ . So $\nabla h (0, 0) = (1, 0)$ . Why this step? We need the only candidate linear map to test the definition against.
Test the differentiability limit. With $h = (h_{1}, h_{2})$ the error is $E = h (h_{1}, h_{2}) - [0 + (1, 0) \cdot (h_{1}, h_{2})] = \frac{h _{1}^{3}}{h _{1}^{2} + h _{2}^{2}} - h_{1} = \frac{h _{1}^{3} - h _{1} ( h _{1}^{2} + h _{2}^{2} )}{h _{1}^{2} + h _{2}^{2}} = \frac{- h _{1} h _{2}^{2}}{h _{1}^{2} + h _{2}^{2}} .$ Divide by $∥ h ∥ = h_{1}^{2} + h_{2}^{2}$ and walk the line $h_{2} = h_{1} > 0$ : $\frac{E}{∥ h ∥} = \frac{- h _{1} h _{2}^{2}}{( h _{1}^{2} + h _{2}^{2} ) ^{3/2}} = \frac{- h _{1} \cdot h _{1}^{2}}{( 2 h _{1}^{2} ) ^{3/2}} = \frac{- h _{1}^{3}}{2 2 h _{1}^{3}} = \frac{- 1}{2 2} \neq = 0.$ Why this step? Differentiability demands this ratio $\to 0$ from every direction. Along the diagonal it sticks at $- \frac{1}{2 2} \approx - 0.354$ . Fails.
Conclude: continuous + partials exist, but the tangent-plane error does not vanish fast enough $\Rightarrow$ not differentiable.

Verify: $- 1/ (22) = - 2 /4 \approx - 0.3536 \neq = 0$ . ✓ Non-zero limit kills differentiability.

Cell D — degenerate / boundary input

Worked example Example 4 (Cell D) — where the formula formally dies

Let $f (x, y) = \frac{sin ( x y )}{y}$ for $y \neq = 0$ , and define $f (x, 0) = x$ . Is $f$ differentiable at the boundary point $(1, 0)$ (where the raw formula divides by zero)?

Forecast: dividing by $y$ looks fatal at $y = 0$ . But $sin (x y) \approx x y$ for small argument — does the zero cancel?

Resolve the degeneracy. For small $y$ , $sin (x y) = x y - \frac{( x y ) ^{3}}{6} + \dots$ , so $\frac{sin ( x y )}{y} = x - \frac{x ^{3} y ^{2}}{6} + \dots$ . Why this step? The apparent $0/0$ is removable — the definition $f (x, 0) = x$ matches the limit, so $f$ is actually the smooth function $x - \frac{x ^{3} y ^{2}}{6} + \dots$ near the axis.
Partials at $(1, 0)$ . From $f \approx x - \frac{x ^{3} y ^{2}}{6}$ : $f_{x} = 1 - \frac{x ^{2} y ^{2}}{2}$ , so $f_{x} (1, 0) = 1$ . And $f_{y} = - \frac{x ^{3} y}{3}$ , so $f_{y} (1, 0) = 0$ . Why this step? Using the resolved expansion avoids the illegal division and gives honest slopes.
$C^{1}$ check. Both $f_{x}, f_{y}$ are continuous at $(1, 0)$ (built from $sin, cos$ , polynomials, no surviving division by zero after the cancellation). So $f$ is differentiable at the boundary point. Why this step? Once the removable singularity is patched, the $C^{1}$ shortcut applies. See C1 and Smooth Functions.
Tangent plane: $z = f (1, 0) + 1 (x - 1) + 0 (y - 0) = 1 + (x - 1) = x$ .

Verify: at $(1, 0)$ the plane gives $z = 1 = f (1, 0)$ . ✓ And $f_{y} (1, 0) = 0$ matches: along $y$ near the axis $f = x - \frac{x ^{3} y ^{2}}{6}$ is flat to first order in $y$ (only $y^{2}$ appears). ✓

Cell E — every directional derivative exists, yet not differentiable

Worked example Example 5 (Cell E) — the strictly-stronger trap

$p (x, y) = ⎩ ⎨ ⎧ \frac{x ^{2} y}{x ^{2} + y ^{2}} 0 (x, y) \neq = (0, 0) (0, 0)$ Show the directional derivative exists in every direction at $(0, 0)$ , yet $p$ is not differentiable.

Forecast: if all directional derivatives exist, surely we are done? Watch whether they obey the linear rule $D_{u} p = \nabla p \cdot u$ .

Directional derivative along unit vector $u = (cos θ, sin θ)$ . Set $x = t cos θ, y = t sin θ$ : $\frac{p ( t u ) - 0}{t} = \frac{1}{t} \cdot \frac{t ^{2} c o s ^{2} θ \cdot t s i n θ}{t ^{2}} = cos^{2} θ sin θ .$ Take $t \to 0$ : $D_{u} p (0, 0) = cos^{2} θ sin θ$ . Exists for every $θ$ . Why this step? We probe all directions, not just the axes, to test the stronger claim. See Gradient and Directional Derivatives.
Read off the partials. $θ = 0$ (East): $cos^{2} 0 sin 0 = 0 = p_{x}$ . $θ = \frac{π}{2}$ (North): $0 \cdot 1 = 0 = p_{y}$ . So $\nabla p (0, 0) = (0, 0)$ .
If $p$ were differentiable, the linear rule would force $D_{u} p = \nabla p \cdot u = (0, 0) \cdot u = 0$ for all $θ$ . But at $θ = \frac{π}{4}$ : $cos^{2} \frac{π}{4} sin \frac{π}{4} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{2 2} \neq = 0$ . Why this step? Differentiability predicts a linear formula for directional derivatives. Reality contradicts it $\Rightarrow$ not differentiable.

Verify: predicted $0$ vs actual $\frac{1}{2 2} = \frac{2}{4} \approx 0.354$ at $45°$ . Mismatch $\Rightarrow$ not differentiable. ✓

Cell F — real-world linearization with signed steps

Worked example Example 6 (Cell F) — a physics word problem

A rectangular metal plate has sides $x$ and $y$ metres; its area is $A = x y$ . A plate measured $x = 8.00 m$ , $y = 5.00 m$ is cooled: $x$ shrinks by $0.03 m$ and $y$ grows by $0.02 m$ . Estimate the new area with linearization and compare to the exact value.

Forecast: two opposite-sign changes. Will the area go up or down? Which side's change matters more?

Base value. $A (8, 5) = 40 m^{2}$ .
Partials. $A_{x} = y = 5$ , $A_{y} = x = 8$ (units: m of area per m of side $=$ m). Why this step? $A_{x}$ is how many m² you gain per metre of $x$ -growth — bigger when $y$ is bigger.
Signed step. $h = (- 0.03, + 0.02)$ . Linear estimate: $A \approx 40 + 5 (- 0.03) + 8 (+ 0.02) = 40 - 0.15 + 0.16 = 40.01 m^{2} .$ Why this step? The tangent-plane formula $A (a + h) \approx A (a) + \nabla A \cdot h$ ; note the negative term for the shrinking side.
Interpret: area rises slightly — the $y$ -growth ( $+ 0.16$ ) barely beats the $x$ -shrink ( $- 0.15$ ).

Verify: exact $A = (8 - 0.03) (5 + 0.02) = 7.97 \times 5.02 = 40.0094 m^{2}$ . Linear estimate $40.01$ differs by $0.0006 m^{2}$ — the dropped $o (∥ h ∥)$ term is $h_{1} h_{2} = (- 0.03) (0.02) = - 0.0006$ . ✓ Units: m² throughout.

Cell G — proving differentiability from the raw limit

Worked example Example 7 (Cell G) — squeeze the error to zero

Prove from the definition (no $C^{1}$ shortcut) that $q (x, y) = x^{2} + y^{2}$ is differentiable at $(0, 0)$ .

Forecast: the partials at the origin are both $0$ . What is the leftover error, and does it die faster than $∥ h ∥$ ?

Candidate gradient. $q_{x} = 2 x, q_{y} = 2 y$ , so $\nabla q (0, 0) = (0, 0)$ , and $q (0, 0) = 0$ .
Build the error. With $h = (h_{1}, h_{2})$ : $E = q (h) - [0 + (0, 0) \cdot h] = h_{1}^{2} + h_{2}^{2} = ∥ h ∥^{2} .$ Why this step? The definition tests exactly this leftover after subtracting the linear part.
Divide by $∥ h ∥$ : $\frac{E}{∥ h ∥} = \frac{∥ h ∥ ^{2}}{∥ h ∥} = ∥ h ∥ \to 0$ . Why this step? This is the $o (∥ h ∥)$ condition. It holds from every direction because $∥ h ∥ \to 0$ regardless of angle.
Conclude: the limit is $0$ , so $q$ is differentiable at the origin with tangent plane $z = 0$ (a flat bottom of the bowl).

Verify: at $∥ h ∥ = 0.1$ the ratio is $0.1$ ; at $0.01$ it is $0.01$ — halving distance halves the ratio, confirming $\to 0$ . ✓

Cell H — exam twist: find the parameter

Worked example Example 8 (Cell H) — solve for continuity/differentiability

For which value of the constant $k$ is $f (x, y) = ⎩ ⎨ ⎧ \frac{x ^{2} + y ^{2}}{∣ x ∣ + ∣ y ∣} k (x, y) \neq = (0, 0) (0, 0)$ continuous at $(0, 0)$ , and is it then differentiable there?

Forecast: guess $k$ first — does the ratio blow up or vanish as you approach the origin?

Find the required $k$ (continuity). Bound: $x^{2} + y^{2} \leq (∣ x ∣ + ∣ y ∣)^{2}$ , so $0 \leq f = \frac{x ^{2} + y ^{2}}{∣ x ∣ + ∣ y ∣} \leq ∣ x ∣ + ∣ y ∣ \to 0$ . Why this step? Squeeze forces the limit to $0$ , so continuity requires $k = 0$ . See Continuity in Multiple Variables.
Set $k = 0$ and test the partials. On $y = 0$ : $f (x, 0) = \frac{x ^{2}}{∣ x ∣} = ∣ x ∣$ . Its derivative at $0$ is $lim_{h \to 0} \frac{∣ h ∣ - 0}{h}$ , which is $+ 1$ from the right and $- 1$ from the left — no partial $f_{x} (0, 0)$ . Why this step? Even the axis slope fails to exist; $∣ x ∣$ has a corner at $0$ .
Conclude: with $k = 0$ , $f$ is continuous but not differentiable — indeed it has no partials at the origin. Why this step? Differentiable $\Rightarrow$ partials exist. No partials $\Rightarrow$ not differentiable, regardless of continuity.

Verify: $k = 0$ makes it continuous (upper bound $∣ x ∣ + ∣ y ∣ \to 0$ ). Along $x$ -axis $f = ∣ x ∣$ , whose one-sided slopes are $+ 1$ and $- 1$ : unequal $\Rightarrow$ partial undefined $\Rightarrow$ not differentiable. ✓

Recall Which cell was which? (self-test)

Partials exist but discontinuous ::: Cell B (Ex 2, parabola path gives $\frac{1}{2}$ ). Continuous + partials exist + still not differentiable ::: Cell C (Ex 3, diagonal ratio $- \frac{1}{2 2}$ ). All directional derivatives exist yet not differentiable ::: Cell E (Ex 5, $D_{u} = cos^{2} θ sin θ$ ). Removable/degenerate input made differentiable ::: Cell D (Ex 4, $sin (x y) / y \to$ patch). Proving differentiability from the raw $o (∥ h ∥)$ limit ::: Cell G (Ex 7, $E = ∥ h ∥^{2}$ ). Solve for parameter $k$ ::: Cell H (Ex 8, $k = 0$ , still no partials).

Connections

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