Differentiability in multiple variables: Question bank

Before you start, keep the pictures in mind: a partial derivative only walks due East or due North; differentiability demands a single flat tilted board (the tangent plane) fits from every direction at once. Most traps below are just that distinction wearing a disguise.

True or false — justify

f

has both partial derivatives at

a

, therefore

f

is continuous at

a

False. Partials only probe the two axis directions; the function can jump along a diagonal, e.g.

x y / (x^{2} + y^{2})

has

f_{x} = f_{y} = 0

at the origin yet limit

\frac{1}{2}

along

y = x

f

is differentiable at

a

, then

f

is continuous at

a

True. From

f (a + h) - f (a) = \nabla f \cdot h + E

, both the linear term and

E = o (∥ h ∥)

vanish as

h \to 0

, so

f (a + h) \to f (a)

f

is continuous at

a

, then

f

is differentiable at

a

False. The arrow only runs one way:

∣ x ∣ + ∣ y ∣

is continuous everywhere but has no tangent plane at the origin (a sharp corner).

If both partials are continuous near

a

, then

f

is differentiable at

a

True. This is exactly the

C^{1}

sufficient condition — the workhorse that lets you skip the raw

o (∥ h ∥)

limit for almost every function you meet.

f

is differentiable at

a

, then its partials are continuous at

a

False. Differentiable is weaker than

C^{1}

. A function can be differentiable at a point while its partials wiggle discontinuously there (e.g. pieced-together

x^{2} sin (1/ x)

-style constructions).

If every directional derivative of

f

exists at

a

, then

f

is differentiable at

a

False. Even "all directions" can fail: the formula

D_{u} f = \nabla f \cdot u

may break, and continuity can still fail, so directional derivatives everywhere is strictly weaker than differentiability.

A polynomial in

x

and

y

is differentiable everywhere.

True. Its partials are again polynomials, hence continuous everywhere, so it is

C^{1}

and therefore differentiable at every point.

f

is differentiable at

a

, the only possible linear approximation is

f (a) + \nabla f (a) \cdot h

True. Approaching along each axis forces the coefficients to equal

f_{x}

and

f_{y}

; no other linear map can satisfy the

o (∥ h ∥)

condition.

E (h) \to 0

h \to 0

, then

a + m \cdot h

is the tangent plane.

False.

E \to 0

is mere touching (continuity of the error). A tangent plane needs

E /∥ h ∥ \to 0

— the error must die faster than the step size.

C^{1}

a

and differentiable at

a

are the same thing.

False.

C^{1}

(continuous partials) implies differentiable but is genuinely stronger; differentiable functions need not have continuous partials.

Spot the error

f_{x} (0, 0) = 0

and

f_{y} (0, 0) = 0

, so

\nabla f (0, 0) = 0

and the tangent plane is

z = 0

." (for

f = x y / (x^{2} + y^{2})

)

The error is assuming a tangent plane exists. Since

f

is discontinuous at the origin it is not differentiable, so "

\nabla f

" is just a pair of numbers, not a valid linear approximation.

f

is differentiable because I checked the limit along the

x

-axis, the

y

-axis, and

y = x

, and the error ratio went to 0 on all three."

Three lines are not every direction. The

o (∥ h ∥)

limit must vanish along all paths simultaneously; a fourth curved path could still betray it.

"To find the tangent plane I need the raw

o (∥ h ∥)

limit, so I'll compute it for

f = x^{2} y + sin y

Overkill and error-prone: the partials

2 x y

and

x^{2} + cos y

are continuous, so

f

C^{1}

and the plane

z = f (a) + \nabla f (a) \cdot h

is automatically valid.

f (x, y) = x^{2} + y^{2}

is a sum and root of nice functions, so it is differentiable everywhere."

Wrong at the origin:

f_{x} = x / x^{2} + y^{2}

is undefined there (a cone point), so

C^{1}

and differentiability both fail at

(0, 0)

even though they hold everywhere else.

"The error ratio approached

0

along every straight line through

a

, so

f

is differentiable."

Straight lines are still not enough. Some functions send the error ratio to 0 on every line but not along parabolic paths like

y = x^{2}

, so this argument is incomplete.

"Since differentiable

\Rightarrow

continuous, and my

f

is continuous,

f

must be differentiable."

This reverses the implication. Continuity is a consequence of differentiability, not a cause;

∣ x ∣ + ∣ y ∣

is the standing counterexample.

Why questions

Why do we divide the error

E (h)

∥ h ∥

instead of just requiring

E \to 0

Dividing forces the error to be higher-order — smaller than the step itself — which is precisely what distinguishes a genuine tangent plane from any random plane that merely touches the point.

Why are partial derivatives "blind to diagonals"?

Each partial is a one-variable limit taken with the other variable frozen, so it only samples motion parallel to an axis and never sees a step like

(h, h)

heading off-axis.

Why is the

C^{1}

condition so useful in practice despite being only sufficient?

Almost every function you meet (polynomials,

sin

exp

, ratios away from their zeros) has continuous partials, so a one-line "

C^{1}

hence differentiable" replaces a painful limit computation 99% of the time.

Why must the linear map, if it exists, equal the gradient

(f_{x}, f_{y})

Restricting the approach to the

x

-axis makes the definition collapse to the single-variable derivative in

x

, pinning

m_{1} = f_{x}

; the

y

-axis pins

m_{2} = f_{y}

Why does differentiability imply all directional derivatives exist and obey

D_{u} f = \nabla f \cdot u

The tangent plane is a single linear map; walking along direction

u

just reads that map's slope in that direction, which is the dot product

\nabla f \cdot u

Why is differentiability called a "2D requirement" while a partial is "1D"?

A partial checks one frozen slice; differentiability demands a single flat board fit the surface from every point of the surrounding disk, coupling all directions at once.

Why can a function be differentiable at a point yet not

C^{1}

there?

Differentiability only asks that a good linear fit exist at that point; it says nothing about the partials matching up smoothly in a neighbourhood, which is the extra demand of

C^{1}

Edge cases

Is a constant function

f (x, y) = c

differentiable? What is its tangent plane?

Yes. Its partials are both

0

(continuous), so it is

C^{1}

; the tangent plane is the flat plane

z = c

everywhere.

At an isolated point where

f

is only defined (no neighbourhood), can differentiability even be asked?

No. The limit

h \to 0

requires approaching from all directions in a surrounding disk, so differentiability needs

f

defined on an open set around the point.

Take

f = x^{2} + y^{2}

at the origin: partials, continuity, differentiability?

Continuous there, but the partials don't exist (the cone tip has no single slope), so it fails differentiability at the origin while remaining differentiable everywhere else.

\nabla f (a) = 0

but

f

is differentiable, what does the tangent plane look like?

A perfectly horizontal plane

z = f (a)

— a critical point where the surface is flat to first order (peak, valley, or saddle).

Does differentiability at

a

tell you anything about

f

away from

a

No. It is a purely local, point-by-point property;

f

can be well-behaved at

a

and wildly discontinuous a short distance away.

Can a function be differentiable at exactly one point and nowhere else?

Yes. Constructions like

f = (x^{2} + y^{2}) \cdot (something jagged)

can force a valid linear fit only at the origin, where the multiplying factor crushes all bad behaviour.

Recall One-line summary of the whole bank

Every trap here is the same trap: mistaking axis-information (partials) or touching-information (continuity, $E \to 0$ ) for the all-directions, faster-than-linear fit that is differentiability.

Question bank — Differentiability in multiple variables

True or false — justify

Spot the error

Why questions

Edge cases

Connections