Partial derivatives — notation, calculation, geometric meaning: Question bank

True or false — justify

The claim is either true or false. Your job is to say why.

TF1.

\frac{\partial f}{\partial x}

and

\frac{df}{d x}

mean the same thing for

f (x, y)

False. The straight

d

demands

f

depend on

x

alone; the curly

\partial

signals other variables (

y

here) are held fixed while

x

wiggles. Using

d

hides that a whole variable was frozen.

TF2. If

f_{x} (a, b) = 0

and

f_{y} (a, b) = 0

, then

f

has a flat tangent plane there.

True. Both slice-slopes are zero, so the tangent plane

z = f (a, b) + 0 \cdot (x - a) + 0 \cdot (y - b)

is horizontal. (But flat

\neq =

maximum — it could be a saddle.)

TF3. A function can have

f_{x}

exist at a point while being discontinuous there.

True.

f_{x}

only inspects behaviour along the $x$ -axis line through the point; the function can misbehave along other directions and still have a fine one-directional slope.

TF4. If

f_{x} = f_{y} = 0

everywhere on a region, then

f

is constant on that region.

True (on a connected region). Zero slope in both the

x

and

y

directions leaves no room to change height as you move anywhere in the plane.

TF5.

\partial_{x} (g (y)) = 0

for any function

g

y

alone.

True. To the

x

-derivative, anything built only from

y

is a frozen constant, and constants have zero derivative — no matter how complicated

g

looks.

TF6. Mixed partials always satisfy

f_{x y} = f_{y x}

False. Only when the second partials are continuous (Clairaut's / Schwarz condition). There are engineered functions where the two mixed partials differ at a single bad point.

TF7. The partial

f_{x}

is itself generally a function of both

x

and

y

True. Freezing

y

to differentiate does not delete

y

; e.g. for

f = x^{2} y

f_{x} = 2 x y

still carries

y

. You froze it during the operation, not forever.

TF8. Rewriting

f = \frac{x z}{y}

x z \cdot y^{- 1}

before taking

f_{y}

is a legal trick.

True. It's the same function, and the power rule

\frac{d}{d y} y^{- 1} = - y^{- 2}

(with

x z

frozen) is cleaner than the quotient rule. Rewriting never changes the derivative.

Spot the error

Each is a wrong worked line. Name the exact mistake.

SE1. "

f = x^{2} y \Rightarrow f_{x} = 2 x

The frozen

y

was dropped. It is a constant multiplier, not an absent term, so

f_{x} = 2 x y

SE2. "

f = x^{3} - 7 y \Rightarrow f_{x} = 3 x^{2} - 7

The

- 7 y

term contains no

x

, so w.r.t.

x

it is a constant and its derivative is

0

, not

- 7

. Correct:

f_{x} = 3 x^{2}

SE3. "To get

f_{x} (1, 2)

for

f = x^{2} y

, first plug in:

f (1, 2) = 2

, then differentiate

2

to get

0

Substituting before differentiating turns the variable into a number and destroys its slope. Differentiate first (

f_{x} = 2 x y

), evaluate last (

f_{x} (1, 2) = 4

SE4. "

f = sin (x y^{2}) \Rightarrow f_{x} = cos (x y^{2})

The chain rule's inner derivative was forgotten. Inner is

x y^{2}

whose

x

-derivative is

y^{2}

, so

f_{x} = cos (x y^{2}) \cdot y^{2}

SE5. "

f = e^{x} cos y \Rightarrow f_{y} = e^{x} cos y

They differentiated the wrong factor. With

e^{x}

frozen,

\frac{d}{d y} cos y = - sin y

, giving

f_{y} = - e^{x} sin y

SE6. "Since

f_{x y}

means 'differentiate w.r.t.

x

then

y

', for

f = x^{2} y^{3}

we get

f_{x y} = \partial_{x} (\partial_{y} f)

Order/reading slip.

f_{x y}

means

x

first then

y

\partial_{y} (\partial_{x} f)

. (Here it equals

f_{y x}

anyway by Clairaut, but the definition still matters.)

SE7. "

f = \frac{x}{y}

near

y = 0

has

f_{y} = - \frac{x}{y ^{2}}

, which is fine everywhere."

The formula is right where $f$ is defined, but at

y = 0

the function itself is undefined, so the partial cannot be evaluated there — you cannot freeze a value the function never takes.

Why questions

WHY1. Why does a surface

z = f (x, y)

have no single "slope" the way

f (x)

does?

At one point you can walk in infinitely many directions, each rising at a different rate. A single number can't describe them all, so we pick the special

x

and

y

directions to measure separately.

WHY2. Why is a partial derivative "just an ordinary derivative in disguise"?

Freezing

y = b

turns

f (x, y)

into a one-variable function

g (x) = f (x, b)

; then

f_{x} (a, b) = g^{'} (a)

, so all single-variable rules from Single-variable derivative apply unchanged.

WHY3. Why do the two partials

f_{x}, f_{y}

suffice to write the whole tangent plane?

The plane must reproduce both slice-slopes; forcing its

x

-slope to be

f_{x}

and its

y

-slope to be

f_{y}

pins it down completely, giving

z = f (a, b) + f_{x} (x - a) + f_{y} (y - b)

WHY4. Why do we bundle

f_{x}, f_{y}

into the gradient

\nabla f

Because knowing slope in the two axis directions lets you reconstruct the slope in any direction (the Directional derivative), and the gradient is the vector that does this via a dot product.

WHY5. Why is the total differential

df = f_{x} d x + f_{y} d y

and not just

df = f_{x} d x

A small move changes both inputs; each contributes its own slope times its own step, and the linear approximation adds these two effects together.

WHY6. Why is Clairaut's symmetry

f_{x y} = f_{y x}

believable geometrically?

Both measure the same "twist" of the surface — how the slope in one direction changes as you shift in the other. Measuring the twist from either order describes the same bend.

WHY7. Why does the multivariable chain rule need all the partials, not one?

x

and

y

each depend on

t

, the total change of

f

collects a contribution through each variable's pathway:

\frac{df}{d t} = f_{x} \frac{d x}{d t} + f_{y} \frac{d y}{d t}

WHY8. Why is the notation

\partial

("curly d") worth its own symbol at all?

It is a standing reminder that other variables exist and are held constant. A plain

d

would silently pretend

f

is one-dimensional, inviting exactly the "I forgot the other variable" error.

Edge cases

EC1. What is

f_{z}

for

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

if we suddenly allow a

z

-slot?

0

. The function has no

z

dependence, so wiggling

z

changes nothing — its partial is identically zero, the honest answer for an absent variable.

EC2. Can

f_{x} (a, b)

exist even if

f_{y} (a, b)

does not?

Yes. The two directions are inspected independently; the slice along

x

can be smooth while the slice along

y

has a corner or jump at the same point.

EC3. What does

f_{x}

look like for a constant function

f (x, y) = 5

f_{x} = 0

and

f_{y} = 0

everywhere. The surface is a flat sheet; no wiggle in any input changes the height.

EC4. For

f (x, y) = ∣ x ∣ + y

(0, 0)

, does

f_{x}

exist?

No. The

x

-slice is

z = ∣ x ∣

, which has a corner at

x = 0

— the left slope

- 1

and right slope

+ 1

disagree, so the limit defining

f_{x}

fails. But

f_{y} = 1

exists fine.

EC5. If

f_{x} (a, b) > 0

, is the whole surface rising near

(a, b)

Not necessarily. It only guarantees rising as you step in the

+ x

direction; the surface could simultaneously fall in the

y

direction (a saddle-like slope pattern).

EC6. What happens to

f_{x}

f = ln x

(viewed as

f (x, y) = ln x

) at

x = 0

Undefined.

f_{x} = 1/ x

blows up as

x \to 0^{+}

and

ln x

isn't even defined for

x \leq 0

, so the partial has no value there — a genuine domain boundary.

EC7. Is

f_{x}

at an isolated point where only

f (a, b)

is known computable?

No. The limit definition needs values

f (a + h, b)

for a whole range of small

h

; a single point gives no neighbouring heights to compare, so no slope can be formed.

Active Recall

Recall One trap that catches everyone at least once

Q: In $f_{x}$ , why is it wrong to keep the constant term's coefficient like $- 7 y$ ? A: Relative to $x$ , the entire term $- 7 y$ holds no $x$ , so it is a constant — and the derivative of any constant is $0$ . Freezing $y$ makes $- 7 y$ behave like a fixed number.

Connections

Gradient vector — bundles the partials; see WHY4.
Directional derivative — the "any direction" answer to WHY1.
Tangent plane and linear approximation — WHY3, TF2.
Chain rule (multivariable) — WHY7.
Total differential — WHY5.
Single-variable derivative — WHY2, the frozen-variable reduction.
Clairaut's theorem — TF6, WHY6.