Partial derivatives: Question bank — AI-ML deep dive

Before we start, two words we lean on:

partial derivative $\partial f / \partial x$ = the slope of $f$ if you move ONLY $x$ and freeze every other variable.
residual $r = w x + b - t$ = "prediction minus target" in a linear-regression loss.

True or false — justify

TRUE / FALSE:

\partial f / \partial x

is itself an ordinary single-variable derivative.

TRUE — once you freeze the other variables they act like fixed numbers, so it collapses to the 1-D derivative of a function of

x

alone.

TRUE / FALSE: If

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

then

\partial f / \partial x

still depends on

y

FALSE — the

y^{2}

term has no

x

, so it vanishes;

\partial f / \partial x = 2 x

, with no

y

in sight.

TRUE / FALSE: For any

f

, the two partials

\partial f / \partial x

and

\partial f / \partial y

are always different functions.

FALSE — e.g.

f = x y

gives

\partial f / \partial x = y

and

\partial f / \partial y = x

(different), but

f = x + y

gives both equal to

1

. Symmetry of

f

can make them coincide.

TRUE / FALSE: A partial derivative can be zero even though

f

genuinely changes when both inputs move.

TRUE —

\partial f / \partial x

only measures the pure-

x

direction;

f

can still change along a diagonal where both

x

and

y

move.

TRUE / FALSE:

\partial (e^{x y}) / \partial x = e^{x y}

FALSE — you forgot the inner factor. The chain rule gives

y e^{x y}

; dropping the

y

is the classic bug that mis-scales your gradient.

TRUE / FALSE: The gradient

\nabla f

is just a list of numbers with no direction.

FALSE — it is a vector; its direction is steepest ascent and its length is the steepness there. See gradients and directional derivatives.

TRUE / FALSE: To minimise a loss you step in the

+ \nabla f

direction.

FALSE —

+ \nabla f

points uphill (steepest ascent). Descent follows

- \nabla f

, so you step opposite the gradient.

TRUE / FALSE: For

L = (w x + b - t)^{2}

with data

x, t

fixed,

\partial L / \partial x

is what training uses.

FALSE —

x

is data, not a parameter; training uses

\partial L / \partial w

and

\partial L / \partial b

. Differentiating w.r.t. the wrong symbol is a real trap.

TRUE / FALSE: If both partials are zero at a point,

f

must be at a minimum there.

FALSE — zero gradient marks a stationary point, which could be a maximum, minimum, or saddle. Flat slope ≠ lowest valley.

Spot the error

Someone writes

\partial (x^{2} y + 3 y) / \partial x = 2 x y + 3

. Find the mistake.

The

3 y

term has no

x

, so its

x

-slope is

0

, not

3

. Correct answer is

2 x y

; they wrongly differentiated

3 y

as if

y

were

x

Someone writes "

f_{x}

means we hold

x

constant and vary the rest." Fix it.

Backwards. The subscript names the moving variable, so

f_{x}

varies

x

and freezes everything else.

For

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

, someone gets

\partial f / \partial y = 3 x^{2} \cdot 2 y = 6 x^{2} y

. What went wrong?

They differentiated the

x^{3}

factor too. Since

y

is the moving variable,

x^{3}

is a frozen constant; answer is

x^{3} \cdot 2 y = 2 x^{3} y

Someone claims "

e^{x y}

has no chain rule because it's already an exponential." Rebut.

The exponent

x y

is an inside function, so the chain rule still applies: multiply by

\partial (x y) / \partial x = y

. "Already exponential" doesn't exempt the inner part. See multivariable chain rule.

A student computes

\partial L / \partial w = 2 (w x + b - t)

for

L = (w x + b - t)^{2}

. What factor is missing?

The inner factor

\partial r / \partial w = x

. Correct is

2 (w x + b - t) x

; forgetting

x

silently rescales every weight update.

Someone says "the limit definition needs both

x

and

y

to change by

h

." Correct them.

Only

x

shifts to

x + h

;

y

appears unchanged in both

f (x + h, y)

and

f (x, y)

. If

y

also moved, you'd no longer have a partial derivative.

Why questions

Why does a term with no

x

contribute nothing to

\partial f / \partial x

Because it doesn't change when only

x

moves — its numerator difference

f (x + h, y) - f (x, y)

0

for that term, so its slope in the

x

-direction is exactly

0

Why is treating

y

as a constant legal rather than a cheat?

The limit literally holds

y

fixed in both terms, so from the

x

-slope's viewpoint

y

genuinely behaves like a number such as

7

π

Why does ML need partials at all instead of one ordinary derivative?

A loss

L (w_{1}, \dots, w_{n}, b)

has many knobs; each partial tells you how to nudge one weight, and stacking them gives the gradient that gradient descent walks.

Why does the geometric picture use a slice of the surface?

Freezing

y =

const cuts the surface

z = f (x, y)

into a 1-D curve; the partial

\partial f / \partial x

is just that curve's ordinary slope — the freeze becomes a slice.

Why is the gradient's direction "steepest ascent" and not some arbitrary direction?

Among all directions you could step, the vector of partials happens to point where

f

rises fastest; its negative is therefore the fastest way down, which is why we minimise along

- \nabla f

Why can the same rule (

2 (w x + b - t) x

) be called "how linear regression learns"?

That partial is the exact quantity gradient descent multiplies by the learning rate to update

w

; iterating it is the training loop of linear regression.

Edge cases

At a point where

f

has a sharp corner (kink) in the

x

-direction, what happens to

\partial f / \partial x

The one-sided slopes disagree, so the limit doesn't exist — the partial is undefined there, even if the

y

-partial is perfectly fine.

For a constant function

f (x, y) = 7

, what are both partials?

Both are

0

everywhere — nothing changes when you nudge any input, so every slope is flat.

f

depends on only one variable, say

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

, what is

\partial f / \partial y

Exactly

0

—

f

ignores

y

, so moving

y

never changes the output.

Can a partial derivative exist at a point while the function is not "smooth" (not fully differentiable) there?

YES — partials existing along the axes does not guarantee the surface is differentiable; a function can have both partials yet still fail to have a proper tangent plane. Existence of partials is weaker than full differentiability.

What does

\partial L / \partial b

reduce to when the residual

r = 0

(a perfect prediction)?

It becomes

2 r \cdot 1 = 0

— a zero gradient, so gradient descent leaves

b

unchanged, exactly what you'd want at a perfect fit.

If two variables are secretly linked (e.g. you think they're independent but they aren't), is the plain partial still valid?

The plain partial assumes the others are truly frozen and independent; if they actually move together you need the chain rule, or your "partial" measures the wrong thing.

Connections

Partial derivatives — parent note these traps stress-test.
1.2.02-Derivatives-single-variable — the 1-D case each partial reduces to.
1.2.04-Gradient-and-directional-derivatives — where partials become a vector.
1.2.05-Chain-rule-multivariable — the missing inner factor lives here.
1.3.01-Gradient-descent — why the sign of $\nabla f$ matters.
Linear-regression — the loss whose partials train it.