Chain rule for multivariable functions — all cases: Question bank

Before you start, keep three pictures in your head:

A dependency tree — output at the top, arrows pointing down to "what it depends on".
Paths add, links multiply — one product per path from output to the target variable.
$d$ vs $\partial$ — $d$ when the end result depends on one variable, $\partial$ when several survive.

True or false — justify

Every statement below is either true or false. The reveal gives the reason, which matters more than the verdict.

For

z = f (x, y)

with

x = x (t), y = y (t)

, we have

\frac{d z}{d t} = \frac{\partial z}{\partial x} \frac{d x}{d t}

False. There are two doors to

t

(through

x

and through

y

), so we must add the second product

\frac{\partial z}{\partial y} \frac{d y}{d t}

. Forgetting a path is the single most common error.

The chain rule adds path-contributions because the total differential

d z = z_{x} d x + z_{y} d y

is linear.

True. Linear approximation means independent inputs contribute independently, so their effects superpose — that "+" in the differential becomes the "+" in the chain rule.

In Case 2 (

z = f (x, y)

x = x (s, t)

y = y (s, t)

) every derivative must be written with

\partial

True. The final output still depends on two surviving variables

s

and

t

, so nothing here is a single-variable derivative — even

\frac{\partial x}{\partial s}

needs

\partial

because

x

also depends on

t

The order of factors in the Jacobian form

J_{u \circ x} = J_{u} J_{x}

can be swapped since multiplication is chain-rule "just multiplying".

False. Matrix multiplication is not commutative, and the shapes only fit in the order

J_{u} J_{x}

(outer map's Jacobian on the left). Swapping usually gives a dimension mismatch or a wrong matrix.

w = f (x, y, t)

and also

x = x (t), y = y (t)

, then

\frac{d w}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t}

False. There is also a direct path from

w

t

, so the term

+ \frac{\partial f}{\partial t}

must be added.

t

reaches

w

both explicitly and through

x, y

The polar radial derivative

\frac{\partial z}{\partial r} = z_{x} cos θ + z_{y} sin θ

is a directional derivative.

True.

(cos θ, sin θ)

is the unit radial direction, so this is

\nabla f \cdot \overset{r}{^}

— the chain rule quietly computes the directional derivative in the outward direction.

Since

\frac{d z}{d t}

uses

d

, all the inner derivatives inside it must also use

d

False.

z

depends on both

x

and

y

, so

\frac{\partial z}{\partial x}

keeps

\partial

; only the final dependence on the lone variable

t

(as in

\frac{d x}{d t}

\frac{d z}{d t}

) earns a

d

x

and

y

do not actually depend on some variable

s

, that path simply contributes zero.

True. A missing dependence gives

\frac{\partial x}{\partial s} = 0

, so that product vanishes automatically — the general formula still works, it just kills the dead paths.

The chain rule requires

f

to be differentiable, not merely to have partial derivatives.

True. The derivation uses

Δ z = f_{x} Δ x + f_{y} Δ y + ε_{1} Δ x + ε_{2} Δ y

with

ε \to 0

; that error-shrinking is differentiability. Mere existence of

f_{x}, f_{y}

is not enough.

Spot the error

Each line states a (flawed) piece of work. The reveal names the mistake and repairs it.

z = x^{2} y

x = cos t

y = sin t

, so

\frac{d z}{d t} = 2 x y \cdot (- sin t)

The

y

-path is dropped. The full answer adds

x^{2} cos t

, giving

\frac{d z}{d t} = 2 x y (- sin t) + x^{2} cos t

"For

z = f (x, y)

with

x = x (s, t)

, I wrote

\frac{d z}{d s} = \frac{d z}{d x} \frac{d x}{d s} + \frac{d z}{d y} \frac{d y}{d s}

Wrong differential symbols. Because

z

depends on two survivors

s, t

, every derivative here is partial:

\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}

w = f (x, y, t)

x = x (t), y = y (t)

; the answer is

z_{x} x^{'} + z_{y} y^{'}

— that's all the paths."

The direct explicit path to

t

is missing. Add

\frac{\partial f}{\partial t}

; the tree has three branches reaching

t

, not two.

"To differentiate

z = r^{2}

via chain rule I got

2 r cos θ + 2 r sin θ

The pieces weren't combined correctly. It should be

2 r cos θ \cdot cos θ + 2 r sin θ \cdot sin θ = 2 r (cos^{2} θ + sin^{2} θ) = 2 r

— each link must be multiplied by its own inner derivative before adding.

"Because chain rule 'multiplies', I wrote

\frac{\partial u}{\partial t _{j}} = \frac{\partial u}{\partial x _{i}} \frac{\partial x _{i}}{\partial t _{j}}

for the general case."

A sum is missing. Every

x_{i}

is a separate path, so

\frac{\partial u}{\partial t _{j}} = \sum_{i = 1}^{n} \frac{\partial u}{\partial x _{i}} \frac{\partial x _{i}}{\partial t _{j}}

— you multiply along a path and add across paths.

"I substituted

x = cos t

before differentiating, then used

\frac{\partial z}{\partial x}

Once you substitute,

z

is already a pure function of

t

; you can no longer speak of

\frac{\partial z}{\partial x}

from the reduced form. Either take partials first (chain rule) or substitute and differentiate as one variable — don't mix the two.

Why questions

Explain the reason, not just the rule.

Why do we ADD path contributions instead of multiplying them?

Because the total differential is a sum of independent linear effects; multiplying would wrongly couple contributions that act independently. We multiply only within a single path (link by link).

Why does the general formula have a

\sum

over

i

(the inner variables) and not over

j

j

labels the one target variable we differentiate against (fixed for each equation); the sum runs over

i

because each inner variable

x_{i}

is a distinct road from the output down to that target.

Why does dividing the total differential by

Δ t

and taking a limit "give the chain rule for free"?

The error terms

ε_{1} \frac{Δ x}{Δ t} + ε_{2} \frac{Δ y}{Δ t}

vanish because

ε \to 0

while

\frac{Δ x}{Δ t}

stays finite; what survives is exactly

z_{x} x^{'} + z_{y} y^{'}

Why does the Jacobian view call the chain rule "just matrix multiplication"?

Each layer of a composition is locally a linear map (its Jacobian); composing maps composes their linear parts, and composing linear maps is matrix multiplication — so

J_{u \circ x} = J_{u} J_{x}

Why must

x

and

y

be continuous (not just differentiable) in the derivation?

The step "

Δ t \to 0

forces

Δ x, Δ y \to 0

" is continuity. Without it, the inner changes might not shrink, and the error terms wouldn't vanish.

Why does Case 2 reduce to Case 1 when you differentiate with respect to

s

Holding

t

fixed turns the map into a function of the single variable

s

; along that frozen slice it's literally the one-independent-variable chain rule applied twice (once per inner variable).

Edge cases

Where the rule meets zeros, degeneracies, and boundaries.

What is

\frac{d z}{d t}

z = f (x, y)

but

x

is constant (does not depend on

t

Then

\frac{d x}{d t} = 0

, killing that path; only

\frac{\partial z}{\partial y} \frac{d y}{d t}

survives. The formula self-corrects — you never delete a term by hand.

At the origin in polar coordinates (

r = 0

), is

\frac{\partial z}{\partial r} = z_{x} cos θ + z_{y} sin θ

still meaningful?

The formula is fine, but

θ

is undefined at

r = 0

, so the radial direction is ambiguous; the value may depend on which

θ

you approach along, signalling a genuine geometric singularity of the coordinate map.

If a variable appears both explicitly and through an inner variable, and its two contributions cancel, what does that mean?

It means the output is locally insensitive to that variable — the explicit and indirect effects offset. The math is correct; the zero is a real feature, e.g. a quantity conserved along the motion.

Can the chain rule apply if

f

has partials that exist but

f

is not differentiable at the point?

No guarantee. Existence of partials is weaker than differentiability; the standard chain rule can fail there, so the differentiability hypothesis is not decorative.

What happens to

J_{u \circ x} = J_{u} J_{x}

when

u

maps to a single number (a scalar field)?

J_{u}

becomes a single row (the gradient

\nabla u

as a row vector), and the product

\nabla u J_{x}

returns a row of the composed partials — the scalar case is just the general matrix rule with one output row.

If two different inner variables happen to be equal functions, say

x = y = t^{2}

, do we still write two path terms?

Yes. The tree structure, not the numerical coincidence, dictates the paths; you write

z_{x} \frac{d x}{d t} + z_{y} \frac{d y}{d t}

and only then substitute the equal values — collapsing early risks losing a term.

Connections

4.4.07 Chain rule for multivariable functions — all cases — the parent this bank stress-tests.
Partial Derivatives — the $d$ vs $\partial$ distinction lives here.
Total Differential — the source of "paths add".
Directional Derivative — the polar edge case.
Gradient and Jacobian — the matrix-order trap.
Implicit Differentiation (multivariable) — the explicit-reappearance trap.
Single-variable Chain Rule — the one-path baseline these traps deviate from.