Chain rule for multivariable functions — all cases: Worked examples

Before anything, one promise: we never write a symbol without saying what it means.

$z = f (x, y)$ means a machine $f$ that eats two numbers $x, y$ and spits out one number $z$ .
$\frac{\partial z}{\partial x}$ (read "partial dee z by dee x") is the rate $z$ changes when you nudge only $x$ and freeze $y$ . See Partial Derivatives.
$\frac{d z}{d t}$ (round-free $d$ ) is the rate when $z$ ultimately depends on the single variable $t$ .

We introduce more symbols ( $J$ for the Jacobian matrix, $\sum$ for summation) later, at the exact moment we first use them.

The scenario matrix

Every problem this topic throws lives in one of these cells:

#	Cell (what makes it tricky)	Which example
A	Case 1: one independent variable, clean	Ex 1
B	Case 2: two independent variables (polar)	Ex 2
C	Sign / quadrant trap (a component goes negative)	Ex 3
D	Zero input: a link derivative is $0$ at a point	Ex 4
E	Degenerate: two inputs collapse to the same variable	Ex 5
F	Explicit reappearance: $t$ appears directly and through $x, y$	Ex 6
G	Limiting behaviour: rate as $t \to$ a special value	Ex 7
H	Real-world word problem (units matter)	Ex 8
I	Exam twist: Jacobian / matrix form	Ex 9

We hit every cell below.

Cell A — one independent variable, clean

Before we compute, look at the dependency tree for this problem. It shows the two roads from the deep variable $t$ up to the output $z$ : one road passes through $x$ , the other through $y$ . That is the picture behind "two paths".

Figure — Chain rule for multivariable functions — all cases

Worked example Ex 1 (Cell A)

$z = x^{2} y$ , $x = cos t$ , $y = sin t$ . Find $\frac{d z}{d t}$ and evaluate at $t = \frac{π}{2}$ .

Forecast: at $t = π /2$ we sit at the top of the unit circle, $x = 0, y = 1$ . Guess: since $x = 0$ there, does $\frac{d z}{d t}$ vanish? Hold that thought.

$\frac{\partial z}{\partial x} = 2 x y$ , $\frac{\partial z}{\partial y} = x^{2}$ . Why this step? Freeze one variable, differentiate the other — that is what $\partial$ means.
$\frac{d x}{d t} = - sin t$ , $\frac{d y}{d t} = cos t$ . Why? Ordinary single-variable derivatives — see Single-variable Chain Rule.
$\frac{d z}{d t} = 2 x y (- sin t) + x^{2} cos t$ . Why? The tree above has exactly two paths ( $t \to x \to z$ and $t \to y \to z$ ): multiply the link-derivatives along each path, then add the two paths.
Substitute $x = cos t, y = sin t$ into step 3: $2 (cos t) (sin t) (- sin t) + (cos t)^{2} cos t = - 2 cos t sin^{2} t + cos^{3} t$ .
At $t = π /2$ : $cos t = 0$ , so both terms are $0$ . $\frac{d z}{d t} = 0$ .

Verify: direct substitution $z = cos^{2} t sin t$ . Differentiate directly: $\frac{d z}{d t} = 2 cos t (- sin t) sin t + cos^{2} t cos t = - 2 cos t sin^{2} t + cos^{3} t$ . ✓ Matches step 4. The forecast was right — the $x = 0$ factor kills both paths.

Cell B — two independent variables (polar), with a figure

The next figure does two jobs at once. On the left is the polar dependency tree: the output $z$ reaches each ultimate variable $r$ and $θ$ through both doors $x$ and $y$ — that is why every polar chain-rule sum has two terms. On the right is the geometric picture: the point $(x, y)$ on a ray at angle $θ$ , and the red radial arrow showing the direction you move when you increase $r$ . Increasing $r$ slides the point straight out along that arrow, which is exactly why $\partial z / \partial r$ turns out to be the Directional Derivative of $f$ in the radial direction.

Worked example Ex 2 (Cell B)

$z = f (x, y)$ with $x = r cos θ$ , $y = r sin θ$ . Take the concrete $f = x^{2} + y^{2}$ and find $\frac{\partial z}{\partial r}$ and $\frac{\partial z}{\partial θ}$ .

Forecast: $z = x^{2} + y^{2}$ is exactly $r^{2}$ , so guess $\partial z / \partial r = 2 r$ and $\partial z / \partial θ = 0$ (angle doesn't change distance-squared).

$\frac{\partial z}{\partial x} = 2 x = 2 r cos θ$ , $\frac{\partial z}{\partial y} = 2 y = 2 r sin θ$ . Why? Partials of $f$ , then rewrite in polar using $x = r cos θ, y = r sin θ$ .
$\frac{\partial x}{\partial r} = cos θ$ , $\frac{\partial y}{\partial r} = sin θ$ . Why? Hold $θ$ fixed, differentiate in $r$ .
$\frac{\partial z}{\partial r} = (2 r cos θ) (cos θ) + (2 r sin θ) (sin θ) = 2 r cos^{2} θ + 2 r sin^{2} θ = 2 r (cos^{2} θ + sin^{2} θ) = 2 r$ . Why? Sum the two paths from the left tree; then use $cos^{2} θ + sin^{2} θ = 1$ .
$\frac{\partial x}{\partial θ} = - r sin θ$ , $\frac{\partial y}{\partial θ} = r cos θ$ ; so $\frac{\partial z}{\partial θ} = (2 r cos θ) (- r sin θ) + (2 r sin θ) (r cos θ) = - 2 r^{2} cos θ sin θ + 2 r^{2} sin θ cos θ = 0$ .

Verify: direct $z = r^{2} \Rightarrow \partial z / \partial r = 2 r$ , $\partial z / \partial θ = 0$ . ✓ Both match the forecast. Step 3 is the directional derivative of $f$ along the red radial arrow in the figure.

Cell C — sign / quadrant trap

The figure below plants the point at $t = 3 π /4$ on the unit circle: it lives in quadrant II, where the horizontal coordinate $x = cos t$ is negative (red, pointing left) but the vertical coordinate $y = sin t$ is positive (mint, pointing up). Keeping these signs straight is the whole game.

Worked example Ex 3 (Cell C)

$z = x y$ , $x = cos t$ , $y = sin t$ . Find $\frac{d z}{d t}$ at $t = \frac{3 π}{4}$ (second quadrant, where $x < 0$ , $y > 0$ ).

Forecast: in quadrant II, $x = cos (3 π /4) < 0$ but $y > 0$ . A student who forgets signs may keep everything positive. Watch the signs carefully.

$\frac{\partial z}{\partial x} = y$ , $\frac{\partial z}{\partial y} = x$ . Why? $\partial (x y) / \partial x = y$ , freezing $y$ ; and $\partial (x y) / \partial y = x$ .
$\frac{d x}{d t} = - sin t$ , $\frac{d y}{d t} = cos t$ . Why? These are the ordinary single-variable derivatives of $cos t$ and $sin t$ ; the minus sign on $\frac{d x}{d t}$ is exactly the trap this cell is about — differentiating cosine flips the sign, so the negative must be carried into the sum.
$\frac{d z}{d t} = y (- sin t) + x (cos t)$ . Now substitute $x = cos t$ and $y = sin t$ : $= (sin t) (- sin t) + (cos t) (cos t) = - sin^{2} t + cos^{2} t = cos 2 t$ . Why the last equality? The identity $cos^{2} t - sin^{2} t = cos 2 t$ .
At $t = 3 π /4$ : $2 t = 3 π /2$ , so $cos (3 π /2) = 0$ . Thus $\frac{d z}{d t} = 0$ .

Verify: $z = cos t sin t = \frac{1}{2} sin 2 t$ , so $\frac{d z}{d t} = cos 2 t$ . ✓ The trap: at $t = 3 π /4$ the term $x cos t = cos^{2} (3 π /4) = + \frac{1}{2}$ (positive because it's $cos^{2}$ ) while $- sin^{2} t = - \frac{1}{2}$ ; they cancel to $0$ . If you'd dropped the negative sign on $sin^{2}$ , you'd wrongly get $+ 1$ .

Cell D — a zero input

Worked example Ex 4 (Cell D)

$z = x^{2} + y^{3}$ , $x = t^{2}$ , $y = t$ . Find $\frac{d z}{d t}$ at $t = 0$ .

Forecast: at $t = 0$ , $\frac{d x}{d t} = 2 t = 0$ . So the $x$ -path derivative is switched off. Guess: only the $y$ -path survives.

$\frac{\partial z}{\partial x} = 2 x$ , $\frac{\partial z}{\partial y} = 3 y^{2}$ . Why? Partial in $x$ freezes $y$ , so $y^{3}$ is a constant and drops, leaving $2 x$ ; partial in $y$ freezes $x$ , so $x^{2}$ drops and $y^{3}$ becomes $3 y^{2}$ by the power rule.
$\frac{d x}{d t} = 2 t$ , $\frac{d y}{d t} = 1$ . Why? These are ordinary single-variable derivatives: $\frac{d}{d t} t^{2} = 2 t$ and $\frac{d}{d t} t = 1$ by the power rule.
$\frac{d z}{d t} = 2 x \cdot 2 t + 3 y^{2} \cdot 1$ . Substitute $x = t^{2}, y = t$ : $= 2 t^{2} \cdot 2 t + 3 t^{2} = 4 t^{3} + 3 t^{2}$ . Why? Sum the two paths, then plug the parametrisation in.
At $t = 0$ : $4 (0) + 3 (0) = 0$ . Both paths give $0$ here because $x = 0$ and $y = 0$ too.

Verify: direct $z = (t^{2})^{2} + t^{3} = t^{4} + t^{3}$ , so $\frac{d z}{d t} = 4 t^{3} + 3 t^{2}$ . ✓ At $t = 0$ this is $0$ . The lesson: when a link derivative is $0$ you don't drop the whole rule — you keep both terms, and that term simply contributes nothing.

Cell E — degenerate (inputs collapse to the same variable)

Worked example Ex 5 (Cell E)

$z = f (x, y) = x^{2} + 3 x y + y^{2}$ but with $x = t$ and $y = t$ (both inputs are the same variable). Find $\frac{d z}{d t}$ .

Forecast: the two doors now open onto the same street. You still write two paths — one via $x$ , one via $y$ — then add. Don't be tempted to "count $t$ once".

$\frac{\partial z}{\partial x} = 2 x + 3 y$ , $\frac{\partial z}{\partial y} = 3 x + 2 y$ . Why? Partial in $x$ treats $y$ constant, and vice versa — even though later both equal $t$ , the partials are taken independently.
$\frac{d x}{d t} = 1$ , $\frac{d y}{d t} = 1$ . Why? Each of $x = t$ and $y = t$ is the identity map in $t$ , and $\frac{d}{d t} t = 1$ ; both links change at unit rate, which is precisely why both paths survive with equal weight.
$\frac{d z}{d t} = (2 x + 3 y) (1) + (3 x + 2 y) (1) = (2 x + 3 y) + (3 x + 2 y) = 5 x + 5 y$ . Why? Sum both paths, then collect like terms.
Substitute $x = y = t$ : $\frac{d z}{d t} = 5 t + 5 t = 10 t$ .

Verify: direct $z = t^{2} + 3 t \cdot t + t^{2} = 5 t^{2}$ , so $\frac{d z}{d t} = 10 t$ . ✓ Had you "counted $t$ once" and used only $\partial z / \partial x = 2 t + 3 t = 5 t$ , you'd get $5 t$ — half the truth. The collapse of inputs does not collapse the number of paths.

Cell F — explicit reappearance ( $t$ direct and through $x, y$ )

Worked example Ex 6 (Cell F)

$w = f (x, y, t) = x y + t^{2}$ , with $x = t$ , $y = t^{2}$ . Find $\frac{d w}{d t}$ .

Forecast: $t$ appears three ways: through $x$ , through $y$ , and sitting right there as $t^{2}$ . Three paths, three products. Miss the explicit one and you'll be off.

To avoid confusion, we keep one output symbol: the machine is $f$ , and $w = f$ is its output value. All partials below are partials of $f$ ; the total derivative on the left is written $\frac{d w}{d t}$ since $w$ is that same output.

$\frac{\partial f}{\partial x} = y$ , $\frac{\partial f}{\partial y} = x$ , $\frac{\partial f}{\partial t} = 2 t$ . Why? Each partial freezes the other two slots; the last one, freezing $x, y$ , captures the explicit $t$ in $t^{2}$ .
$\frac{d x}{d t} = 1$ , $\frac{d y}{d t} = 2 t$ . Why? These are ordinary single-variable derivatives of the parametrisation: $\frac{d}{d t} t = 1$ and $\frac{d}{d t} t^{2} = 2 t$ by the power rule — they measure how fast each intermediate variable moves as $t$ moves.
$\frac{d w}{d t} = \frac{\partial f}{\partial x} \frac{d x}{d t} + \frac{\partial f}{\partial y} \frac{d y}{d t} + \frac{\partial f}{\partial t}$ . Why? The tree has a direct branch to $t$ plus two indirect branches — see Implicit Differentiation (multivariable).
$= y (1) + x (2 t) + 2 t$ . Substitute $x = t, y = t^{2}$ : $= t^{2} + t \cdot 2 t + 2 t = t^{2} + 2 t^{2} + 2 t = 3 t^{2} + 2 t$ .

Verify: direct $w = t \cdot t^{2} + t^{2} = t^{3} + t^{2}$ , so $\frac{d w}{d t} = 3 t^{2} + 2 t$ . ✓ Drop the $+ 2 t$ explicit term and you'd wrongly get $3 t^{2}$ .

Cell G — limiting behaviour

Worked example Ex 7 (Cell G)

$z = e^{x y}$ , $x = t$ , $y = \frac{1}{t}$ (for $t > 0$ ). Find $\frac{d z}{d t}$ and its limit as $t \to \infty$ .

Forecast: here $x y = t \cdot \frac{1}{t} = 1$ always, so $z = e^{1} = e$ is constant. A constant has zero rate. Guess $\frac{d z}{d t} = 0$ everywhere, so the limit is $0$ .

To differentiate $z = e^{x y}$ we treat the exponent as a single lump: let $u = x y$ , so $z = e^{u}$ . The single-variable fact $\frac{d}{d u} e^{u} = e^{u}$ (the exponential is its own derivative) then gives, by the ordinary chain rule, $\frac{\partial z}{\partial x} = e^{u} \frac{\partial u}{\partial x} = e^{x y} \cdot y$ and $\frac{\partial z}{\partial y} = e^{u} \frac{\partial u}{\partial y} = e^{x y} \cdot x$ . Why $e$ here? Because the variable sits in the exponent, and $e^{u}$ is the one function whose derivative is itself — that is the tool that answers "how does $e^{something}$ change?"
$\frac{d x}{d t} = 1$ , $\frac{d y}{d t} = - \frac{1}{t ^{2}}$ . Why the minus? $\frac{1}{t} = t^{- 1}$ has derivative $- t^{- 2}$ .
$\frac{d z}{d t} = y e^{x y} (1) + x e^{x y} (- \frac{1}{t ^{2}})$ . Why this step? Two paths reach $t$ (via $x$ and via $y$ ), so we multiply each path's links and add — this is the chain rule sum of Cell A. Now substitute $x = t, y = 1/ t, x y = 1$ (the constraint that makes the exponent constant): $= \frac{1}{t} e - t \cdot e \cdot \frac{1}{t ^{2}} = \frac{e}{t} - \frac{e}{t} = 0$ .
Limit as $t \to \infty$ : it's identically $0$ , so the limit is $0$ .

Verify: direct $z = e^{t \cdot (1/ t)} = e^{1} = e$ , constant, derivative $0$ . ✓ The two paths exactly cancel — a beautiful check that the chain rule respects hidden constraints.

Cell H — real-world word problem (units)

Worked example Ex 8 (Cell H)

A metal plate's temperature is $T (x, y) = x^{2} + 2 y$ degrees Celsius, where $x, y$ are in metres. A bug walks along $x = 1 + t$ , $y = 2 - t$ ( $t$ in seconds). How fast does the bug feel the temperature changing at $t = 0$ ?

Forecast: moving in $+ x$ warms it (since $T$ grows with $x^{2}$ ), moving in $- y$ cools it (since $T$ grows with $y$ ). The net could be either sign — compute.

$\frac{\partial T}{\partial x} = 2 x (°C/m)$ , $\frac{\partial T}{\partial y} = 2 (°C/m)$ . Why? Partial in $x$ freezes $y$ : $2 y$ is constant and drops, leaving $2 x$ ; partial in $y$ freezes $x$ : $x^{2}$ drops and $2 y$ becomes $2$ . The units are °C per metre because $T$ is in °C and $x, y$ in metres.
$\frac{d x}{d t} = 1 (m/s)$ , $\frac{d y}{d t} = - 1 (m/s)$ . Why? These are ordinary derivatives of the bug's path: $\frac{d}{d t} (1 + t) = 1$ and $\frac{d}{d t} (2 - t) = - 1$ ; the minus says the bug moves in the $- y$ direction, and the units are metres per second.
$\frac{d T}{d t} = 2 x (1) + 2 (- 1) = 2 x - 2 (°C/s)$ . Why? Sum the two paths; note the units $\frac{°C}{m} \cdot \frac{m}{s} = \frac{°C}{s}$ line up.
At $t = 0$ : $x = 1 + 0 = 1$ , so $\frac{d T}{d t} = 2 (1) - 2 = 0$ °C/s.

Verify: direct $T = (1 + t)^{2} + 2 (2 - t) = 1 + 2 t + t^{2} + 4 - 2 t = t^{2} + 5$ , so $\frac{d T}{d t} = 2 t$ ; at $t = 0$ that's $0$ °C/s. ✓ Warming and cooling exactly balance at the start.

Cell I — exam twist: Jacobian / matrix form

First, two pieces of notation, each defined before use.

The summation symbol $\sum$ . The sign $k = 1 \sum n a_{k}$ is shorthand for "add up the terms $a_{k}$ as the counter $k$ runs from $1$ to $n$ ": it just means $a_{1} + a_{2} + \dots + a_{n}$ . We use it because the chain-rule "add over all paths" is exactly such a running sum, one term per intermediate variable $k$ .

The Jacobian matrix $J$ . A Jacobian matrix is a table that collects all the first partials of a vector-valued map into a grid: row $i$ is "the $i$ -th output", column $k$ is "with respect to the $k$ -th input". We write it $J$ . For a map $(x, y) \mapsto (u, v)$ ,

$J_{(u, v)} = \frac{\partial u}{\partial x} \frac{\partial v}{\partial x} \frac{\partial u}{\partial y} \frac{\partial v}{\partial y} .$

The deep fact (proved in the parent note): composing two maps chains their Jacobians by matrix multiplication, $J_{outer \circ inner} = J_{outer} J_{inner}$ . And matrix multiplication is exactly "sum over paths": entry $(i, j)$ of the product is $k \sum (J_{outer})_{ik} (J_{inner})_{k j}$ — the sum, over every intermediate variable $k$ , of "outer rate through $k$ " times "inner rate to $k$ ". That sum-over- $k$ is the tree's paths added up. The figure shows this: the inner map pushes an input change into the intermediate slots, and the outer map's rows gather those pushes.

Worked example Ex 9 (Cell I)

$u = x + y$ , $v = x - y$ , with $x = r cos θ$ , $y = r sin θ$ . Find the Jacobian $\frac{\partial ( u , v )}{\partial ( r , θ )}$ using the matrix product $J_{(u, v)} J_{(x, y)}$ .

Forecast: the general chain rule is $J_{u \circ x} = J_{u} J_{x}$ — matrix multiply. See Gradient and Jacobian. Guess the outer Jacobian is a constant matrix.

Inner map partials: $\frac{\partial x}{\partial r} = cos θ, \frac{\partial x}{\partial θ} = - r sin θ, \frac{\partial y}{\partial r} = sin θ, \frac{\partial y}{\partial θ} = r cos θ$ . Why? Each is an ordinary partial of $x = r cos θ$ or $y = r sin θ$ , freezing the other variable. So $J_{(x, y)} = (cos θ sin θ - r sin θ r cos θ) .$
Outer map: $u = x + y$ , $v = x - y$ gives $J_{(u, v)} = (11 1 - 1)$ . Why constant? $u, v$ are linear in $x, y$ , so their partials are numbers ( $\partial u / \partial x = 1$ , $\partial u / \partial y = 1$ , $\partial v / \partial x = 1$ , $\partial v / \partial y = - 1$ ).
Multiply — each entry is a sum over the intermediate variable, i.e. a path sum $\sum_{k}$ : $J = (11 1 - 1) (cos θ sin θ - r sin θ r cos θ) = (cos θ + sin θ cos θ - sin θ - r sin θ + r cos θ - r sin θ - r cos θ) .$ Why this step? Row $i$ of the outer matrix times column $j$ of the inner matrix adds "outer rate through $x$ " and "outer rate through $y$ " — exactly the two paths to output $i$ from input $j$ .

Verify: substitute directly. $u = r cos θ + r sin θ$ , so $\partial u / \partial r = cos θ + sin θ$ ✓ and $\partial u / \partial θ = - r sin θ + r cos θ$ ✓. $v = r cos θ - r sin θ$ , so $\partial v / \partial r = cos θ - sin θ$ ✓ and $\partial v / \partial θ = - r sin θ - r cos θ$ ✓. The matrix product is the chain rule.

Recap

Recall Which cell was hardest, and why?

Cell E (degenerate) and Cell F (explicit reappearance) trip most people. The number of paths in the tree never shrinks just because two inputs equal the same variable ::: true — sum all paths regardless In $w = f (x, y, t)$ with $x, y$ depending on $t$ , how many terms? ::: three — two indirect plus one explicit $\partial f / \partial t$

Connections

Parent topic
Partial Derivatives — the $\partial z / \partial x$ links.
Total Differential — why terms add.
Directional Derivative — Cell B radial form.
Gradient and Jacobian — Cell I matrix product.
Implicit Differentiation (multivariable) — Cell F explicit term.
Single-variable Chain Rule — the one-path base case.