Multivariable-Calculus · Unit 10 · Video 2 · Interactive Practice

The Multivariable Chain Rule: df/dt, Why It Works, and the Product Rule

IKey Formulas

Formula	Name	What it says
$\frac{df}{dt} = f_x\,\frac{dx}{dt} + f_y\,\frac{dy}{dt} + f_z\,\frac{dz}{dt}$	Chain rule	One term per path $t \to x \to f$ , $t \to y \to f$ , $t \to z \to f$ ; the contributions add
$df = f_x\,dx + f_y\,dy + f_z\,dz$ with $dx = x'(t)\,dt$ , …	Reason 1: substitute differentials	Every term carries a factor $dt$ ; divide it out
$\frac{\Delta f}{\Delta t} \approx f_x\,\frac{\Delta x}{\Delta t} + f_y\,\frac{\Delta y}{\Delta t} + f_z\,\frac{\Delta z}{\Delta t}$	Reason 2: divide by $\Delta t$	Ordinary numbers; as $\Delta t \to 0$ the $\approx$ becomes $=$
$(uv)' = v\,u' + u\,v'$ , $\ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$	Product and quotient rules	$f_u = v$ , $f_v = u$ ; $\ g_u = \frac{1}{v}$ , $g_v = -\frac{u}{v^2}$

Key Insight: the partial derivatives are taken at the starting point and stay fixed as $\Delta t$ shrinks; only the ratios $\frac{\Delta x}{\Delta t}$ , $\frac{\Delta y}{\Delta t}$ , $\frac{\Delta z}{\Delta t}$ move, and they tend to $\frac{dx}{dt}$ , $\frac{dy}{dt}$ , $\frac{dz}{dt}$ .

IIVisualization 1 — One Term per Path

For $w = x^2y + z$ with $x = t$ , $y = e^t$ , $z = \sin t$ , each path contributes (partial) $\times$ (rate), and the three add.

t

−1

IIIVisualization 2 — Dividing the Approximation by $\Delta t$

At $t = -1$ , both $\frac{\Delta w}{\Delta t}$ and $w_x\frac{\Delta x}{\Delta t} + w_y\frac{\Delta y}{\Delta t} + w_z\frac{\Delta z}{\Delta t}$ tend to the same limit $\frac{dw}{dt}$ .

\Delta t

IVVisualization 3 — The Product Rule as a Two-Path Chain Rule

For $f = uv$ , the strips are $f_u\,\Delta u = v\,\Delta u$ and $f_v\,\Delta v = u\,\Delta v$ ; the corner $\Delta u\,\Delta v$ vanishes after dividing by $\Delta t$ .

\Delta t

VQuiz Questions

Problem 1 · Add Up the Paths

Given: $f(x, y, z) = xy + yz$ with $x = t$ , $y = t^2$ , $z = t^3$ . Find $\frac{df}{dt}$ at $t = 1$ using the chain rule.

✅ Correct!

f_x x' + f_y y' + f_z z' = (1)(1) + (2)(2) + (1)(3) = 8

, and direct substitution

f = t^3 + t^5

agrees.

❌ Not quite.

f_x + f_y + f_z = 1 + 2 + 1 = 4

adds the partials alone. Each partial must be multiplied by the rate of its own variable.

❌ Close, but…

f

depends on

z

through the term

yz

, so the path through

z

contributes

f_z\,z' = (1)(3) = 3

as well.

❌ Not quite.

x' + y' + z' = 1 + 2 + 3 = 6

adds the rates alone; each rate must be weighted by

f

's response to that variable.

❌ Not quite. At

t = 1

x = y = z = 1

, so

f_x = y = 1

f_y = x + z = 2

f_z = y = 1

Show solution

Partial derivatives: $f_x = y$ , $\ f_y = x + z$ , $\ f_z = y$ . Rates: $\frac{dx}{dt} = 1$ , $\ \frac{dy}{dt} = 2t$ , $\ \frac{dz}{dt} = 3t^2$ .

At $t = 1$ the point is $(x, y, z) = (1, 1, 1)$ , so $f_x = 1$ , $f_y = 2$ , $f_z = 1$ and the rates are $1$ , $2$ , $3$ :

\frac{df}{dt} = f_x\,\frac{dx}{dt} + f_y\,\frac{dy}{dt} + f_z\,\frac{dz}{dt} = (1)(1) + (2)(2) + (1)(3) = 8.

Check by substitution: $f = t \cdot t^2 + t^2 \cdot t^3 = t^3 + t^5$ , so $\frac{df}{dt} = 3t^2 + 5t^4 = 8$ at $t = 1$ .

Problem 2 · Where the Partials Are Evaluated

Given: $f(x, y) = x^2y$ with $x = 2t$ , $y = t + 1$ . Find $\frac{df}{dt}$ at $t = 1$ .

✅ Correct! At

t = 1

the point is

(x, y) = (2, 2)

f_x\,x' + f_y\,y' = (8)(2) + (4)(1) = 20

❌ Close, but… the partials are evaluated at the point

(x, y) = (x(1), y(1)) = (2, 2)

, not at

x = y = 1

. Then

f_x = 2xy = 8

and

f_y = x^2 = 4

❌ Not quite.

f_x + f_y = 8 + 4 = 12

leaves out the rates

\frac{dx}{dt} = 2

and

\frac{dy}{dt} = 1

❌ Close, but…

(8)(2) = 16

is only the path through

x

. The path through

y

adds

f_y\,\frac{dy}{dt} = (4)(1) = 4

❌ Not quite. Compute

f_x = 2xy

and

f_y = x^2

(2, 2)

, then weight them by

\frac{dx}{dt} = 2

and

\frac{dy}{dt} = 1

Show solution

At $t = 1$ : $x = 2$ , $y = 2$ . Partial derivatives there: $f_x = 2xy = 8$ , $\ f_y = x^2 = 4$ . Rates: $\frac{dx}{dt} = 2$ , $\ \frac{dy}{dt} = 1$ .

\frac{df}{dt} = f_x\,\frac{dx}{dt} + f_y\,\frac{dy}{dt} = (8)(2) + (4)(1) = 20.

Check by substitution: $f = (2t)^2(t + 1) = 4t^3 + 4t^2$ , so $\frac{df}{dt} = 12t^2 + 8t = 20$ at $t = 1$ .

Problem 3 · No Formula, Only Partials

Given: a drone flies through air whose temperature is $T(x, y, z)$ (in $\degree\mathrm{C}$ , distances in metres). At its current position $T_x = 0.4$ , $T_y = -0.3$ , $T_z = -0.01$ (in $\degree\mathrm{C}$ per metre), and its velocity is $\frac{dx}{dt} = 5$ , $\frac{dy}{dt} = 2$ , $\frac{dz}{dt} = 20$ (metres per second). No formula for $T$ is known. Find the rate $\frac{dT}{dt}$ at which the drone's thermometer reading changes.

✅ Correct!

(0.4)(5) + (-0.3)(2) + (-0.01)(20) = 2 - 0.6 - 0.2 = 1.2

. Substitution is impossible here; the chain rule only needs the partials and the rates.

❌ Not quite.

2 + 0.6 + 0.2 = 2.8

drops the signs. Moving in

y

and in

z

carries the drone toward cooler air, so those contributions are negative.

❌ Close, but…

T_z = -0.01

is small, but the drone climbs fast:

T_z\,\frac{dz}{dt} = (-0.01)(20) = -0.2

is not negligible.

❌ Not quite.

0.4 - 0.3 - 0.01 = 0.09

adds the partials without the velocity components; each must be multiplied by its rate.

❌ Not quite. Multiply each partial by the matching velocity component and add the three products.

Show solution

The reading is $T(x(t), y(t), z(t))$ , a function of $t$ alone. Its derivative, by the chain rule:

\frac{dT}{dt} = T_x\,\frac{dx}{dt} + T_y\,\frac{dy}{dt} + T_z\,\frac{dz}{dt} = (0.4)(5) + (-0.3)(2) + (-0.01)(20).

\frac{dT}{dt} = 2 - 0.6 - 0.2 = 1.2\ \degree\mathrm{C/s}.

With no formula for $T$ , substituting is not possible; the partial derivatives at the current point and the velocity are all the chain rule needs.

Problem 4 · The Quotient Rule from the Chain Rule

Given: $g(u, v) = \frac{u}{v}$ , where at $t = 2$ : $u = 6$ , $v = 3$ , $\frac{du}{dt} = 1$ , $\frac{dv}{dt} = 2$ . Find $g_v$ at this point and $\frac{dg}{dt}$ at $t = 2$ .

What is $g_v = \frac{\partial g}{\partial v}$ at $(u, v) = (6, 3)$ ?

What is $\frac{dg}{dt}$ at $t = 2$ ?

✅ Correct!

g_u\,u' + g_v\,v' = \tfrac{1}{3}(1) - \tfrac{2}{3}(2) = -1

, matching the quotient rule

\frac{u'v - uv'}{v^2} = \frac{3 - 12}{9} = -1

❌ Close, but…

\frac{\partial}{\partial v}\left(u \cdot v^{-1}\right) = -u\,v^{-2}

keeps the factor

u

-\frac{6}{9} = -\frac{2}{3}

❌ Check the sign. Increasing the denominator

v

decreases

\frac{u}{v}

when

u > 0

, so

g_v = -\frac{u}{v^2}

is negative.

❌ Not quite.

-\frac{u}{v} = -2

divides by

v

only once: the derivative of

v^{-1}

-v^{-2}

, so

g_v = -\frac{u}{v^2} = -\frac{2}{3}

❌ Check the order.

\frac{uv' - u'v}{v^2} = 1

reverses the numerator; the chain rule gives

g_u\,u' + g_v\,v' = \frac{u'}{v} - \frac{u\,v'}{v^2}

❌ Close, but…

\tfrac{1}{3} + \tfrac{4}{3}

uses

g_v = +\tfrac{2}{3}

; with

g_v = -\tfrac{2}{3}

the second term is

-\tfrac{4}{3}

❌ Not quite.

-3 = \frac{u'v - uv'}{v}

divides by

v

once; over the common denominator

v^2 = 9

the result is

\frac{-9}{9} = -1

❌ Check $g_v$ . Hold

u

fixed and differentiate

u\,v^{-1}

with respect to

v

❌ Check $\frac{dg}{dt}$ . Use

\frac{dg}{dt} = g_u\,\frac{du}{dt} + g_v\,\frac{dv}{dt}

with

g_u = \frac{1}{v}

Show solution

Partial derivatives of $g = \frac{u}{v}$ : $\ g_u = \frac{1}{v}$ , $\ g_v = -\frac{u}{v^2}$ . At $(u, v) = (6, 3)$ :

g_u = \frac{1}{3}, \qquad g_v = -\frac{6}{9} = -\frac{2}{3}.

Two paths, $t \to u \to g$ and $t \to v \to g$ :

\frac{dg}{dt} = g_u\,\frac{du}{dt} + g_v\,\frac{dv}{dt} = \frac{1}{3}(1) - \frac{2}{3}(2) = \frac{1}{3} - \frac{4}{3} = -1.

Check with the quotient rule: $\frac{u'v - uv'}{v^2} = \frac{(1)(3) - (6)(2)}{3^2} = \frac{-9}{9} = -1$ .

Solved: 0 / 4

The Multivariable Chain Rule: df/dt, Why It Works, and the Product Rule

IKey Formulas

IIVisualization 1 — One Term per Path

IIIVisualization 2 — Dividing the Approximation by Δt\Delta tΔt

IVVisualization 3 — The Product Rule as a Two-Path Chain Rule

VQuiz Questions

IIIVisualization 2 — Dividing the Approximation by $\Delta t$