MATH U101 · Multivariable Calculus · Module 4

Partial Derivatives & the Chain Rule

Thomas §14.3–14.4, lectures L8–L9. The differentiation engine of the whole second half of the course — Modules 5 and 6 are applications of exactly these two sections.

🎧 Narrated walkthroughs for this module

Partial derivatives & implicit differentiation (6.7 min, T6 Q1–Q2) · The chain rule as a tree (7.4 min, T6 Q3–Q5) · Rates along a curve (6.2 min, T6 Q6–Q7). Each one is sound-on and pauses for your next move. The chain-rule walkthrough labels the bottom edges of the tree without working them out; that step (freeze s, then the school chain rule on cos(r + s)) is written out in the tutorial companion, T6 Q3 step 2.

Start here

The computing in this module is one-variable differentiation with a discipline attached (freeze the other letter) — the daily drill already built the hard part. The two genuinely new items, for everyone in the room: the ∂ concept itself and the §14.4 route-counting rule. No 12th-class background covers either.

First time? Start with the Module 4 lesson (slow build, slice-the-surface widget); this page is the compressed map for revision.

1 · Partial derivatives: definition and notation §14.3

Intuition: a surface has a slope per direction; f_x is the slope walking east (y frozen), f_y walking north (x frozen). Formal:

f_x(x₀, y₀) = lim_h→0 f(x₀+h, y₀) − f(x₀, y₀)h

Equivalent notations, all exam-legal: ∂f/∂x = f_x = ∂z/∂x = z_x; at a point, f_x(a, b) or ∂f/∂x |_(a,b). Use ∂ (never d) whenever the function has two or more inputs.

2 · Computing: freeze, then use the old rules

For f_x: treat every y as a literal constant; apply power/product/quotient/chain exactly as in one variable. A term with no x in it differentiates to 0.
Three-variable functions work identically: f_z freezes both x and y.
At-a-point questions: differentiate first, substitute last. (Substituting first kills the variable you're differentiating in.)

Reference results (both derived slowly in the lesson): for f = x²y + sin(xy), f_x = 2xy + y cos(xy), f_y = x² + x cos(xy); for g = xy/(x+y), g_x = y²/(x+y)², g_y = x²/(x+y)².

3 · Second partials and the Mixed Derivative Theorem

Four second partials: f_xx, f_yy, f_xy = (f_x)_y, f_yx = (f_y)_x. Clairaut: if the mixed partials are continuous, f_xy = f_yx — true for every function in this course. Uses: pick the easier order; and compute both as a built-in error check.

Worked example 1All four second partials of f(x, y) = x³y² − e^xy

First partials. f_x = 3x²y² − y e^xy (chain rule on e^xy: inner derivative y). f_y = 2x³y − x e^xy.
f_xx: differentiate f_x in x (y frozen): 6xy² − y² e^xy.
f_yy: differentiate f_y in y: 2x³ − x² e^xy.
f_xy: differentiate f_x = 3x²y² − y e^xy in y — product rule on y e^xy: e^xy + xy e^xy. Total: 6x²y − e^xy − xy e^xy.
f_yx: differentiate f_y = 2x³y − x e^xy in x: 6x²y − (e^xy + xy e^xy) — identical to step 4 ✓ (Clairaut, as it must).

4 · Differentiability §14.3

Meaning: near the point, the surface is well approximated by a plane (differentiable ⟹ continuous). The catch: partials existing is not enough — xy/(x²+y²) has f_x(0,0) = f_y(0,0) = 0 yet isn't even continuous at the origin. The rescue theorem (quote this): if f_x, f_y exist near the point and are continuous at it, f is differentiable there. Standard functions pass automatically. Depth beyond this (increment form with error terms) — confirm with the instructor.

5 · The chain rule §14.4

The one sentence: draw the dependency diagram; one product per route from the top variable to the one you're differentiating by; multiply along a route, add the routes. The two exam cases:

The diagram for z = f(x, y), x(t), y(t): two routes → two terms. Three moving inputs → three routes → three terms.

x(t), y(t): dzdt = f_x dxdt + f_y dydt x(s,t), y(s,t): ∂z∂t = f_x ∂x∂t + f_y ∂y∂t

Symbols carry marks: d where a quantity has one input, ∂ where it has several. Three inputs on top (w = f(x, y, z), all moving with t) → three terms; the diagram never lies.

Worked example 2Full chain-rule workout: z = x²y, x = cos t, y = sin t — find dz/dt, evaluate at t = π/4

Diagram: z on top; x, y below; t at the bottom. Two routes → two terms.
Pieces: f_x = 2xy, f_y = x²; dx/dt = −sin t, dy/dt = cos t.
Assemble: dz/dt = (2xy)(−sin t) + x² cos t = −2 cos t sin²t + cos³t (after substituting x = cos t, y = sin t).
Check by substituting first: z = cos²t sin t; product + chain gives −2 cos t sin²t + cos³t — same ✓.
At t = π/4 (cos = sin = √2/2): −2·(√2/2)·(1/2) + (√2/2)³ = −√2/2 + √2/4 = −√2/4 ≈ −0.354.

6 · Implicit differentiation via partials §14.4

Curve given as F(x, y) = 0, need dy/dx: wherever F_y ≠ 0,

dydx = − F_xF_y

(Derivation: chain rule on F(x, y(x)) = 0 — in the lesson.) Routine: move everything to one side to make F → compute the two partials → ratio with the minus sign → substitute the point after checking it lies on the curve.

Worked example 3Slope of the folium x³ + y³ = 6xy at (3, 3)

On the curve? 27 + 27 = 54 = 6·3·3 ✓.
F = x³ + y³ − 6xy: F_x = 3x² − 6y, F_y = 3y² − 6x.
dy/dx = −(3x² − 6y)/(3y² − 6x) = −(x² − 2y)/(y² − 2x).
At (3, 3): −(9 − 6)/(9 − 6) = −1. Geometry check: the folium is symmetric about y = x, so at a point on that mirror line the tangent must be perpendicular to it — slope −1 ✓.

Classic traps · the module's five mark-losers

1) d for ∂ (or the reverse) — symbols are graded. 2) The frozen letter thaws: in the chain rule every moving input gets a term; count routes first. 3) Substituting the point before differentiating — differentiate first, evaluate last. 4) Forgetting the minus sign in −F_x/F_y (and mixing up which partial goes on top: x on top, y below — the opposite of what dy/dx's letters suggest, which is exactly why it's a trap). 5) Mixed partials that disagree: by Clairaut they must match for these functions — a mismatch means a computational slip, find it.

Your minimal prerequisite kit for this module

One-variable differentiation, fluent: power, product, quotient, chain; sin, cos, e^u, ln u. This module is 80% that — the daily drill is the patch.
12th-class implicit differentiation (for recognising what −F_x/F_y replaces).
Module 3's level-curve picture and the xy/(x²+y²) two-path fact (used verbatim in section 4).
Exact values of sin, cos at the standard angles (evaluation questions love π/4 and π/6) — unit-circle flashcards.

What to practise in Thomas

Ranges are approximate for the 15th edition — check against your copy. Prefer odd numbers (answers in the back).

Skill	Where	How many
First partials, two and three variables	§14.3 exercises ~1–40	10–12 odd, until fast
Second partials + verifying f_xy = f_yx	§14.3 exercises ~41–56	4–5 odd
Chain rule, one independent variable	§14.4 exercises ~1–12	4–6 odd, always double-checking by substitution
Chain rule, two independent variables / diagrams	§14.4 exercises ~13–24	3–4 odd
Implicit differentiation via −F_x/F_y	§14.4 exercises ~25–34	3–4 odd