Partial Derivatives and the Chain Rule, taught from zero

1 · A surface has no single slope — so pick a direction §14.3

For a one-variable function, "the derivative at a point" made sense because a curve has one slope at each point. A surface z = f(x, y) doesn't. Stand on a hillside: facing east the ground may climb steeply, facing north it may be level, facing north-east something in between. "The slope of the hill" is not a number — the slope of the hill in a chosen direction is.

§14.3 picks the two simplest directions, the ones along the axes:

• Walk east (increase x, hold y fixed): the rate the ground rises is the partial derivative of f with respect to x, written ∂f/∂x or f_x.
• Walk north (increase y, hold x fixed): that rate is ∂f/∂y, or f_y.

Formally each one is the old limit definition with the other variable frozen:

Notice y₀ never moves in that formula — that is the entire content of the word "partial". The symbol ∂ (a curled d, read "partial") exists purely to flag this: other variables are being held still. Connect it to Module 3: walking east across the contour map, f_x is how fast you cross the level curves.

2 · Computing them: the old rules, one letter frozen

The recipe: to find f_x, differentiate with respect to x exactly as in one-variable calculus, treating every y as if it were a number like 7. And symmetrically for f_y. Watch it once, slowly, on

For f_x (y is "7"): the term x²y is "x² times a constant", derivative 2xy. The term sin(xy) is sin of ("constant" times x) — chain rule, inner derivative y — giving y cos(xy). So:

For f_y (now x is "7"): x²y is "constant times y", derivative x²; and sin(xy) gives x cos(xy):

One with a quotient, because exams like them: g(x, y) = xyx + y. Quotient rule in x (y frozen): g_x = y(x+y) − xy·1(x+y)² = y²(x+y)², and by the mirror argument g_y = x²/(x+y)².

3 · Seeing it: slice the surface

Where does "freeze y" live geometrically? Cut the surface with the vertical plane y = y₀. The cut edge — the trace — is an ordinary one-variable curve, and f_x(x₀, y₀) is its ordinary slope at x₀. The widget slices the dome z = 9 − x² − y²: slide y₀ and the whole trace drops (a slice further north cuts a lower part of the dome); slide x₀ and watch the tangent tilt — slope f_x = −2x₀, positive on the left of the dome, zero at the crest, negative on the right.

Two readings of the widget worth saying aloud: the tangent's slope never depends on y₀ (for this surface, f_x = −2x has no y in it — moving the slice slides the parabola down without reshaping it), and at the crest of every slice the tangent is level, which is where Module 6's max/min story will begin.

4 · Second partials — and a small miracle §14.3

Partial derivatives are functions of (x, y) themselves, so differentiate again — four second partials:

The miracle: compute the two mixed partials for section 2's f = x²y + sin(xy). Differentiate f_x = 2xy + y cos(xy) with respect to y (product rule on the second term): 2x + cos(xy) − xy sin(xy). Now go the other way: differentiate f_y = x² + x cos(xy) with respect to x: 2x + cos(xy) − xy sin(xy). Identical. That is the Mixed Derivative (Clairaut) Theorem: if f_xy and f_yx are continuous, then f_xy = f_yx — the order of differentiation doesn't matter. Every function this course throws at you meets the continuity condition, so in practice: compute whichever order is easier, and use the equality as a free error-check — if your two mixed partials disagree, one computation is wrong.

5 · "Differentiable" means more than "the partials exist" §14.3

In one variable, having a derivative at a point was the gold standard of smoothness. In two variables the bar sits higher, and the handout's word "differentiability" (lecture L9) is about exactly this. Differentiable at a point means: near that point the surface is well approximated by a flat plane — zoom in far enough and the landscape looks like a tilted sheet (that sheet is the tangent plane, which Module 5 will build). The two partials are just the plane's east-slope and north-slope.

Here is the surprise: the two partials can exist while the surface is nothing like a plane there. The standard example is Module 3's troublemaker f(x, y) = xy/(x² + y²) with f(0, 0) = 0. Along the entire x-axis the function is 0 (numerator has a factor y), so walking east the ground is flat: f_x(0, 0) = 0. Same north: f_y(0, 0) = 0. Both partials exist — yet the two-path test showed this function isn't even continuous at the origin (along y = x it sits at ½). Two flat directions, and a cliff between them. Partials only sample two directions; differentiability is a promise about all of them.

What rescues everyday computations is the theorem you'll actually cite:

Polynomials, sin, cos, e^x and their sane combinations have continuous partials everywhere they're defined — so differentiability holds automatically for the course's standard functions, and differentiable ⟹ continuous (no differentiable function has a cliff). That's the honest, examable core of L9's "differentiability"; if your instructor pushes deeper (the increment definition, ε₁/ε₂ error terms), confirm the required depth with them — the handout line names the topic without fixing it.

6 · The chain rule: when the inputs move together §14.4

Motivation first. Temperature over a map is T(x, y); you drive along a road, position (x(t), y(t)). The temperature you feel depends on one variable — time — and its rate of change ought to combine "how temperature varies over the ground" with "how fast you're moving". It does, one term per input:

Read it as travel bookkeeping: your eastward drift changes the temperature at rate (east-slope) × (east-speed); your northward drift adds (north-slope) × (north-speed); total rate = sum. Note the symbols carefully — ∂ on the slopes (T has two inputs) and d on the speeds and on the left (each is a function of t alone). Getting the symbols right is understanding the formula.

The diagram that organises everything. Draw z at the top, its direct inputs x, y below, and t at the bottom, with a line for each dependence. The rule: one product per route from z down to t, multiply the derivatives along the route, add the routes. Two routes here (z→x→t, z→y→t) — two terms. Every §14.4 formula, however many variables, is this one sentence applied to a diagram.

Worked, verified both ways. Let z = x² + y² (the bowl) with x = cos t, y = sin t (walking the unit circle). Chain rule:

Zero — and it must be: substituting directly, z = cos²t + sin²t = 1, a constant. Walking a circle around the bowl keeps your altitude fixed; the chain rule knows it. When both routes are available (substitute-then-differentiate vs chain rule), they always agree — on an exam, use whichever is faster and the other as a check.

7 · More routes: two independent variables, and implicit curves §14.4

Two independent variables. Suppose z = f(x, y) where x = x(s, t), y = y(s, t). Nothing new — same diagram, now with s and t on the bottom row. For ∂z/∂t, trace only the routes ending at t:

(all ∂ now — every quantity has two inputs). The skill being examined is diagram discipline: draw it, count routes, one product per route.

Implicit differentiation, upgraded. In 12th class you differentiated x² + y² = 25 term by term, wrestling y′ out by algebra. §14.4 turns that into a formula. Write the curve as F(x, y) = 0 (here F = x² + y² − 25). Then wherever F_y ≠ 0:

For the circle: F_x = 2x, F_y = 2y, so dy/dx = −x/y — at (3, 4), slope −3/4. The old method (2x + 2y y′ = 0) gives the same, but the formula wins on speed the moment the equation is ugly: two partial derivatives, one ratio, done. The notes page runs it on the folium x³ + y³ = 6xy.

8 · You're ready — what to do next

The whole module in four sentences: a partial derivative is an old derivative with the other letter frozen; mixed partials match (free error-check); "differentiable" is a promise about every direction, guaranteed by continuous partials; and when inputs move together, one product per route, summed. To turn it into marks:

• Drill the mechanics — the notes page's exercise table maps §14.3–14.4 (odd numbers, answers in the back). Aim for partials computed as fast as ordinary derivatives.
• Rehearse the three exam patterns in the notes' worked examples: all four second partials, a full chain-rule with numeric check, an implicit slope via −F_x/F_y.
• Revise from the notes page, not this one — before the quiz you want the compressed map.