Partial Derivatives

Differentiate a multivariable function with respect to one variable while holding the others constant.

The idea. A partial derivative asks: how fast does f change if I nudge one variable and hold the rest still? That's it. Everything else is bookkeeping.

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Limit definition — the honest one, which you'll be asked to write once and then never use again:

f_x(a,b) = \lim_{h \to 0} \frac{f(a+h,\, b) - f(a,b)}{h}

Partial derivative, limit definition

Notice b never changes. You're walking in the x-direction only.

Notation. All of these mean the same thing, and your professor will use several without warning:

\frac{\partial f}{\partial x} \;=\; f_x \;=\; \partial_x f \;=\; \frac{\partial z}{\partial x}

Equivalent notations for the partial derivative

The ∂ is a "rounded d" — Thompson's a little bit of, but flagged to say other variables are being held frozen.

The working rule. To find f_x, treat y as a constant and differentiate normally. Every single-variable rule still applies — power, product, quotient, chain. A y sitting in a term is just a number that happens to be spelled with a letter.

Geometric meaning, and this is where it connects to your 3D viz: slice the surface z = f(x, y) with the vertical plane y = b. You get a curve. f_x(a, b) is the slope of that curve at x = a. Slicing with x = a instead gives you f_y. Same slicing move that makes Fubini visible.

Second-order partials. Differentiate twice:

f_{xx} = \frac{\partial^2 f}{\partial x^2}, \qquad f_{xy} = \frac{\partial^2 f}{\partial y\, \partial x}

Second-order partials

Read the order carefully — this is a real trap. f_xy means x first, then y: left to right. But ∂²f/∂y∂x means the same thing read right to left, inner to outer. The two notations run in opposite directions.

Clairaut's theorem. If f_xy and f_yx are both continuous near (a, b), then:

f_{xy}(a,b) = f_{yx}(a,b)

Clairaut's theorem

Mixed partials don't care about order. Same flavor as Fubini — a hypothesis of continuity buying you permission to swap. Worth pointing that parallel out; it's not a coincidence, and it's the kind of thing that makes a course cohere. (And like Fubini, it can fail without continuity. Rare, real, not on your exam.)

Worked example

f(x,y) = x^3y^2 - 4xy + 7y

$f_x = 3x^2y^2 - 4y$
f_x — hold y constant. In x^3y^2, the y^2 is a constant multiplier, so it survives untouched while x^3 becomes 3x^2. In -4xy, the -4y is a constant, and x becomes 1. The 7y has no x at all, so it's a constant, and its derivative is zero.
$f_y = 2x^3y - 4x + 7$
f_y — now hold x constant. In x^3y^2, the x^3 rides along while y^2 becomes 2y. In -4xy, the -4x is constant. And 7y becomes 7.
$f_{xy} = \dfrac{\partial}{\partial y}\left(3x^2y^2 - 4y\right) = 6x^2y - 4$
Mixed partials, both orders. Differentiate f_x with respect to y.
$f_{yx} = \dfrac{\partial}{\partial x}\left(2x^3y - 4x + 7\right) = 6x^2y - 4$
Differentiate f_y with respect to x. Equal — Clairaut, confirmed on a live example.

Answer:

f_{xy} = f_{yx} = 6x^2y - 4

How to solve one of these

1.
Identify which variable you're differentiating with respect to. The subscript or the $\partial$ denominator tells you. Everything else is frozen.
2.
Scan each term for the active variable. A term with no $x$ in it contributes nothing to $f_x$ — its derivative is zero.
3.
Pull constants out. Any factor made only of the frozen variables is a numeric coefficient. $y^2$ in $x^3y^2$ behaves exactly like a 5 would.
4.
Apply the ordinary single-variable rules to what's left. Power rule, product, chain — nothing new.
5.
Reassemble term by term. Don't try to do the whole expression at once.
6.
For higher-order partials, work inside-out. One differentiation at a time, and write down which variable you used at each step.
7.
Check with Clairaut's theorem. Compute $f_{xy}$ and $f_{yx}$ independently. If they disagree, you made an arithmetic error — the theorem holds for anything you'll see in this course.

\dfrac{\partial f}{\partial x}

How fast f changes as x alone nudges up, with every other variable frozen in place.

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