Partial Derivatives: one knob at a time

The function below has two inputs ($x$ and $y$). The two panels show what happens when you freeze one and slide the other.

Drag the $x$ slider with $y$ frozen. The left panel’s curve is the slice of $f$ at the current $y$: a one-variable function. It has a slope. That slope is a perfectly ordinary derivative, except its name has the word “partial” in front.

Same trick on the right, with $x$ frozen. Same rule.

That’s the whole concept. Calculus already had a tool for one-variable slopes; partial derivatives just say “freeze everything you’re not currently asking about, then use the tool you already have.”

A function of two variables lives in a 3D picture: height $z = f(x, y)$ is a surface sitting over the $(x, y)$ plane. Pretty, but bad at high dimensions.

A more useful picture is the contour view: a topographic map. Draw curves of constant height $f(x, y) = c$ for a few values of $c$. Tighter contours = steeper ground. We’ll spend most of our time here, because contours collapse the height axis and leave a 2D picture in input space, where the action is. (Neural-network losses have millions of inputs and one output. There’s no surface to draw. Contours are the only picture that scales.)

Three spellings of the same thing:

• $\dfrac{\partial f}{\partial x}$: explicit, textbook-y, the one you’ll see most often. The curly $\partial$ (“partial”) distinguishes it from $d/dx$.
• $f_x$: terse subscript form. The subscript is the variable being differentiated.
• $\partial_x f$ or $\partial_i f$: compact, common in physics and ML matrix-calculus writeups.

All three mean “derivative of $f$ with everything except $x$ held constant.” Pick whichever is clearest in context. We’ll lean on $\dfrac{\partial f}{\partial x}$ for definitions and $f_x$ for quick calculation.

To compute $\dfrac{\partial f}{\partial x}$ for $f(x, y) = x^2 y + 3 y^2$:

• Treat $y$ as a constant. $y$ is a block of lead sitting on the table.
• Differentiate as if $y$ were just, say, the number 7. $\dfrac{\partial}{\partial x}(x^2 \cdot y) = 2 x y$. And $3 y^2$ doesn’t contain $x$, so it’s constant with respect to $x$: its partial is $0$.
• Answer: $\dfrac{\partial f}{\partial x} = 2 x y$.

To get $\dfrac{\partial f}{\partial y}$: swap roles. Freeze $x$, differentiate the $y$-parts. $\dfrac{\partial}{\partial y}(x^2 y) = x^2$. $\dfrac{\partial}{\partial y}(3 y^2) = 6 y$. Sum: $\dfrac{\partial f}{\partial y} = x^2 + 6 y$.

No new machinery. The power rule, the sum rule, and the constant-multiple rule all still apply. The only new habit is noticing which symbol is the variable you’re currently paying attention to.

Take the partial of a partial. $\dfrac{\partial}{\partial y}\left(\dfrac{\partial f}{\partial x}\right)$: differentiate $f$ with respect to $x$ first, then with respect to $y$. Call that $f_{xy}$.

You could also go the other way: differentiate with respect to $y$ first, then $x$. Call that $f_{yx}$.

Common sense says those might differ: different order of operations, different answers. Common sense is wrong. For any reasonably smooth function, they’re equal. This is Clairaut’s theorem (sometimes called Schwarz’s theorem):

It takes a real proof to establish but is easy to verify on examples. Verify on one.