什么是 partial derivative？

Example 25.10 (Level curves and gradients, on one function)

Take $f(x, y) = x^2 - y^2$. Its level sets: $\{f = c\}$ is a hyperbola opening left-right for $c > 0$, up-down for $c < 0$, and the crossed pair of lines $y = \pm x$ for $c = 0$ — the contour map of a mountain pass, with the saddle point at the origin where the two zero-level lines cross. Gradient: $\nabla f = (2x, -2y)$. At the point $(2, 1)$ (on the level $c = 3$): $\nabla f = (4, -2)$, while the tangent vector of the level curve, parametrized near that point by $\bigl(t, \sqrt{t^2 - 3}\bigr)$, is $\bigl(1, \frac{t}{\sqrt{t^2 - 3}}\bigr) = (1, 2)$ at $t = 2$ — and indeed

gradient perpendicular to the contour, pointing toward higher values of $f$ (here: away from the $y$-axis). Two more readings: the gradient vanishes exactly at the saddle, where the contour map pinches; and the tangent line to the level curve at $(2,1)$ is $4(x - 2) - 2(y - 1) = 0$, i.e. $2x - y = 3$ — the equation “$\langle \nabla f,\ M - M_0\rangle = 0$” that generalizes the ellipse tangent of Exercise 24.11.

Example 25.16 (A four-point study, in full)

$f(x, y) = xy\,(3 - x - y) = 3xy - x^2y - xy^2$. Gradient:

Critical points: if $y = 0$, the second equation gives $x \in \{0, 3\}$; if $x = 0$, the first gives $y \in \{0, 3\}$; if $xy \neq 0$, solve $2x + y = 3$, $x + 2y = 3$: $x = y = 1$. Four points: $(0,0)$, $(3,0)$, $(0,3)$, $(1,1)$. Second derivatives: $r = -2y$, $s = 3 - 2x - 2y$, $t = -2x$.

• $(1,1)$: $r = -2$, $s = -1$, $t = -2$: $rt - s^2 = 3 > 0$, $r < 0$: local maximum, $f(1,1) = 1$.
• $(0,0)$: $r = t = 0$, $s = 3$: $rt - s^2 = -9 < 0$: saddle; likewise $(3, 0)$ ($s = -3$) and $(0, 3)$: three saddles.

The maximum is only local: $f(-T, -T) = T^2(3 + 2T) \to +\infty$. Symmetry check: $f(x, y) = f(y, x)$, and indeed the critical set and the classification are symmetric in $x \leftrightarrow y$. Interpretation: among rectangles-with-slack $x, y \geq 0$, $x + y \leq 3$, the product $xy(3 - x - y)$ of the three “parts” of $3$ is largest when the parts are equal — a two-variable shadow of the AM–GM principle.