Mathematics · Glossary

What is partial derivative?

Also known as: gradient

Definition 25.4 University Mathematics — Year 1 · Chapter 25 — Functions of Two Variables

The partial derivatives of ff at (a,b)(a, b) are the one-variable derivatives along the axes:

fx(a,b)=limh0f(a+h,b)f(a,b)h,fy(a,b)=limk0f(a,b+k)f(a,b)k.\frac{\partial f}{\partial x}(a,b) = \lim_{h \to 0} \frac{f(a + h, b) - f(a,b)}{h}, \qquad \frac{\partial f}{\partial y}(a,b) = \lim_{k \to 0} \frac{f(a, b + k) - f(a,b)}{k}.

ff is of class C1C^1 on UU when both exist and are continuous on UU. The gradient is f(a,b)=(fx,fy)(a,b)\nabla f(a,b) = \bigl(\frac{\partial f}{\partial x},\, \frac{\partial f}{\partial y}\bigr)(a,b).

Examples

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

Take f(x,y)=x2y2f(x, y) = x^2 - y^2. Its level sets: {f=c}\{f = c\} is a hyperbola opening left-right for c>0c > 0, up-down for c<0c < 0, and the crossed pair of lines y=±xy = \pm x for c=0c = 0 — the contour map of a mountain pass, with the saddle point at the origin where the two zero-level lines cross. Gradient: f=(2x,2y)\nabla f = (2x, -2y). At the point (2,1)(2, 1) (on the level c=3c = 3): f=(4,2)\nabla f = (4, -2), while the tangent vector of the level curve, parametrized near that point by (t,t23)\bigl(t, \sqrt{t^2 - 3}\bigr), is (1,tt23)=(1,2)\bigl(1, \frac{t}{\sqrt{t^2 - 3}}\bigr) = (1, 2) at t=2t = 2 — and indeed

(4,2), (1,2)=44=0:\langle (4, -2),\ (1, 2)\rangle = 4 - 4 = 0 :

gradient perpendicular to the contour, pointing toward higher values of ff (here: away from the yy-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)(2,1) is 4(x2)2(y1)=04(x - 2) - 2(y - 1) = 0, i.e. 2xy=32x - y = 3 — the equation “f, MM0=0\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(3xy)=3xyx2yxy2f(x, y) = xy\,(3 - x - y) = 3xy - x^2y - xy^2. Gradient:

fx=y(32xy),fy=x(3x2y).\frac{\partial f}{\partial x} = y\,(3 - 2x - y), \qquad \frac{\partial f}{\partial y} = x\,(3 - x - 2y).

Critical points: if y=0y = 0, the second equation gives x{0,3}x \in \{0, 3\}; if x=0x = 0, the first gives y{0,3}y \in \{0, 3\}; if xy0xy \neq 0, solve 2x+y=32x + y = 3, x+2y=3x + 2y = 3: x=y=1x = y = 1. Four points: (0,0)(0,0), (3,0)(3,0), (0,3)(0,3), (1,1)(1,1). Second derivatives: r=2yr = -2y, s=32x2ys = 3 - 2x - 2y, t=2xt = -2x.

  • (1,1)(1,1): r=2r = -2, s=1s = -1, t=2t = -2: rts2=3>0rt - s^2 = 3 > 0, r<0r < 0: local maximum, f(1,1)=1f(1,1) = 1.
  • (0,0)(0,0): r=t=0r = t = 0, s=3s = 3: rts2=9<0rt - s^2 = -9 < 0: saddle; likewise (3,0)(3, 0) (s=3s = -3) and (0,3)(0, 3): three saddles.

The maximum is only local: f(T,T)=T2(3+2T)+f(-T, -T) = T^2(3 + 2T) \to +\infty. Symmetry check: f(x,y)=f(y,x)f(x, y) = f(y, x), and indeed the critical set and the classification are symmetric in xyx \leftrightarrow y. Interpretation: among rectangles-with-slack x,y0x, y \geq 0, x+y3x + y \leq 3, the product xy(3xy)xy(3 - x - y) of the three “parts” of 33 is largest when the parts are equal — a two-variable shadow of the AM–GM principle.

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