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Partial Derivatives of Functions of Three or More Variables

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Core Definition

For a scalar-valued function of three variables w=f(x,y,z)w = f(x, y, z) defined on a domain D⊂R3D \subset \mathbb{R}^3, the first-order partial derivatives are defined by holding two variables constant while taking the standard single-variable derivative with respect to the third:

∂f∂x(x,y,z)=fx=lim⁡h→0f(x+h,y,z)−f(x,y,z)h\frac{\partial f}{\partial x}(x, y, z) = f_x = \lim_{h \to 0} \frac{f(x+h, y, z) - f(x, y, z)}{h}
∂f∂y(x,y,z)=fy=lim⁡h→0f(x,y+h,z)−f(x,y,z)h\frac{\partial f}{\partial y}(x, y, z) = f_y = \lim_{h \to 0} \frac{f(x, y+h, z) - f(x, y, z)}{h}
∂f∂z(x,y,z)=fz=lim⁡h→0f(x,y,z+h)−f(x,y,z)h\frac{\partial f}{\partial z}(x, y, z) = f_z = \lim_{h \to 0} \frac{f(x, y, z+h) - f(x, y, z)}{h}
More generally, for f(x1,x2,…,xn)f(x_1, x_2, \dots, x_n), ∂f∂xi\frac{\partial f}{\partial x_i} is computed by treating all xjx_j (j≠ij \ne i) as constants.

Concept Explanation

The Operational Rule in Higher Dimensions

When extending from two variables to three or more:

  • To find ∂f∂x\frac{\partial f}{\partial x}: Treat both yy and zz as constant numbers.
  • To find ∂f∂y\frac{\partial f}{\partial y}: Treat both xx and zz as constant numbers.
  • To find ∂f∂z\frac{\partial f}{\partial z}: Treat both xx and yy as constant numbers.
3D Spatial Domain (x, y, z) in R^3
   Step along x (parallel to x-axis, y and z held fixed) ===> Rate of change = f_x
   Step along y (parallel to y-axis, x and z held fixed) ===> Rate of change = f_y
   Step along z (parallel to z-axis, x and y held fixed) ===> Rate of change = f_z

Physical Interpretation in 3D Scalar Fields

Unlike functions of two variables whose graphs form 2D surfaces in 3D space, the graph of w=f(x,y,z)w = f(x, y, z) exists in 4D space R4\mathbb{R}^4, making direct graphical visualization impossible. Instead, engineers visualize f(x,y,z)f(x, y, z) as a scalar field distributed throughout 3D space:

  • T(x,y,z)T(x, y, z): Temperature distribution inside an engine combustion chamber.
  • P(x,y,z)P(x, y, z): Acoustic pressure field inside an auditorium.
  • ϕ(x,y,z)\phi(x, y, z): Electrostatic potential between capacitor plates.

In this context, the partial derivatives represent the directional spatial rates of change:

  • ∂T∂x\frac{\partial T}{\partial x}: Temperature change per unit distance moved east (xx-axis).
  • ∂T∂y\frac{\partial T}{\partial y}: Temperature change per unit distance moved north (yy-axis).
  • ∂T∂z\frac{\partial T}{\partial z}: Temperature change per unit distance moved upward (zz-axis).

Together, these three components form the gradient vector field:

∇f=⟨∂f∂x,∂f∂y,∂f∂z⟩\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle
which points in the direction of maximum spatial increase and is orthogonal to the 3D Level Surfaces f(x,y,z)=cf(x, y, z) = c.

Engineering Applications

  • Combustion & Chemical Process Engineering: In rocket combustion chambers, spatial gradients of gas species concentrations ∂C∂x,∂C∂y,∂C∂z\frac{\partial C}{\partial x}, \frac{\partial C}{\partial y}, \frac{\partial C}{\partial z} govern molecular diffusion via Fick's laws.
  • Biomedical Imaging & Radiation Therapy: 3D radiation dose distributions D(x,y,z)D(x, y, z) require evaluating partial derivatives across tumor boundaries to spare adjacent critical organs.
  • Electromagnetics & RF Engineering: Maxwell's equations evaluate spatial field derivatives (∂x,∂y,∂z)(\partial_x, \partial_y, \partial_z) of electric and magnetic fields to predict waveguide modes and antenna radiation patterns.

Example Problems

Example 1: Three-Variable Logarithmic Function with Radial Argument

Compute ∂f∂x\frac{\partial f}{\partial x}, ∂f∂y\frac{\partial f}{\partial y}, and ∂f∂z\frac{\partial f}{\partial z} for:

f(x,y,z)=ln⁡(x2+y2+z2)f(x, y, z) = \ln(x^2 + y^2 + z^2)

Solution

  1. Differentiate with respect to xx:

    ∂f∂x=1x2+y2+z2⋅∂∂x(x2+y2+z2)\frac{\partial f}{\partial x} = \frac{1}{x^2+y^2+z^2} \cdot \frac{\partial}{\partial x}(x^2+y^2+z^2)
    Since yy and zz are constants, ∂∂x(x2+y2+z2)=2x+0+0=2x\frac{\partial}{\partial x}(x^2+y^2+z^2) = 2x + 0 + 0 = 2x:
    ∂f∂x=2xx2+y2+z2\frac{\partial f}{\partial x} = \frac{2x}{x^2+y^2+z^2}

  2. Differentiate with respect to yy and zz by symmetry:

    ∂f∂y=2yx2+y2+z2\frac{\partial f}{\partial y} = \frac{2y}{x^2+y^2+z^2}
    ∂f∂z=2zx2+y2+z2\frac{\partial f}{\partial z} = \frac{2z}{x^2+y^2+z^2}

Connections & References