Directional Derivatives Study Notes
Institution: California State University, Northridge（Northridge, CA）
Course: Math 250（Vector Calculus）
Instructor: David Klein
Instructor Time: Last Tuesday
I. Course Introduction: The Core Significance of Directional Derivatives
In single-variable calculus, the derivative f(x) describes the
instantaneous rate of change of the function along the x axis. However,
for multivariable functions z = f (x, y) or w = f (x, y, z), only partial
derivatives ∂f/∂x and ∂f/∂y can reflect the rate of change along the
coordinate axes, which cannot meet the need for the rate of change in
any direction in practical problems.
Professor David Klein emphasized in class that directional derivatives
(DirectionalDerivatives) are a natural extension of partial derivatives.
They can precisely describe the trend of change of multivariable
functions in any specified direction in space, serving as the foundation
for subsequent learning of gradients, optimization problems, and vector
field-related theorems, with widespread applications in fields such as
physics, engineering, and economics. For example, in physics, they can
be used to calculate the rate of change of velocity in the direction of
force, and in engineering, they can analyze the slope change of a
surface in a specific direction.
II. Strict Definition of Directional Derivatives
2.1 Definition of Directional Derivatives for Binary Functions
Let the function z = f (x, y) be defined in a neighborhood of the point
P₀(a, b) and let the vector u = (cosθ, sinθ) be a unit vector in the plane
(where θ is the angle between the vector u and the positive direction of
the x axis). If the limit:
Dᵤf (a, b) = lim (h→0) [f (a + h・cosθ, b + h・sinθ) - f (a, b)] /h
exists, then this limit is called the directional derivative of the function f
(x, y) at the point P₀(a, b) in the direction of the vector u .
2.2 Directional derivative of a ternary function
, Extending to a ternary function w = f (x, y, z), let the vector u = (cosα,
cosβ, cosγ) be a unit vector in space (where α, β, γ are the angles
between the vector u and the positive directions of the x, y, z axes,
known as direction cosines). Then the directional derivative of the
function at the point P₀(x₀, y₀, z₀) in the direction of u is:
Dᵤf (x₀, y₀, z₀) = lim (h→0) [f (x₀ + h・cosα, y₀ + h・cosβ, z₀ + h・cosγ) - f
(x₀, y₀, z₀)] /h
2.3 Geometric Meaning of the Definition
Professor David Klein intuitively explains through three-dimensional
surface graphs: For a binary function z = f (x, y), its graph is a surface in
space. Through the point P₀(a, b, f (a, b)) draw a plane parallel to the
direction u and perpendicular to the xy plane. The intersection line of
this plane with the surface is a space curve. The directional derivative
Dᵤf (a, b) is the slope of the tangent line to this curve at the point P₀ .
2.4 Key Considerations in the Definition
Direction vector u must be a unit vector: If the given direction vector is
not a unit vector, it must first be normalized (divided by the magnitude
of the vector).
Existence of Directional Derivatives: The existence of a directional
derivative of a function at a point in a certain direction does not imply
that the function is differentiable at that point, nor does it imply the
existence of partial derivatives; however, conversely, if the function is
differentiable at that point, then the directional derivative in any
direction must exist (a subsequent theorem will prove this).
Partial derivatives are a special case of directional derivatives:
When u = (1, 0) (positive x-axis direction), Dᵤf = ∂f/∂x；
When u = (0, 1) (positive y-axis direction), Dᵤf = ∂f/∂y。
III. Methods for Calculating Directional Derivatives
3.1 Direct Calculation Using the Definition (Basic Method)
Example 1: Let the function f (x, y) = x² + xy, find the f (x, y) directional
derivative at the point (1, 2) in the direction u = (cos (π/3), sin (π/3)) =
(1/2, √3/2) direction.
Solution: According to the definition of the directional derivative:
Dᵤf (1, 2) = lim (h→0) [f (1 + h/2, 2 + (√3 h)/2) - f (1, 2)] /h
Step 1: Calculate f (1 + h/2, 2 + (√3 h)/2)：
f (1 + h/2, 2 + (√3 h)/2) = (1 + h/2)² + (1 + h/2)(2 + (√3 h)/2)
Institution: California State University, Northridge（Northridge, CA）
Course: Math 250（Vector Calculus）
Instructor: David Klein
Instructor Time: Last Tuesday
I. Course Introduction: The Core Significance of Directional Derivatives
In single-variable calculus, the derivative f(x) describes the
instantaneous rate of change of the function along the x axis. However,
for multivariable functions z = f (x, y) or w = f (x, y, z), only partial
derivatives ∂f/∂x and ∂f/∂y can reflect the rate of change along the
coordinate axes, which cannot meet the need for the rate of change in
any direction in practical problems.
Professor David Klein emphasized in class that directional derivatives
(DirectionalDerivatives) are a natural extension of partial derivatives.
They can precisely describe the trend of change of multivariable
functions in any specified direction in space, serving as the foundation
for subsequent learning of gradients, optimization problems, and vector
field-related theorems, with widespread applications in fields such as
physics, engineering, and economics. For example, in physics, they can
be used to calculate the rate of change of velocity in the direction of
force, and in engineering, they can analyze the slope change of a
surface in a specific direction.
II. Strict Definition of Directional Derivatives
2.1 Definition of Directional Derivatives for Binary Functions
Let the function z = f (x, y) be defined in a neighborhood of the point
P₀(a, b) and let the vector u = (cosθ, sinθ) be a unit vector in the plane
(where θ is the angle between the vector u and the positive direction of
the x axis). If the limit:
Dᵤf (a, b) = lim (h→0) [f (a + h・cosθ, b + h・sinθ) - f (a, b)] /h
exists, then this limit is called the directional derivative of the function f
(x, y) at the point P₀(a, b) in the direction of the vector u .
2.2 Directional derivative of a ternary function
, Extending to a ternary function w = f (x, y, z), let the vector u = (cosα,
cosβ, cosγ) be a unit vector in space (where α, β, γ are the angles
between the vector u and the positive directions of the x, y, z axes,
known as direction cosines). Then the directional derivative of the
function at the point P₀(x₀, y₀, z₀) in the direction of u is:
Dᵤf (x₀, y₀, z₀) = lim (h→0) [f (x₀ + h・cosα, y₀ + h・cosβ, z₀ + h・cosγ) - f
(x₀, y₀, z₀)] /h
2.3 Geometric Meaning of the Definition
Professor David Klein intuitively explains through three-dimensional
surface graphs: For a binary function z = f (x, y), its graph is a surface in
space. Through the point P₀(a, b, f (a, b)) draw a plane parallel to the
direction u and perpendicular to the xy plane. The intersection line of
this plane with the surface is a space curve. The directional derivative
Dᵤf (a, b) is the slope of the tangent line to this curve at the point P₀ .
2.4 Key Considerations in the Definition
Direction vector u must be a unit vector: If the given direction vector is
not a unit vector, it must first be normalized (divided by the magnitude
of the vector).
Existence of Directional Derivatives: The existence of a directional
derivative of a function at a point in a certain direction does not imply
that the function is differentiable at that point, nor does it imply the
existence of partial derivatives; however, conversely, if the function is
differentiable at that point, then the directional derivative in any
direction must exist (a subsequent theorem will prove this).
Partial derivatives are a special case of directional derivatives:
When u = (1, 0) (positive x-axis direction), Dᵤf = ∂f/∂x；
When u = (0, 1) (positive y-axis direction), Dᵤf = ∂f/∂y。
III. Methods for Calculating Directional Derivatives
3.1 Direct Calculation Using the Definition (Basic Method)
Example 1: Let the function f (x, y) = x² + xy, find the f (x, y) directional
derivative at the point (1, 2) in the direction u = (cos (π/3), sin (π/3)) =
(1/2, √3/2) direction.
Solution: According to the definition of the directional derivative:
Dᵤf (1, 2) = lim (h→0) [f (1 + h/2, 2 + (√3 h)/2) - f (1, 2)] /h
Step 1: Calculate f (1 + h/2, 2 + (√3 h)/2)：
f (1 + h/2, 2 + (√3 h)/2) = (1 + h/2)² + (1 + h/2)(2 + (√3 h)/2)
