Derivatives of Vector-Valued Functions
Outline
1. Components
Consider a function general vector-valued function f : Rm → Rn . Such a function can be
written as
f(x1 , . . . , xm ) = f1 (x1 , . . . , xm ), . . . , fn (x1 , . . . , xm ) ,
where each fi : Rm → R. The real-valued functions f1 , . . . , fn are called the components
of f.
For example, if f : R2 → R3 is the function
f(x, y) = (x2 + y 2 , x2 − y 2 , 2xy)
then f has components f1 (x, y) = x2 + y 2 , f2 (x, y) = x2 − y 2 , and f3 (x, y) = 2xy.
2. Partial Derivatives
If f : Rm → Rn , the partial derivative of f with respect to xi is the vector
∂f1
∂fn
∂f
=
, ...,
.
∂xi
∂xi
∂xi
We sometimes use subscripts to denote partial derivatives. For example, if f(x, y, z) is vectorvalued function on R3 , then fy would denote the partial derivative of f with respect to y.
3. Parameter Curves
Given a function f : Rm → Rn , a parameter curve for f is a curve γ : R → Rn of the form
γ(t) = f(p + tei )
= f(p1 , . . . , pi−1 , pi + t, pi+1 , . . . , pm ).
where p is a point in Rm , and ei is a unit vector in the xi direction. Note that γ is obtained
obtained from f(x1 , . . . , xm ) by varying xi and holding the other x’s constant. Thus parameter
curves can be thought of as the images of the “gridlines” under the function f.
The partial derivatives of a function f at a point p can be interpreted as the tangent
vectors to the parameter curves through f(p). Specifically, if γ is the parameter curve defined
above, then
∂f
(p).
γ̇(0) =
∂xi
4. The Derivative Matrix
If f : Rm → Rn the derivative of f at a point p

∂f1
(p)
 ∂x1
 .
.
Dp f = 
 .
 ∂f
n
(p)
∂x1
is the matrix
···
...
···

∂f1
(p)

∂xm

..

.


∂fn
(p)
∂xm
∂f
∂f
(p), . . . ,
(p).
∂x1
∂xm
For example, if f : R2 → R3 is the function f(x, y) = (x2 + y 2 , x2 − y 2 , 2xy), then
That is, Dp f is the matrix whose columns are the partial derivatives

2x 2y
D(x,y) f =  2x −2y 
2y 2x

Note that the derivative is actually a function that takes a point (x, y) as input and outputs
a matrix of numbers.
In some books, the derivative Dp f is denoted Dfp , Df(p), dfp , or f0 (p). In addition, the
derivative is sometimes referred to as the Jacobian, in which case it may be denoted with a
J instead of a D. We will use the notation Dp f to mean the derivative of f at p, and Dp f(v)
to mean the product of the matrix Dp f with a vector v.
5. Special Cases of the Derivative
For a parametric curve γ : R → Rn , the derivative of γ is the same as the tangent vector γ̇:


γ̇1 (t)


Dt γ = γ̇(t) =  ...  .
γ̇n (t)
For a real-valued function f : Rm → R, the derivative Dp f is the transpose of the gradient
vector ∇f :
T
∂f
(p)
 ∂x1



∂f
..

 = ∇f (p)T .
(p) = 
.

∂xm
 ∂f

(p)
∂xm

Dp f =
∂f
(p) · · ·
∂x1
6. Directional Derivatives
Let f : Rm → Rn be a differentiable function, and let ei be a unit vector in the xi direction.
Then:
∂f
Dp f(ei ) =
(p).
∂xi
That is, the i’th column of Dp f is the partial derivative on the right.
More generally, if v is any vector in Rm , then the product
Dp f(v)
is called the directional derivative of f in the direction of v. This is something like a
“partial derivative” in the direction of the vector v.
The directional derivative Dp (v) can be interpreted as a tangent vector to a certain parametric curve. Specifically, let γ : R → Rn be the curve
γ(t) = f(p + tv).
That is, γ is the image under f of a straight line in the direction of v. Then
γ̇(0) = Dp f(v).
7. Differentials
The derivative of a function f : Rm → Rn can also be thought of in terms of differentials.
Specifically, let p be a point in Rm , with corresponding value f(p). Now, suppose we move
from p to a nearby point p + dp, and let df denote the corresponding change in the value of f
df = f(p + dp) − f(p).
Then the vectors df and dp are related by the formula
df ≈ Dp f(dp).
That is, the change in f is roughly the product of the matrix Dp f with the vector dp. Note
that, since we cannot divide vectors, we cannot interpret Dp f as the “ratio” of df to dp.
8. The Chain Rule
Suppose we have a pair of differentiable functions f : Rm → Rn and g : Rk → Rm . Since the
codomain of g is the same as the domain of f, we can form the composition f ◦ g : Rk → Rn .
In this case, the Chain Rule gives us a formula for the derivative of f ◦ g. According to
the rule, if p ∈ Rk and q = g(p), then
Dp (f ◦ g) = (Dq f)(Dp g).
That is, the derivative matrix for f ◦ g at p is the product of the derivative matrix for f at
q and the derivative matrix for g at p. This is simply a matrix form of the Chain Rule for
partial derivatives.
As a special case, let f : Rm → Rn , let γ : R → Rm is a parametric curve in Rm , and let
p = γ(0). Then the composition γ̃ = f ◦ γ is a parametric curve in Rn , and
˙
γ̃(0)
= Dp f γ̇(0)
That is, the tangent vector to γ̃ is the product of the derivative matrix Dp f with the tangent
vector to γ.
Practice Problems
1. Let f : R2 → R2 be the polar coordinates transformation f(r, θ) = (r cos θ, r sin θ).
(a) Make a drawing showing the parameter curves θ = C for C ∈ {0, π/4, π/2, 3π/4, π}, as
well as the curves r = C for C ∈ {1, 2, 3}.
(b) Compute D(2,3π/4) f. Add the the vectors fr (2, 3π/4) and fθ (2, 3π/4) to your drawing as
tangent vectors to parameter curves.
(c) Let γ : R → R2 be a regular curve, and suppose that γ(0) = (2, 3π/4) and γ̇(0) = (3, 1).
Find the tangent vector to the curve f ◦ γ at the point (f ◦ γ)(0).
2. Let f : R2 → R3 be a differentiable function, and suppose that f(3, 9) = (5, 3, 1) and


2 1
D(3,9) f = 3 5 .
1 2
(a) Estimate f(3.02, 9.05).
(b) Compute the directional derivative of f in the direction of the vector (1, 1).
(c) Let γ : R → R3 be the curve γ(t) = f(t, t2 ). Compute γ̇(3).
3. Let f : R2 → R3 be the map f (θ, φ) = (cos θ cos φ, sin θ cos φ, sin φ).
(a) Compute the matrix D(θ,φ) f.
b be a unit tangent vector pointing in the direction of
(b) At each point on the unit sphere, let θ
b be a unit tangent vector pointing in the direction of increasing φ.
increasing θ, and let φ
b and φ.
b
Use your answer to part (a) to find formulas for θ
4. Let ψ : R3 → R, and suppose that ψ(1, 2, 5) = 3 and ∇ψ(1, 2, 5) = (1, 3, 2). Let γ : R → R3
be a regular curve, and suppose that γ(3) = (1, 2, 5) and γ̇(3) = (4, 5, 2).
(a) Let f : R → R be the function f (t) = ψ(γ(t)). Compute f 0 (3).
(b) Let g : R3 → R3 be the function g(x, y, z) = γ(ψ(x, y, z)). Compute D(1,2,5) g.