THE DERIVATIVE MAP IS THE MATRIX OF PARTIALS
Let f : Rm → Rn be a differentiable map, written in coordinates as
(x1 , x2 , . . . , xm ) 7→ (f1 (x1 , x2 , . . . , xm ), f2 (x1 , x2 , . . . , xm ), . . . , fn (x1 , x2 , . . . , xm )).
Let Df be the derivative of f at some point ~v in Rm . We have defined Df as an abstract linear
map Rm → Rn . Linear maps of finite dimensional vector spaces are given by matrices.
The point of this note is to prove the following result:
Theorem The matrix of Df is given by


∂f1
∂f1
∂f1
∂f1
·
·
·
∂xm 
 ∂x1 ∂x2 ∂x3
 ∂f2 ∂f2 ∂f2
∂f2 
 ∂x1 ∂x2 ∂x2 · · · ∂x

m

..
..
.. 
 ..
..
 .
.
.
.
. 


∂fn
∂f1
∂fn
∂fn
∂x1
∂x2
∂x3 · · ·
∂xm
Let A be the matrix of Df , meaning that
(Df )(0 0 · · · t · · · 0)T = (A1i A2i · · · Ani )T ,
where the 1 entry is in the i-th position. (The superscript T ’s are transposes, in order to turn rows
vectors into column vectors; just so that this equation isn’t two inches tall.)
Proof: By hypothesis,
f (~v + ~h) − f (~v ) − (Df )~h
lim
=0
~h→0
|~h|
as ~h approaches 0 along any path. In particular,
f (v + (0, 0, . . . , t, . . . , 0)T ) − f (v) − (Df )(0, 0, . . . , t, . . . , 0)T .
0 = lim
t→0
|(0 0 . . . , t . . . 0)T |
Written out a bit more explicitly,
f (v1 , v2 , . . . , vi + t, . . . , vm ) − f (v1 , v2 , . . . , vi , . . . , vm ) − t(Df )(0 0 . . . 1 . . . 0)T 0 = lim t→0
t
f (v1 , v2 , . . . , vi + t, . . . , vm ) − f (v1 , v2 , . . . , vi , . . . , vm ) − t(A1i A2i · · · Ani )T = lim t→0
t
f (v1 , v2 , . . . , vi + t, . . . , vm ) − f (v1 , v2 , . . . , vi , . . . , vm )
= lim − (A1i A2i · · · Ani )T t→0
t
Saying that the norm of a vector approaches 0 is the same as saying that all of its components
approach 0. Looking at the j-th component of the previous limit,
fj (v1 , v2 , . . . , vi + t, . . . , vm ) − fj (v1 , v2 , . . . , vi , . . . , vm )
0 = lim
− Aji .
t→0
t
So
fj (v1 , v2 , . . . , vi + t, . . . , vm ) − fj (v1 , v2 , . . . , vi , . . . , vm )
lim
= Aji .
t→0
t
We have
∂fj
Aji =
∂ti
as promised.