Now it is clear this is just a generalization of the notion of the derivative of a function of
one variable because in this more specialized situation,
lim |f-(x-+-v)−-f (x)−-f′(x)v|= 0,
|v|→0 |v|
due to the definition which says
f′(x) = lim f-(x+-v)−-f (x).
v→0 v
For functions of n variables, you can’t define the derivative as the limit of a difference
quotient like you can for a function of one variable because you can’t divide by a vector.
That is why there is a need for a more general definition.
The term o
(v)
is notation that is descriptive of the behavior in 5.12.14 and it is only
this behavior that is of interest. Thus, if t and k are constants,
o(v) = o(v)+ o (v ),o (tv) = o(v),ko(v) = o(v)
and other similar observations hold. The sloppiness built in to this notation is useful
because it ignores details which are not important. It may help to think of o
(v )
as an adjective describing what is left over after approximating f
(x +v )
by
f
(x)
+ Df
(x)
v.
Theorem 5.12.2The derivative is well defined.
Proof:First note that for a fixed vector, v,o
(tv )
= o
(t)
. Now suppose both L1 and
L2 work in the above definition. Then let v be any vector and let t be a real scalar which
is chosen small enough that tv + x ∈ U. Then
f (x+ tv) = f (x )+ L1tv + o(tv),f (x+ tv) = f (x) +L2tv + o(tv ).
Therefore, subtracting these two yields
(L − L )
2 1
(tv)
= o
(tv)
= o
(t)
. Therefore,
dividing by t yields
(L2 − L1)
(v)
=
o(t)
t
. Now let t → 0 to conclude that
(L2 − L1)
(v)
= 0. Since this is true for all v, it follows L2 = L1. This proves the
theorem.
Lemma 5.12.3Let f be differentiable at x. Then f is continuous at x and in fact,there exists K > 0 such that whenever
Theorem 5.12.4(The chain rule)Let U and V be open sets, U ⊆ Fnand V ⊆ Fm.Suppose f : U → V is differentiable at x ∈ U and suppose g : V → Fqis differentiable atf
(x)
∈ V . Then g ∘ f is differentiable at x and
D (g ∘f)(x) = Dg (f (x))Df (x ).
Proof:This follows from a computation. Let B
(x,r)
⊆ U and let r also be small
enough that for
|v|
≤ r, it follows that f
(x + v)
∈ V . Such an r exists because f
is continuous at x. For
|v|
< r, the definition of differentiability of g and f
implies
g (f (x+ v))− g (f (x)) = Dg (f (x))Df (x)v+ o (v )
which proves the theorem.
The derivative is a linear transformation. What is the matrix of this linear
transformation taken with respect to the usual basis vectors? Let ei denote the vector of
Fn which has a one in the ith entry and zeroes elsewhere. Then the matrix of the linear
transformation is the matrix whose ith column is Df
(x)
ei. What is this? Let t ∈ ℝ such
that
|t|
is sufficiently small.
f (x + tei)− f (x) = Df (x)tei + o(tei)
= Df (x)te + o(t) .
i
Then dividing by t and taking a limit,
Df (x)ei = lim f (x+-tei)-− f-(x-)≡-∂f (x ).
t→0 t ∂xi
Thus the matrix of Df
(x)
with respect to the usual basis vectors is the matrix of the
form
( )
f1,x1 (x) f1,x2 (x) ⋅⋅⋅ f1,xn (x)
|( ... ... ... |) .
f (x) f (x) ⋅⋅⋅ f (x)
m,x1 m,x2 m,xn
As mentioned before, there is no harm in referring to this matrix as Df
(x)
but it may
also be referred to as Jf
(x)
.
This is summarized in the following theorem.
Theorem 5.12.5Let f : Fn→ Fmand suppose f is differentiable at x. Thenall the partial derivatives
∂f∂i(xx)
j
exist and if Jf
(x)
is the matrix of the lineartransformation with respect to the standard basis vectors, then the ijthentry is givenby fi,jor
∂fi
∂xj
(x)
.
What if all the partial derivatives of f exist? Does it follow that f is differentiable?
Consider the following function.
{ xy
f (x,y) = x2+y2 if (x,y) ⁄= (0,0) .
0 if (x,y) = (0,0)
Then from the definition of partial derivatives,
lim f (h,0)−-f-(0,0) = lim 0-−-0= 0
h→0 h h→0 h
and
f (0,h)− f (0,0) 0 − 0
lim -------------- = lim -----= 0
h→0 h h→0 h
However f is not even continuous at
(0,0)
which may be seen by considering
the behavior of the function along the line y = x and along the line x = 0.
By Lemma 5.12.3 this implies f is not differentiable. Therefore, it is necessary
to consider the correct definition of the derivative given above if you want to
get a notion which generalizes the concept of the derivative of a function of
one variable in such a way as to preserve continuity whenever the function is
differentiable.