Given a function , the change from to is . For example, if , then the change in going from to is

This can be seen as a linear function plus another part which is small in the sense that

**Definition 1:** If
is a
function, then a derivative
of
at
is a
linear function from
to
such that

**Proposition 1:** There is at most one derivative of
at
.

**Proof:** Suppose that
and
are two derivatives.
Then each satisfies the limit condition of the definition. Subtracting
these gives:

For , let for . Then

since and are linear. So

**Exercise 1:** If
is
linear, then
for all
. If
is a constant
function, then
for all
.

**Proposition 2:** If
and
, then
is differentiable at
if and only if
the coordinate functions
are
differentiable for all
. If either is true, then
.

**Proof:** If
is differentiable at
, then

In particular each of the terms of the sum have limit zero. So the coordinate functions have derivatives and they are the coordinate functions of the derivative of . Conversely, if the coordinate functions are differentiable, then putting their derivatives in the above formula in the place of the coordinate functions of shows that is differentiable at .

The main computational tool for derivatives is:

**Theorem 1:** (Chain Rule) If
is differentiable at
and if
is differentiable at
, then
is differentiable at
and
.

**Proof:** Define
and
by

and

We need to show that where

Now, letting and , we have by the definition of ,

Using the definition of , we get

By linearity of , this simplifies to:

We must show that . Since is linear, there is an such that

and the right hand side approaches zero.

For the other term, for any , there is a such that

whenever . Note that we can remove the condition that the middle term be non-zero -- if it were zero, then the condition becomes which is true since . Now is bounded for small . So

This completes the proof.

**Corollary 1:** If
are
differentiable at
, then

**Proof:** The sum
(respectively the product
) function can be
written as the composition of the
defined by
and the function
(respectively
) defined by
(respectively
). Now
is linear and
so its derivative is itself. Further, it is easy to check that:

and so since

The formulas now follow by the chain rule.

**Definition 2:** Let
be a function and
.
Then, for each j, one can define
by
. The
partial
derivative
of
at
is defined to be
.

**Proposition 3:** Let
be a function with domain
and
be in the interior of
and assume that
is either a local minimum or local maximum of
. Then
provided that the partial derivative exists.

**Proof:**
is a local extremum of the function of one variable
.

**Theorem 2:** Let
be a function and
. If
is differentiable at
, then
is the
linear transformation whose matrix is the
array of
partial derivatives
of the coordinate functions
of
.
Conversely, if The partial derivatives of the coordinate functions of
all have partials defined in an open neighborhood of
and are continuous
at
, then
is differentiable at
.

**Proof:** By Proposition 2, it is enough to check the case where
.
The first assertion is immediate from the definitions. For the
second assertion break up the change in
into a sum of changes in which
one varies only one parameter at a time:

One can apply the Mean Value Theorem to each of the differences on the right:

where . It follows that

because the are continuous at .

**Corollary 2:** (Chain Rule) If
for
are
continuously differentiable at
and if
is
differentiable at (g_1(a),...,g_m(a)), then
has
partial derivatives

**Proof:** Apply the Chain Rule and Theorem 2.

**Theorem 3:** If
is a function defined on
an open subset
and if
and its first and
second order partials are exist throughout
and are continuous there,
then
for all
.

**Proof:** It is clearly enough to prove the result in the case where
. Let
and
be such that the rectangle
with diagonal from
to
is contained in
.
Let
be defined by
. First apply the
Mean Value Theorem to find a
such that

Now, . Apply the Mean Value Theorem again to find between and such that

Combining results we have

But

which is symmetric in the two variables. So, one can obtain another expression for the same by repeating the same construction swapping the roles of the two variables: Let and get and such that

and

So . Now take the limit as and use the continuity of the second partials to conclude that the mixed partials are equal.