**Lemma 1:** Let
be defined and differentiable
on a convex open set A. Then for all
, one has

where is chosen to be an upper bound on all the for all , , and all .

**Proof:** Let
. For each
, the Mean Value
Theorem says that there is a
between 0 and 1 such that

By the chain rule,

So,

**Theorem 1:** (Inverse Function Theorem) Let
be continuously differentiable on an open subset containing
.
If
, then there are open subsets
containing
and
containing
such that
and there is an inverse
which is differentiable with derivative satisfying:

**Exercise 1:** Show that Theorem 1 is true for linear functions.

**Proof: (of Theorem 1)** By replacing
with
, we
can assume that
is the identity map (Why?).

Apply Lemma 1 to : for and in some open rectangle containing . But then,

Rearranging gives

Since , and so by choosing the rectangle small enough, we can assume that and so

In particular, it follows that f is one-to-one when restricted to this rectangle, and the inverse will be continuous if it exists. Replacing the rectangle with a smaller one, we can assume the same is true when f is restricted to the closure of the rectangle. Now the boundary of the rectangle is compact and so is also compact and does not contain . Let be the minimum of for . Let be the set .

Now, for every , there is at least one in the rectangle with . In fact, consider the function . The image of under the closure of the rectangle. The minimum of this function does not occur on because . So it must occur where the derivative is zero, i.e. one has:

for . But by taking the rectangle sufficiently small, we can assume that for all in the rectangle. But then the only solution of this system of linear equations is .

If , then maps the open set one-to-one and onto the open set . It remains to check differentiability of the inverse. Since is differentiable, one has for ,

where . Letting and , we get after substitution:

Rearranging gives:

It remains to show that

Since the derivative is just a linear function, it is enough to show that . As , we have because is continuous. So, . But we know that So, the limit of the product is zero, as desired.

**Note:** It follows from the formula for the derivative of the inverse
that the inverse is also continuously differentiable.

**Corollary 1:** (The Implicit Function Theorem) Let
be continuously differentiable in an open set containing the point
which satisfies
. Suppose that
.
Then there is a continuously differentiable function
mapping
an open set
containing
to an open set
containing
such that
for all
.

**Proof:** We can extend
to a function
by
and apply the Inverse Function Theorem to get
an inverse
defined in an open subset containing
and mapping onto
an open subset containing
. Let
where
is the projection onto the last
coordinates of
.
Then one has
.

**Exercise 2:** Show that the Inverse Function Theorem is a Corollary
of the Implicit Function Theorem.