If
is differentiable at
, then
. So, we need only show that
,
but this follows immediately from Problem 1-10.
The first assertion is trivial: If
is independent of the second variable,
you can let
be defined by
. Conversely, if
, then
.
If
is independent of the second variable, then
because:
Note: Actually,
is the Jacobian, i.e. a 1 x 2 matrix. So,
it would be more proper to say that
, but I will often
confound
with
, even though one is a linear transformation
and the other is a matrix.
The function
is independent of the first variable if and only if
for all
. Just as before,
this is equivalent to their being a function
such that
for all
. An argument similar
to that of the previous problem shows that
.
One has
when
and
otherwise. In
both cases,
is linear and hence differentiable.
Suppose
is differentiable at
with, say,
.
Then one must have:
.
But
and so
. Similarly, one gets
.
More generally, using the definition of derivative, we get for fixed
:
. But
, and so we see that this just says that
for all
. Thus
is identically zero.
Show that
is a function of the kind considered in Problem 2-4, so that
is not differentiable at
.
Define
by
for all
. Then it is trivial to show that
satisfies all the
properties of Problem 2-4 and that the function
obtained from this
is as in the statement of this problem.
Just as in the proof of Problem 2-4, one can show that, if
were
differentiable at 0, then
would be the zero map. On the other hand,
by approaching zero along the 45 degree line in the first quadrant, one
would then have:
in spite
of the fact that the limit is clearly 1.
In fact,
by the squeeze principle using
.
Suppose that
. Then one
has the inequality:
. So, by the squeeze principle,
must be differentiable at
with
.
On the other hand, if the
are differentiable at
, then use the
inequality derived from Problem 1-1:
and the squeeze principle to conclude that
is differentiable at
with the desired derivative.
If
is differentiable at
, then the function
works by the definition of derivative.
The converse is not true. Indeed, you can change the value of
at
without changing whether or not
and
are equal up to first order. But
clearly changing the value of
at
changes whether or not
is
differentiable at
.
To make the converse true, add the assumption that
be continuous at
:
If there is a
of the
specified form with
and
equal up to first order, then
. Multiplying
this by
, we see that
. Since
is continuous, this means that
. But then the condition
is equivalent to the assertion that
is differentiable at
with
.
are equal up to
order at a.
Apply L'Hôpital's Rule n - 1 times to the limit
to see that the value of the limit is
. On the other hand, one has:
Subtracting these two results gives shows that
and
are equal up to
order at
.