,
,
.
if it exists, is denoted
and called the directional derivative
of
at
, in the direction
.
This is obvious from the definitions.
One has
which shows
the result whenever
. The case when
is trivially true.
The last assertion follows from the additivity of the function
.
With the notation of Problem 2-4, part (a) of that problem says that
exists for all
. Now suppose
. Then
, But
.
By Problem 1-26 (a),
for all
.
Show that
is differentiable at 0 but
is not continuous at 0.
Clearly,
is differentiable at
. At
, one
has
since
.
For
, one has
. The first term has
limit 0 as
approaches 0. But the second term takes on all values between
-1 and 1 in every open neighborhood of
. So,
does not even exist.
Show that
is differentiable at (0,0) but that
is not continuous
at
.
The derivative at (0, 0) is the zero linear transformation because
, just as in part (a).
However,
for
where
is as in part (a). It follows from the differentiability of
,
that
and
are defined for
. (The argument
given above also shows that they are defined and 0 at
.) Further
the partials are equal to
up to a sign, and so they cannot be
continuous at 0.
Proceed as in the proof of Theorem 2-8 for all
. In the
case,
it suffices to note that
follows from the definition of
. This is all that is needed in
the rest of the proof.
Applying Theorem 2-9 to
gives
.
On the other hand,
and so
.
Substituting
in these two formulas show the result.
Following the hint, let
. Then
. On the other hand, Theorem 2-9 gives
. So, we have the result with
.