We have
and so
by the chain rule, one has:
, i.e.
.
Using Theorem 2-3 (3) and part (a), one has:
.
One has
, and so by the chain rule:
If
is the function of part (c), then
. Using the chain rule, we get:
If
, then
and
we know the derivative of
from part (a). The chain rule gives:
If
, then
. So one gets:
.
If
, then
. So one gets:
.
The chain rule gives:
.
Using the last part:
.
Using parts (h), (c), and (a), one gets
If
, then
, and so:
.
If
is as in part (a), then
, and so:
.
One has
where
and
have
derivatives as given in parts (d) and (a) above.
Let
have a 1 in the
place only.
Then we have
by
an obvious induction using bilinearity. It follows that there is an
depending only on
such that:
. Since
, we see that it
suffices to show the result in the case where
and the bilinear
function is the product function. But, in this case, it was verified in
the proof of Theorem 2-3 (5).
One has
by bilinearity and part (a).
This follows by applying (b) to the bilinear function
.
By Problem 2-12 and the fact that
is bilinear, one has
. So
.
(Note that
is an
matrix; its transpose
is an
matrix, which we consider as a member of
.)
Since
, one can apply the chain rule to get the
assertion.
Use part (b) applied to
to get
. This shows the result.
Trivially, one could let
. Then
is not differentiable at 0.
This is an immediate consequence of Problem 2-12 (b).
This can be argued similarly to Problem 2-12. Just apply the definition
expanding the numerator out using multilinearity; the remainder looks like
a sum of terms as in part (a) except that there may be more than two
type
arguments. These can be expanded out as in the proof of the bilinear case to
get a sum of terms that look like constant multiples of
where
is at least two and the
are distinct. Just as
in the bilinear case, this limit is zero. This shows the result.
This is an immediate consequence of Problem 2-14 (b) and the multilinearity of the determinant function.
This follows by the chain rule and part (a).
Without writing all the details, recall that Cramer's Rule allows you
to write down explicit formulas for the
where
is the matrix of the coefficients
and the
are obtained
from
by replacing the
column with the column of the
.
We can take transposes since the determinant of the transpose is the same
as the determinant of the original matrix; this makes the formulas simpler
because the formula for derivative of a determinant involved rows and so
now you are replacing the
row rather than the
column.
Anyway, one can use the quotient formula and part (b) to give a formula
for the derivative of the
.
This follows immediately by the chain rule applied to
.