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 .