1
$\begingroup$

Consider

  • a bilinear map $\langle \cdot, \cdot\rangle : \mathbb{R}^m \times \mathbb{R}^n \to \mathbb{R}^k$,
  • an open set $U \subseteq \mathbb{R}^j$,
  • a pair of maps $f: U \to \mathbb{R}^m$ and $g: U \to \mathbb R^n$, and
  • the composite map $F(x) = \langle f(x), g(x) \rangle$.

Then, is it necessarily true that $$ dF_{a}(b) = \langle df_{a}(b), g(a) \rangle + \langle f(a), dg_{a}(b) \rangle, $$ and if not, is there a similar product rule for $dF$?

$\endgroup$
1
$\begingroup$

Yes the equality

$$ dF_{a}(b) = \langle df_{a}(b), g(a) \rangle + \langle f(a), dg_{a}(b) \rangle, $$ always holds. This is a consequence of the chain rule as the derivative of the map $G(u,v) = \langle u, v \rangle$ is

$$G^\prime(u,v)(h,k) = \langle u, k \rangle + \langle h, v \rangle.$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.