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→Дифференцирование «произведений»
{{Лемма
|statement=
Пусть <tex> F, G: \ E \subset \mathbb{R}^m \to \mathbb{R}^l </tex>, <tex> \lambda: E \to \mathbb{R} </tex>, <tex> a \in \operatorname{Int} E </tex>, ; <tex> F, G, \lambda </tex> — дифференцируемые в <tex> a </tex>. тогда:
1) <tex> (\lambda F)' (a) h = ( \lambda'(a), h ) F(a) + \lambda(a) (F'(a) h) </tex>
(здесь <tex> \left \langle a, b \right \rangle </tex> — скалярное произведение <tex> a </tex> и <tex> b </tex>)
|proof=
1. Введём координатную ф-ю <tex> F = (f_1 \ldots f_l) </tex>
<tex> (\lambda f_i)'(a)h = (\lambda'(a)(h))f_i(a) + \lambda(a)(f'_i(a)h) </tex> — <tex>i</tex>-ая коорд. док. ф-лы; <tex> ]f_i \leftrightarrow f </tex>
<tex> \lambda(a + h)f(a + h) - \lambda(a)f(a) = (\lambda(a + h) - \lambda(a))f(a + h) + \lambda(a)f(a + b) - f(a)) =
(\lambda'(a)h + o(h))f(a + h) + \lambda(a)(f'(a)h + o(h)) =
(\lambda'(a)h) \cdot f(a) + \lambda(a)f'(a)h + (\lambda'(a)h)(f(a + h) - f(a)) + o(h)f(a + h) + \lambda(a) \cdot o(h) </tex>
<tex> || \frac{1 slag.}{||h||} || = \frac{|\lambda'(a)h|\cdot||f(a + h) - f(a)||}{||h||} \le \frac{||\lambda'(a)||\cdot||h||\cdot||f(a + h) - f(a)||}{||h||} \rightarrow 0 </tex>
<tex> ||2 slag.|| = |o(h)| \cdot ||f(a + h)|| = o(h); \ \ ||f(a + h)|| </tex> — ограничена.
<tex> ||3 slag.|| = ||\lambda(a) \cdot o(h)|| = |\lambda(a)| \cdot ||o(h)|| = o(h) </tex>
2. <tex> \left \langle F, G \right \rangle ' (a)h = (\sum_{i = 1}^{l}f_i g_i)'(a)h = </tex> лин. дифф. <tex> \sum(f_i g_i)'(a)h =
\sum(f'_i(a)h)g_i(a) + f_i(a)(g'_i(a)h) = \left \langle F'(a)h, G(a) \right \rangle + \left \langle F(a), G'(a)h \right \rangle </tex>
Замечание: <tex>m = 1; \ F, G : \Re \rightarrow \Re^l </tex>
<tex> \left \langle F, G \right \rangle ' (a) = \left \langle F'(a), G(a) \right \rangle + \left \langle F(a), G'(a) \right \rangle </tex>
}}
