English
If f,g are differentiable within s at x with derivatives f', g', then the within-derivative of y ↦ B(f(y), g(y)) is (fderivWithin g)(t) precomposed with f and (fderivWithin f)(s) precomposed with g.
Русский
Если f,g внутри множества s дифференцируемы, то внутри-производная билинейного отображения задаётся аналогично.
LaTeX
$$$fderivWithin 𝕜 (y\mapsto B(f(y),g(y))) s x = (fderivWithin 𝕜 g t (f x)) \; \circ f + (fderivWithin 𝕜 f s x) \circ g$$$
Lean4
@[fun_prop]
theorem comp {g : F → G} {g' : F →L[𝕜] G} {t : Set F} (hg : HasFDerivWithinAt g g' t (f x))
(hf : HasFDerivWithinAt f f' s x) (hst : MapsTo f s t) : HasFDerivWithinAt (g ∘ f) (g'.comp f') s x :=
HasFDerivAtFilter.comp x hg hf <| hf.continuousWithinAt.tendsto_nhdsWithin hst
