English
If HasDerivAt f f' x, then conj ∘ f ∘ conj has derivative conj f' at conj x (equiv to HasDerivAt conj_conj).
Русский
Если есть HasDerivAt f f' x, то conj ∘ f ∘ conj имеет производную conj f' в conj x.
LaTeX
$$theorem conj_conj {f : 𝕜 → 𝕜} {f' : 𝕜} (hf : HasDerivAt f f' x) : HasDerivAt (Function.comp (starRingEnd 𝕜) (Function.comp f (starRingEnd 𝕜))) (star f') (star x)$$
Lean4
/-- If `f` has derivative `f'` at `z`, then `conj ∘ f ∘ conj` has derivative `conj f'` at
`conj z`. -/
theorem conj_conj {f : 𝕜 → 𝕜} {f' : 𝕜} (hf : HasDerivAt f f' x) : HasDerivAt (conj ∘ f ∘ conj) (conj f') (conj x) :=
hf.star_conj