English
If hc is HasDerivWithinAt for c with derivative c' and d is a constant, then HasDerivWithinAt for y ↦ c(y) · d has derivative c' · d.
Русский
Если hc задаёт производную внутри для c и d константа, то производная внутри для y ↦ c(y) · d равна c'(x) · d.
LaTeX
$$$\text{HasDerivWithinAt}(y \mapsto c(y) \cdot d) (s) x = c'(x) \cdot d$$$
Lean4
theorem mul_const (hc : HasDerivAt c c' x) (d : 𝔸) : HasDerivAt (fun y => c y * d) (c' * d) x :=
by
rw [← hasDerivWithinAt_univ] at *
exact hc.mul_const d