Metamath Proof Explorer


Theorem dvxpaek

Description: Derivative of the polynomial ( x + A ) ^ K . (Contributed by Glauco Siliprandi, 5-Apr-2020)

Ref Expression
Hypotheses dvxpaek.s ⊢ φ → S ∈ ℝ ℂ
dvxpaek.x ⊢ φ → X ∈ TopOpen ⁡ ℂ fld ↾ 𝑡 S
dvxpaek.a ⊢ φ → A ∈ ℂ
dvxpaek.k ⊢ φ → K ∈ ℕ
Assertion dvxpaek ⊢ φ → dx ∈ X x + A K dS x = x ∈ X ⟼ K ⁢ x + A K − 1

Proof

Step Hyp Ref Expression
1 dvxpaek.s ⊢ φ → S ∈ ℝ ℂ
2 dvxpaek.x ⊢ φ → X ∈ TopOpen ⁡ ℂ fld ↾ 𝑡 S
3 dvxpaek.a ⊢ φ → A ∈ ℂ
4 dvxpaek.k ⊢ φ → K ∈ ℕ
5 cnelprrecn ⊢ ℂ ∈ ℝ ℂ
6 5 a1i ⊢ φ → ℂ ∈ ℝ ℂ
7 1 2 dvdmsscn ⊢ φ → X ⊆ ℂ
8 7 adantr ⊢ φ ∧ x ∈ X → X ⊆ ℂ
9 simpr ⊢ φ ∧ x ∈ X → x ∈ X
10 8 9 sseldd ⊢ φ ∧ x ∈ X → x ∈ ℂ
11 3 adantr ⊢ φ ∧ x ∈ X → A ∈ ℂ
12 10 11 addcld ⊢ φ ∧ x ∈ X → x + A ∈ ℂ
13 1red ⊢ φ ∧ x ∈ X → 1 ∈ ℝ
14 0red ⊢ φ ∧ x ∈ X → 0 ∈ ℝ
15 13 14 readdcld ⊢ φ ∧ x ∈ X → 1 + 0 ∈ ℝ
16 simpr ⊢ φ ∧ y ∈ ℂ → y ∈ ℂ
17 4 nnnn0d ⊢ φ → K ∈ ℕ 0
18 17 adantr ⊢ φ ∧ y ∈ ℂ → K ∈ ℕ 0
19 16 18 expcld ⊢ φ ∧ y ∈ ℂ → y K ∈ ℂ
20 18 nn0cnd ⊢ φ ∧ y ∈ ℂ → K ∈ ℂ
21 nnm1nn0 ⊢ K ∈ ℕ → K − 1 ∈ ℕ 0
22 4 21 syl ⊢ φ → K − 1 ∈ ℕ 0
23 22 adantr ⊢ φ ∧ y ∈ ℂ → K − 1 ∈ ℕ 0
24 16 23 expcld ⊢ φ ∧ y ∈ ℂ → y K − 1 ∈ ℂ
25 20 24 mulcld ⊢ φ ∧ y ∈ ℂ → K ⁢ y K − 1 ∈ ℂ
26 1 2 dvmptidg ⊢ φ → dx ∈ X x dS x = x ∈ X ⟼ 1
27 1 2 3 dvmptconst ⊢ φ → dx ∈ X A dS x = x ∈ X ⟼ 0
28 1 10 13 26 11 14 27 dvmptadd ⊢ φ → dx ∈ X x + A dS x = x ∈ X ⟼ 1 + 0
29 dvexp ⊢ K ∈ ℕ → dy ∈ ℂ y K d ℂ y = y ∈ ℂ ⟼ K ⁢ y K − 1
30 4 29 syl ⊢ φ → dy ∈ ℂ y K d ℂ y = y ∈ ℂ ⟼ K ⁢ y K − 1
31 oveq1 ⊢ y = x + A → y K = x + A K
32 oveq1 ⊢ y = x + A → y K − 1 = x + A K − 1
33 32 oveq2d ⊢ y = x + A → K ⁢ y K − 1 = K ⁢ x + A K − 1
34 1 6 12 15 19 25 28 30 31 33 dvmptco ⊢ φ → dx ∈ X x + A K dS x = x ∈ X ⟼ K ⁢ x + A K − 1 ⁢ 1 + 0
35 1p0e1 ⊢ 1 + 0 = 1
36 35 oveq2i ⊢ K ⁢ x + A K − 1 ⁢ 1 + 0 = K ⁢ x + A K − 1 ⋅ 1
37 36 a1i ⊢ φ ∧ x ∈ X → K ⁢ x + A K − 1 ⁢ 1 + 0 = K ⁢ x + A K − 1 ⋅ 1
38 4 nncnd ⊢ φ → K ∈ ℂ
39 38 adantr ⊢ φ ∧ x ∈ X → K ∈ ℂ
40 22 adantr ⊢ φ ∧ x ∈ X → K − 1 ∈ ℕ 0
41 12 40 expcld ⊢ φ ∧ x ∈ X → x + A K − 1 ∈ ℂ
42 39 41 mulcld ⊢ φ ∧ x ∈ X → K ⁢ x + A K − 1 ∈ ℂ
43 42 mulridd ⊢ φ ∧ x ∈ X → K ⁢ x + A K − 1 ⋅ 1 = K ⁢ x + A K − 1
44 37 43 eqtrd ⊢ φ ∧ x ∈ X → K ⁢ x + A K − 1 ⁢ 1 + 0 = K ⁢ x + A K − 1
45 44 mpteq2dva ⊢ φ → x ∈ X ⟼ K ⁢ x + A K − 1 ⁢ 1 + 0 = x ∈ X ⟼ K ⁢ x + A K − 1
46 34 45 eqtrd ⊢ φ → dx ∈ X x + A K dS x = x ∈ X ⟼ K ⁢ x + A K − 1