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 φS
dvxpaek.x φXTopOpenfld𝑡S
dvxpaek.a φA
dvxpaek.k φK
Assertion dvxpaek φdxXx+AKdSx=xXKx+AK1

Proof

Step Hyp Ref Expression
1 dvxpaek.s φS
2 dvxpaek.x φXTopOpenfld𝑡S
3 dvxpaek.a φA
4 dvxpaek.k φK
5 cnelprrecn
6 5 a1i φ
7 1 2 dvdmsscn φX
8 7 adantr φxXX
9 simpr φxXxX
10 8 9 sseldd φxXx
11 3 adantr φxXA
12 10 11 addcld φxXx+A
13 1red φxX1
14 0red φxX0
15 13 14 readdcld φxX1+0
16 simpr φyy
17 4 nnnn0d φK0
18 17 adantr φyK0
19 16 18 expcld φyyK
20 18 nn0cnd φyK
21 nnm1nn0 KK10
22 4 21 syl φK10
23 22 adantr φyK10
24 16 23 expcld φyyK1
25 20 24 mulcld φyKyK1
26 1 2 dvmptidg φdxXxdSx=xX1
27 1 2 3 dvmptconst φdxXAdSx=xX0
28 1 10 13 26 11 14 27 dvmptadd φdxXx+AdSx=xX1+0
29 dvexp KdyyKdy=yKyK1
30 4 29 syl φdyyKdy=yKyK1
31 oveq1 y=x+AyK=x+AK
32 oveq1 y=x+AyK1=x+AK1
33 32 oveq2d y=x+AKyK1=Kx+AK1
34 1 6 12 15 19 25 28 30 31 33 dvmptco φdxXx+AKdSx=xXKx+AK11+0
35 1p0e1 1+0=1
36 35 oveq2i Kx+AK11+0=Kx+AK11
37 36 a1i φxXKx+AK11+0=Kx+AK11
38 4 nncnd φK
39 38 adantr φxXK
40 22 adantr φxXK10
41 12 40 expcld φxXx+AK1
42 39 41 mulcld φxXKx+AK1
43 42 mulridd φxXKx+AK11=Kx+AK1
44 37 43 eqtrd φxXKx+AK11+0=Kx+AK1
45 44 mpteq2dva φxXKx+AK11+0=xXKx+AK1
46 34 45 eqtrd φdxXx+AKdSx=xXKx+AK1