Metamath Proof Explorer


Theorem dvsinexp

Description: The derivative of sin^N . (Contributed by Glauco Siliprandi, 29-Jun-2017)

Ref Expression
Hypothesis dvsinexp.5 φN
Assertion dvsinexp φdxsinxNdx=xNsinxN1cosx

Proof

Step Hyp Ref Expression
1 dvsinexp.5 φN
2 cnelprrecn
3 2 a1i φ
4 sinf sin:
5 4 a1i φsin:
6 5 ffvelcdmda φxsinx
7 cosf cos:
8 7 a1i φcos:
9 8 ffvelcdmda φxcosx
10 simpr φyy
11 1 nnnn0d φN0
12 11 adantr φyN0
13 10 12 expcld φyyN
14 1 nncnd φN
15 14 adantr φyN
16 nnm1nn0 NN10
17 1 16 syl φN10
18 17 adantr φyN10
19 10 18 expcld φyyN1
20 15 19 mulcld φyNyN1
21 dvsin Dsin=cos
22 5 feqmptd φsin=xsinx
23 22 oveq2d φDsin=dxsinxdx
24 8 feqmptd φcos=xcosx
25 21 23 24 3eqtr3a φdxsinxdx=xcosx
26 dvexp NdyyNdy=yNyN1
27 1 26 syl φdyyNdy=yNyN1
28 oveq1 y=sinxyN=sinxN
29 oveq1 y=sinxyN1=sinxN1
30 29 oveq2d y=sinxNyN1=NsinxN1
31 3 3 6 9 13 20 25 27 28 30 dvmptco φdxsinxNdx=xNsinxN1cosx