let x be Real; for Z being open Subset of REAL st x in Z holds
((diff cos ,Z) . 2) . x = - (cos . x)
let Z be open Subset of REAL ; ( x in Z implies ((diff cos ,Z) . 2) . x = - (cos . x) )
assume A1:
x in Z
; ((diff cos ,Z) . 2) . x = - (cos . x)
A2:
cos is_differentiable_on Z
by FDIFF_1:34, SIN_COS:72;
dom (((- 1) (#) sin ) | Z) =
(dom ((- 1) (#) sin )) /\ Z
by RELAT_1:90
.=
(dom sin ) /\ Z
by VALUED_1:def 5
.=
Z
by SIN_COS:27, XBOOLE_1:28
;
then A3:
Z c= dom ((- 1) (#) sin )
by RELAT_1:89;
A4:
sin is_differentiable_on Z
by FDIFF_1:34, SIN_COS:73;
then A5:
(- 1) (#) sin is_differentiable_on Z
by FDIFF_2:19;
((diff cos ,Z) . 2) . x =
((diff cos ,Z) . (1 + 1)) . x
.=
(((diff cos ,Z) . (1 + 0 )) `| Z) . x
by TAYLOR_1:def 5
.=
((((diff cos ,Z) . 0 ) `| Z) `| Z) . x
by TAYLOR_1:def 5
.=
(((cos | Z) `| Z) `| Z) . x
by TAYLOR_1:def 5
.=
((cos `| Z) `| Z) . x
by A2, FDIFF_2:16
.=
(((- sin ) | Z) `| Z) . x
by TAYLOR_2:17
.=
(((- 1) (#) sin ) `| Z) . x
by A5, FDIFF_2:16
.=
(- 1) * (diff sin ,x)
by A1, A4, A3, FDIFF_1:28
.=
(- 1) * (cos . x)
by SIN_COS:69
.=
- (cos . x)
;
hence
((diff cos ,Z) . 2) . x = - (cos . x)
; verum