let Z be open Subset of REAL ; ( Z c= dom (sin * (arctan + arccot )) & Z c= ].(- 1),1.[ implies ( sin * (arctan + arccot ) is_differentiable_on Z & ( for x being Real st x in Z holds
((sin * (arctan + arccot )) `| Z) . x = 0 ) ) )
assume that
A1:
Z c= dom (sin * (arctan + arccot ))
and
A2:
Z c= ].(- 1),1.[
; ( sin * (arctan + arccot ) is_differentiable_on Z & ( for x being Real st x in Z holds
((sin * (arctan + arccot )) `| Z) . x = 0 ) )
A3:
arctan + arccot is_differentiable_on Z
by A2, Th37;
A4:
for x being Real st x in Z holds
sin * (arctan + arccot ) is_differentiable_in x
then A6:
sin * (arctan + arccot ) is_differentiable_on Z
by A1, FDIFF_1:16;
for x being Real st x in Z holds
((sin * (arctan + arccot )) `| Z) . x = 0
proof
let x be
Real;
( x in Z implies ((sin * (arctan + arccot )) `| Z) . x = 0 )
A7:
sin is_differentiable_in (arctan + arccot ) . x
by SIN_COS:69;
assume A8:
x in Z
;
((sin * (arctan + arccot )) `| Z) . x = 0
then A9:
arctan + arccot is_differentiable_in x
by A3, FDIFF_1:16;
((sin * (arctan + arccot )) `| Z) . x =
diff (sin * (arctan + arccot )),
x
by A6, A8, FDIFF_1:def 8
.=
(diff sin ,((arctan + arccot ) . x)) * (diff (arctan + arccot ),x)
by A9, A7, FDIFF_2:13
.=
(cos . ((arctan + arccot ) . x)) * (diff (arctan + arccot ),x)
by SIN_COS:69
.=
(cos . ((arctan + arccot ) . x)) * (((arctan + arccot ) `| Z) . x)
by A3, A8, FDIFF_1:def 8
.=
(cos . ((arctan + arccot ) . x)) * 0
by A2, A8, Th37
.=
0
;
hence
((sin * (arctan + arccot )) `| Z) . x = 0
;
verum
end;
hence
( sin * (arctan + arccot ) is_differentiable_on Z & ( for x being Real st x in Z holds
((sin * (arctan + arccot )) `| Z) . x = 0 ) )
by A1, A4, FDIFF_1:16; verum