reserve y for set,
  x,a,b for Real,
  n for Element of NAT,
  Z for open Subset of REAL,
  f,f1,f2,g for PartFunc of REAL,REAL;

theorem
  not 0 in Z & Z c= dom (((id Z)^)(#)cos) implies ((id Z)^)(#)cos
is_differentiable_on Z & for x st x in Z holds ((((id Z)^)(#)cos)`|Z).x =-(1/x)
  *sin.x-(1/x^2)*cos.x
proof
  set f = id Z;
  assume that
A1: not 0 in Z and
A2: Z c= dom (((id Z)^)(#)cos);
A3: f^ is_differentiable_on Z by A1,Th4;
A4: cos is_differentiable_on Z by FDIFF_1:26,SIN_COS:67;
  Z c= dom (f^) /\ dom (cos) by A2,VALUED_1:def 4;
  then
A5: Z c= dom (f^) by XBOOLE_1:18;
  now
    let x;
    assume
A6: x in Z;
    hence
    (((f^)(#)cos)`|Z).x = (cos.x)*diff((f^),x)+((f^).x)*diff(cos,x) by A2,A3,A4
,FDIFF_1:21
      .=(cos.x)*((f^)`|Z).x+ ((f^).x)*diff(cos,x) by A3,A6,FDIFF_1:def 7
      .=(cos.x)*(-1/x^2)+ ((f^).x)*diff(cos,x) by A1,A6,Th4
      .=(cos.x)*(-1/x^2)+((f^).x)*(-sin.x) by SIN_COS:63
      .=(cos.x)*(-1/x^2)+((f.x)")*(-sin.x) by A5,A6,RFUNCT_1:def 2
      .=(cos.x)*(-1/x^2)+(1*x")*(-sin.x) by A6,FUNCT_1:18
      .=-(1/x^2)*(cos.x)+(1/x)*(-sin.x) by XCMPLX_0:def 9
      .=-(1/x)*sin.x-(1/x^2)*cos.x;
  end;
  hence thesis by A2,A3,A4,FDIFF_1:21;
end;
