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

theorem
  Z c= dom (((1/3)(#)(( #Z 3)*cos))-cos) implies (1/3)(#)(( #Z 3)*cos)-
  cos is_differentiable_on Z & for x st x in Z holds ((((1/3)(#)(( #Z 3)*cos))-
  cos)`|Z).x =(sin.x)|^3
proof
  assume
A1: Z c= dom (((1/3)(#)(( #Z 3)*cos))-cos);
  then Z c= dom ((1/3)(#)(( #Z 3)*cos)) /\ dom cos by VALUED_1:12;
  then
A2: Z c= dom ((1/3)(#)(( #Z 3)*cos)) by XBOOLE_1:18;
  then
A3: ((1/3)(#)(( #Z 3)*cos)) is_differentiable_on Z by Th46;
A4: cos is_differentiable_on Z by FDIFF_1:26,SIN_COS:67;
  now
    let x;
    assume
A5: x in Z;
    then
    ((((1/3)(#)(( #Z 3)*cos))-cos)`|Z).x = diff(((1/3)(#)(( #Z 3)*cos)),x
    ) - diff(cos,x) by A1,A3,A4,FDIFF_1:19
      .=diff(((1/3)(#)(( #Z 3)*cos)),x)-(-sin.x) by SIN_COS:63
      .=(((1/3)(#)(( #Z 3)*cos))`|Z).x-(-sin.x) by A3,A5,FDIFF_1:def 7
      .=-(cos.x) #Z (3-1)*sin.x--sin.x by A2,A5,Th46
      .=sin.x*(1-(cos.x) #Z 2 )
      .=sin.x*(1-(cos.x) |^ |.2.| ) by PREPOWER:def 3
      .=sin.x*(1-(cos.x) |^ 2 ) by ABSVALUE:def 1
      .=sin.x*(1-(cos.x)*(cos.x)) by WSIERP_1:1
      .=sin.x*((cos.x)*(cos.x)+(sin.x)*(sin.x)-(cos.x)*(cos.x)) by SIN_COS:28
      .=sin.x*((sin.x)|^2) by WSIERP_1:1
      .=((sin.x)|^(2+1)) by NEWTON:6
      .=(sin.x)|^3;
    hence ((((1/3)(#)(( #Z 3)*cos))-cos)`|Z).x =(sin.x)|^3;
  end;
  hence thesis by A1,A3,A4,FDIFF_1:19;
end;
