reserve n,m for Element of NAT;
reserve h,k,r,r1,r2,x,x0,x1,x2,x3 for Real;
reserve f,f1,f2 for Function of REAL,REAL;
reserve S for Seq_Sequence;

theorem
  (for x holds f.x=1/(cos(x))^2) & cos(x)<>0 & cos(x+h)<>0
  implies fD(f,h).x = (-16)*sin((2*x+h)/2)*sin((-h)/2)*cos((2*x+h)/2)
  *cos((-h)/2)/((cos(2*x+h)+cos(h))^2)
proof
  assume that
A1:for x holds f.x=1/(cos(x))^2 and
A2:cos(x)<>0 & cos(x+h)<>0;
  fD(f,h).x = f.(x+h) - f.x by DIFF_1:3
    .= 1/(cos(x+h))^2 - f.x by A1
    .= 1/(cos(x+h))^2 - 1/(cos(x))^2 by A1
    .= (1*(cos(x))^2-1*(cos(x+h))^2)/((cos(x+h))^2*(cos(x))^2)
                                           by A2,XCMPLX_1:130
    .= ((cos(x))^2-(cos(x+h))^2)/((cos(x+h)*cos(x))^2)
    .= ((cos(x))^2-(cos(x+h))^2)/(((1/2)*(cos((x+h)+x)+cos((x+h)-x)))^2)
                                                             by SIN_COS4:32
    .= ((cos(x))^2-(cos(x+h))^2)/((1/4)*(cos(2*x+h)+cos(h))^2)
    .= ((cos(x))^2-(cos(x+h))^2)/(1/4)/((cos(2*x+h)+cos(h))^2) by XCMPLX_1:78
    .= 4*((((cos(x))-(cos(x+h)))*((cos(x))+(cos(x+h))))
       /((cos(2*x+h)+cos(h))^2))
    .= 4*(((-2*(sin((x+(x+h))/2)*sin((x-(x+h))/2)))*((cos(x))+(cos(x+h))))
       /((cos(2*x+h)+cos(h))^2)) by SIN_COS4:18
    .= 4*(((-2*(sin((2*x+h)/2)*sin((-h)/2)))*(2*(cos((2*x+h)/2)*cos((-h)/2))))
       /((cos(2*x+h)+cos(h))^2)) by SIN_COS4:17
    .= (-16)*sin((2*x+h)/2)*sin((-h)/2)*cos((2*x+h)/2)*cos((-h)/2)
       /((cos(2*x+h)+cos(h))^2);
  hence thesis;
end;
