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+h/2)<>0 & cos(x-h/2)<>0
  implies cD(f,h).x = (-16)*sin(x)*sin((-h)/2)*cos(x)*cos((-h)/2)
  /((cos(2*x)+cos(h))^2)
proof
  assume that
A1:for x holds f.x=1/(cos(x))^2 and
A2:cos(x+h/2)<>0 & cos(x-h/2)<>0;
  cD(f,h).x = f.(x+h/2) - f.(x-h/2) by DIFF_1:5
    .= 1/(cos(x+h/2))^2 - f.(x-h/2) by A1
    .= 1/(cos(x+h/2))^2 - 1/(cos(x-h/2))^2 by A1
    .= (1*(cos(x-h/2))^2-1*(cos(x+h/2))^2)/((cos(x+h/2))^2*(cos(x-h/2))^2)
                                                by A2,XCMPLX_1:130
    .= ((cos(x-h/2))^2-(cos(x+h/2))^2)/((cos(x+h/2)*cos(x-h/2))^2)
    .= ((cos(x-h/2))^2-(cos(x+h/2))^2)/(((1/2)
       *(cos((x+h/2)+(x-h/2))+cos((x+h/2)-(x-h/2))))^2) by SIN_COS4:32
    .= ((cos(x-h/2))^2-(cos(x+h/2))^2)/((1/4)*(cos(2*x)+cos(h))^2)
    .= ((cos(x-h/2))^2-(cos(x+h/2))^2)/(1/4)/((cos(2*x)+cos(h))^2)
                                                                by XCMPLX_1:78
    .= 4*((((cos(x-h/2))-(cos(x+h/2)))*((cos(x-h/2))+(cos(x+h/2))))
       /((cos(2*x)+cos(h))^2))
    .= 4*(((-2*(sin(((x-h/2)+(x+h/2))/2)*sin(((x-h/2)-(x+h/2))/2)))
       *((cos(x-h/2))+(cos(x+h/2))))/((cos(2*x)+cos(h))^2)) by SIN_COS4:18
    .= 4*(((-2*(sin((2*x)/2)*sin((-h)/2)))*(2*(cos((2*x)/2)*cos((-h)/2))))
       /((cos(2*x)+cos(h))^2)) by SIN_COS4:17
    .= (-16)*sin(x)*sin((-h)/2)*cos(x)*cos((-h)/2)/((cos(2*x)+cos(h))^2);
  hence thesis;
end;
