theorem
  (for x holds f.x = sqrt x) & x>0 & x+h>0 implies
  fD(f,h).x = sqrt (x+h) - sqrt x
proof
  assume
A1:for x holds f.x = sqrt x;
  fD(f,h).x = f.(x+h) - f.x by DIFF_1:3
    .= sqrt (x+h) - f.x by A1
    .= sqrt (x+h) - sqrt x by A1;
  hence thesis;
end;
