theorem
  (for x holds f.x = sqrt x) & x>0 & x-h>0 implies
  bD(f,h).x = sqrt x - sqrt (x-h)
proof
  assume
A1:for x holds f.x = sqrt x;
  bD(f,h).x = f.x - f.(x-h) by DIFF_1:4
    .= sqrt x - f.(x-h) by A1
    .= sqrt x - sqrt (x-h) by A1;
  hence thesis;
end;
