$y'=e^x$ => $y''=e^x$ => $y^{(3)}=e^x$
$y'=-\frac{1}{(x-1)^2}$ $y''=\frac{2}{(x-1)^3}$ $y^{(3)}=-\frac{6}{(x-1)^4}$
$y'=\frac{1}{2\sqrt{x}}$ $y''=-\frac{1}{4x\sqrt{x}}$ $y^{(3)}=\frac{3}{8x^2\sqrt{x}}$
$y=\frac{1}{4-x^2}=\frac{1}{4}(\frac{1}{x-2}-\frac{1}{x+2})$ rồi nó quay lại tương tự câu b thôi