Rút gọn $y$ trước khi đạo hàm:
$y=\sin^4x+\cos^4x$
$=(\sin^2x)^2+(\cos^2x)^2$
$=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$
$=1-\dfrac{1}{2}.4\sin^2x\cos^2x$
$=1-\dfrac{1}{2}(2\sin x\cos x)^2$
$=1-\dfrac{1}{2}.\sin^22x$
$=1-\dfrac{1}{2}.\dfrac{1-\cos4x}{2}$
$=\dfrac{1}{4}\cos4x+\dfrac{3}{4}$