$f(x)=\sin^6x+\cos^6x$
$=(\sin^2x+\cos^2x)(\sin^4x+\cos^4x-\sin^2x\cos^2x)$
$=(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x$
$=1-3\sin^2x\cos^2x$
$=1-\dfrac{3}{4}.\sin^22x$
$=1-\dfrac{3}{4}.\dfrac{1-\cos4x}{2}$
$=\dfrac{3}{8}\cos4x+\dfrac{5}{8}$
$\to f'(x)=-\dfrac{3}{8}.\sin4x.(4x)'=\dfrac{-3}{2}\sin4x$
$g(x)=3\sin^2x\cos^2x$
$=\dfrac{3}{4}.\sin^22x$
$=\dfrac{3}{4}.\dfrac{1-\cos4x}{2}$
$=-\dfrac{3}{8}\cos4x+\dfrac{3}{8}$
$\to g'(x)=\dfrac{3}{8}.\sin4x.(4x)'=\dfrac{3}{2}\sin4x$
Vậy $f'(x)+g'(x)=0$
