Đáp án đúng: D
Giải chi tiết:Ta có: \(f\left( x \right) = {\sin ^6}x + {\cos ^6}x\)
\(\begin{array}{l} \Rightarrow f'\left( x \right) = 6{\sin ^5}x\cos x - 6{\cos ^5}x\sin x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6\sin x\cos x\left( {{{\sin }^4}x - {{\cos }^4}x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6\sin x\cos x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}x} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6\sin x\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right).\\g\left( x \right) = 3{\sin ^2}x{\cos ^2}x\\ \Rightarrow g'\left( x \right) = 6\sin x{\cos ^3}x - 6\cos x{\sin ^3}x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 6\sin x\cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right).\end{array}\)
\( \Rightarrow f'\left( x \right) + g'\left( x \right) = 6\sin x\cos x\left( {{{\sin }^2}x - {{\cos }^2}x} \right)\)\( - 6\sin x\cos x\left( {{{\cos }^2}x - {{\sin }^2}x} \right) = 0.\)
Chọn D.