a) Ta có:
\(y' = 6{\sin ^5}x.\cos x - 6{\cos ^5}x.\sin x + 6\sin x.\cos^3x\) \( - 6{\sin ^3}x.\cos x\)
\(= 6{\sin ^3}x.\cos x(\sin^2 x - 1) \)\(+ 6\sin x.\cos^3 x(1 - {\cos ^2}x)\)
\(= - 6{\sin ^3}x.\cos^3 x + 6{\sin ^3}x.\cos^3 x = 0\).
Vậy \(y' = 0\) với mọi \(x\), tức là \(y'\) không phụ thuộc vào \(x\).
b)
\(y = {{1 + \cos \left( {{{2\pi } \over 3} - 2x} \right)} \over 2} + {{1 + \cos \left( {{{2\pi } \over 3} + 2x} \right)} \over 2} + {{1 + \cos \left( {{{4\pi } \over 3} - 2x} \right)} \over 2} \)
\(+ {{1 + \cos \left( {{{4\pi } \over 3} + 2x} \right)} \over 2} - 2{\sin ^2}x\)
\( = \dfrac{1}{2} + \dfrac{1}{2}\cos \left( {\dfrac{{2\pi }}{3} - 2x} \right)\) \( + \dfrac{1}{2} + \dfrac{1}{2}\cos \left( {\dfrac{{2\pi }}{3} + 2x} \right)\) \( + \dfrac{1}{2} + \dfrac{1}{2}\cos \left( {\dfrac{{4\pi }}{3} - 2x} \right)\) \( + \dfrac{1}{2} + \dfrac{1}{2}\cos \left( {\dfrac{{4\pi }}{3} + 2x} \right)\) \( - 2.\dfrac{{1 - \cos 2x}}{2}\)
\( = 1 + \dfrac{1}{2}\cos \left( {\dfrac{{2\pi }}{3} - 2x} \right)\) \( + \dfrac{1}{2}\cos \left( {\dfrac{{2\pi }}{3} + 2x} \right)\) \( + \dfrac{1}{2}\cos \left( {\dfrac{{4\pi }}{3} - 2x} \right)\) \( + \dfrac{1}{2}\cos \left( {\dfrac{{4\pi }}{3} + 2x} \right)\) \( + \cos 2x\)
Do đó \(y' = \dfrac{1}{2}.\left( { - 2} \right).\left[ { - \sin \left( {\dfrac{{2\pi }}{3} - 2x} \right)} \right]\) \( + \dfrac{1}{2}.2.\left[ { - \sin \left( {\dfrac{{2\pi }}{3} + 2x} \right)} \right]\) \( + \dfrac{1}{2}.\left( { - 2} \right).\left[ { - \sin \left( {\dfrac{{4\pi }}{3} - 2x} \right)} \right]\) \( + \dfrac{1}{2}.2.\left[ { - \sin \left( {\dfrac{{4\pi }}{3} + 2x} \right)} \right]\) \( - 2\sin 2x\)
\(=\sin \left ( \dfrac{2\pi }{3}-2x \right ) - \sin \left ( \dfrac{2\pi }{3}+2x \right )+ \sin \left ( \dfrac{4\pi }{3}-2x \right )\) \(- \sin \left ( \dfrac{4\pi }{3}+2x \right )- 2\sin 2x \)
\(= 2\cos \dfrac{2\pi }{3}.\sin(-2x) + 2\cos \dfrac{4\pi }{3}. \sin (-2x) - 2\sin 2x \)
\(= \sin 2x + \sin 2x - 2\sin 2x = 0\),
(Vì \(\cos \dfrac{2\pi }{3}\) = \(\cos \dfrac{4\pi }{3}\) = \( -\dfrac{1}{2}\).)
Vậy \(y' = 0\) với mọi \(x\), do đó \(y'\) không phụ thuộc vào \(x\).
