Cho hàm số \(g\left( t \right) = {\cos ^2}2t\). Tính \(g'''\left( { - \frac{\pi }{2}} \right);g'''\left( { - \frac{\pi }{{24}}} \right);g'''\left( {\frac{{2\pi }}{3}} \right)\)
Ta có: \(g\left( t \right) = {\cos ^2}2t = \frac{{1 + \cos 4t}}{2}\)
Suy ra:
\(\begin{array}{l}
g'\left( t \right) = \frac{1}{2}.\left( { - 4\sin 4t} \right) = - 2\sin 4t\\
g''\left( t \right) = - 8\cos 4t\\
g'''\left( t \right) = 32\sin 4t
\end{array}\)
Do đó:
\(\begin{array}{l}
g'''\left( { - \frac{\pi }{2}} \right) = 32\sin \left( { - \frac{{4\pi }}{2}} \right) = 0\\
g'''\left( { - \frac{\pi }{{24}}} \right) = 32\sin \left( { - \frac{{4\pi }}{{24}}} \right) = - 16\\
g'''\left( {\frac{{2\pi }}{3}} \right) = 32\sin \left( {\frac{{8\pi }}{3}} \right) = 16\sqrt 3
\end{array}\)
-- Mod Toán 11