Cho hàm số \(f(x) = \sin 3x\). Tính \(f''\left( { - \frac{\pi }{2}} \right),f''\left( 0 \right),f''\left( {\frac{\pi }{{18}}} \right)\)
Ta có: \(f'\left( x \right) = 3\cos 3x \Rightarrow f''\left( x \right) = - 9\sin 3x\)
Do đó:
\(f''\left( { - \frac{\pi }{2}} \right) = - 9\sin \left( { - \frac{{3\pi }}{2}} \right) = - 9\)
\(f''\left( 0 \right) = - 9\sin \left( 0 \right) = 0\)
\(f\prime \prime \left( {\frac{\pi }{{18}}} \right) = - 9\sin \left( {\frac{\pi }{6}} \right) = \frac{{ - 9}}{2}\)
-- Mod Toán 11