Nếu \(f(x)=sin^3x+x^2\) thì \(f''(\frac{\pi }{2})\) bằng:
(A) 0
(B) 1
(C) -2
(D) 5
Ta có \(f'(x)=3sin^2x.cosx+2x.\)
\(\Rightarrow f''(x)=(3sin^2x.cosx)'+2\)
\(=3(2.sinx.cos^2x-sin^3x)+2\)
\(\Rightarrow f''\left ( -\frac{\pi }{2} \right )=3\left ( 2.sin\left ( -\frac{\pi }{2} \right ) .cos^2\left ( -\frac{\pi }{2} \right )-sin^3\left ( -\frac{\pi }{2} \right ) \right )+2\)
\(=3+2=5.\)
Vậy (D) là đáp án cần tìm.
-- Mod Toán 11
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