Bài 1. a) Cho f(x)=(x+10)6. Tính f”(2).
b) Cho f(x) = \sin 3x. Tính f” \left ( -\frac{\pi }{2} \right ) , f”(0), f” \left ( \frac{\pi }{18} \right ).
a) Ta có f'(x) = 6(x + 10)’.(x + 10)^5,
f”(x) = 6.5(x + 10)’.(x + 10)^4= 30.(x + 10)^4
Suy ra f”(2) = 30.(2 + 10)^4 = 622 080.
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b) Ta có f'(x) = (3x)’.\cos 3x = 3\cos 3x,
f”(x) = 3.[-(3x)’.\sin 3x] = -9\sin 3x.
Suy ra f”\left ( -\frac{\pi }{2} \right ) = -9\sin \left ( -\frac{3\pi }{2} \right ) = -9;
f”(0) = -9sin0 = 0;
f” \left ( \frac{\pi }{18} \right ) = -9\sin\left ( \frac{\pi }{6} \right ) = -\frac{9}{2}.