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 \(f”(2) = 30.(2 + 10)^4 = 622 080\).
b) Ta có \(f'(x) = (3x)’.\cos 3x = 3\cos 3x\),
\(f”(x) = 3.[-(3x)’.\sin 3x] = -9\sin 3x\).
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Suy ra \(f”\left ( -\frac{\pi }{2} \right ) = -9\sin \left ( -\frac{3\pi }{2} \right ) = -9\);
\(f”(0) = -9sin0 = 0\);
\(f” \left ( \frac{\pi }{18} \right ) = -9\sin\left ( \frac{\pi }{6} \right ) = -\frac{9}{2}\).
Bài 2: Tìm đạo hàm cấp hai của các hàm số sau:
a) \(y = \frac{1}{1-x}\);
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b) \(y = \frac{1}{\sqrt{1-x}}\);
c) \(y = \tan x\);
d) \(y = \cos^2x\) .
a) \(y’ = -\frac{(1-x)’}{(1-x)^{2}}\) = \( \frac{1}{(1-x)^{2}}\), \(y” = -\frac{[(1-x)^{2}]’}{(1-x)^{4}} = – \frac{2.(-1)(1-x)}{(1-x)^{4}}\) = \( \frac{2}{(1-x)^{3}}\).
b) \(y’ = -\frac{(\sqrt{1-x})’}{1-x}\) = \( \frac{1}{2(1-x)\sqrt{1-x}}\);
\(y” = -\frac{1}{2}\frac{[(1-x)\sqrt{1-x}]’}{(1-x)^{3}}\) = \( -\frac{1}{2}\frac{-\sqrt{1-x}+(1-x)\frac{-1}{2\sqrt{1-x}}}{(1-x)^{3}}\) = \( \frac{3}{4(1-x)^{2}\sqrt{1-x}}\).
c) \(y’ = \frac{1}{cos^{2}x}\); \(y” = -\frac{(cos^{2}x)’}{cos^{4}x} = \frac{2cosx.sinx}{cos^{4}x}\) = \( \frac{2sinx}{cos^{3}x}\).
d) \(y’ = 2cosx.(cosx)’ = 2cosx.(-sinx) \)
\(= – 2sinx.cosx = -sin2x\),
\(y” = -(2x)’.cos2x = -2cos2x\).
