Bài 33: Đạo hàm cấp hai

• $y' = 8x^3 - 15x^2 + 6x$.

• $y'' = (8x^3 - 15x^2 + 6x)' = 24x^2 - 30x + 6$.

• $y' = -\dfrac{1}{x^2} = -x^{-2}$.

• $y'' = -(-2) x^{-3} = \dfrac{2}{x^3}$.

$y' = 5x^4 - 6x \Rightarrow y'' = 20x^3 - 6$.

$y' = \dfrac{-3}{(x-1)^2} = -3(x-1)^{-2} \Rightarrow y'' = 6(x-1)^{-3} = \dfrac{6}{(x-1)^3}$.

• $y' = 2 \cos(2x)$.

• $y'' = 2 (-2\sin(2x)) = -4 \sin(2x)$.

• $y' = 2 \cos x (-\sin x) = -\sin(2x)$.

• $y'' = -2 \cos(2x)$.

$y' = -3\sin(3x) \Rightarrow y'' = -9\cos(3x)$.

$y' = \cos x - \sin x \Rightarrow y'' = -\sin x - \cos x = -y$.

• $v(t) = s'(t) = 3t^2 - 12t + 15$.

• $a(t) = s''(t) = 6t - 12$.

• Gia tốc tại $t = 4$ s: $a(4) = 6(4) - 12 = 12 \text{ m/s}^2$.

$a(t) = 6t - 12 = 0 \Leftrightarrow t = 2$ s.

$a(t) = -16\pi^2 \cos(2\pi t + \pi/3) \Rightarrow a(0.5) = -16\pi^2 \cos(4\pi/3) = 8\pi^2$.

$v(t) = 6t^2 - 18t + 12 = 0 \Leftrightarrow t = 1$ hoặc $t = 2$. $a(t) = 12t - 18$. TH1 ($t=1$): $a = -6$. TH2 ($t=2$): $a = 6$.

• $y' = \cos x$.

• $y'' = -\sin x = -y$.

• Suy ra $y'' + y = -y + y = 0$ (đpcm).

• $y' = \dfrac{1-x}{\sqrt{2x-x^2}} = \dfrac{1-x}{y}$.

• $y'' = \dfrac{-y - (1-x)y'}{y^2} = \dfrac{-y - \dfrac{(1-x)^2}{y}}{y^2} = \dfrac{-y^2 - (1-2x+x^2)}{y^3} = \dfrac{-(2x-x^2) - 1 + 2x - x^2}{y^3} = -\dfrac{1}{y^3}$.

• Suy ra $y^3 y'' = -1 \Rightarrow y^3 y'' + 1 = 0$ (đpcm).

$y' = -\sin x \Rightarrow y'' = -\cos x = -y \Rightarrow y'' + y = 0$.

$y' = 3x^2 - 3 \Rightarrow y'' = 6x \Rightarrow y'' - 6x = 0$.