与えられた関数 $x(t)$ について、その一階微分 $\frac{dx}{dt}$ と二階微分 $\frac{d^2x}{dt^2}$ を求める問題です。$a, b, c, d$ は全て定数です。

1. 問題の内容

与えられた関数 $x(t)$ について、その一階微分 $\frac{dx}{dt}$ と二階微分 $\frac{d^2x}{dt^2}$ を求める問題です。$a, b, c, d$ は全て定数です。

2. 解き方の手順

(1) $x = at^2 + bt + c + \frac{d}{t}$

一階微分：

$\frac{dx}{dt} = \frac{d}{dt}(at^2 + bt + c + dt^{-1}) = 2at + b - dt^{-2} = 2at + b - \frac{d}{t^2}$

二階微分：

$\frac{d^2x}{dt^2} = \frac{d}{dt}(2at + b - dt^{-2}) = 2a + 2dt^{-3} = 2a + \frac{2d}{t^3}$

(2) $x = a \sin(bt + c) + d \cos(bt + c)$

一階微分：

$\frac{dx}{dt} = \frac{d}{dt}(a \sin(bt + c) + d \cos(bt + c)) = ab \cos(bt + c) - bd \sin(bt + c)$

二階微分：

$\frac{d^2x}{dt^2} = \frac{d}{dt}(ab \cos(bt + c) - bd \sin(bt + c)) = -ab^2 \sin(bt + c) - b^2d \cos(bt + c) = -b^2(a \sin(bt + c) + d \cos(bt + c)) = -b^2x$

(3) $x = ae^{bt+c} + ce^{-(bt+c)}$

一階微分：

$\frac{dx}{dt} = \frac{d}{dt}(ae^{bt+c} + ce^{-(bt+c)}) = abe^{bt+c} - bce^{-(bt+c)}$

二階微分：

$\frac{d^2x}{dt^2} = \frac{d}{dt}(abe^{bt+c} - bce^{-(bt+c)}) = ab^2e^{bt+c} + b^2ce^{-(bt+c)} = b^2(ae^{bt+c} + ce^{-(bt+c)}) = b^2x$

(4) $x = a \log(bt + c)$

一階微分：

$\frac{dx}{dt} = \frac{d}{dt}(a \log(bt + c)) = a \cdot \frac{1}{bt + c} \cdot b = \frac{ab}{bt + c}$

二階微分：

$\frac{d^2x}{dt^2} = \frac{d}{dt}(\frac{ab}{bt + c}) = ab \cdot \frac{d}{dt}(bt + c)^{-1} = ab \cdot (-1)(bt + c)^{-2} \cdot b = -\frac{ab^2}{(bt + c)^2}$

3. 最終的な答え

(1)

$\frac{dx}{dt} = 2at + b - \frac{d}{t^2}$

$\frac{d^2x}{dt^2} = 2a + \frac{2d}{t^3}$

(2)

$\frac{dx}{dt} = ab \cos(bt + c) - bd \sin(bt + c)$

$\frac{d^2x}{dt^2} = -b^2(a \sin(bt + c) + d \cos(bt + c)) = -b^2x$

(3)

$\frac{dx}{dt} = abe^{bt+c} - bce^{-(bt+c)}$

$\frac{d^2x}{dt^2} = b^2(ae^{bt+c} + ce^{-(bt+c)}) = b^2x$

(4)

$\frac{dx}{dt} = \frac{ab}{bt + c}$

$\frac{d^2x}{dt^2} = -\frac{ab^2}{(bt + c)^2}$