次の微分を計算します。 $\frac{d}{dt} \left[ \left( 5 + \frac{d}{dt} \right) \left( \sin(5t-2) - \cos(5t-2) \right) \right]$

1. 問題の内容

次の微分を計算します。

$\frac{d}{dt} \left[ \left( 5 + \frac{d}{dt} \right) \left( \sin(5t-2) - \cos(5t-2) \right) \right]$

2. 解き方の手順

まず、内側の微分 $\frac{d}{dt} \left( \sin(5t-2) - \cos(5t-2) \right)$ を計算します。

$\frac{d}{dt} \sin(5t-2) = \cos(5t-2) \cdot \frac{d}{dt}(5t-2) = 5\cos(5t-2)$

$\frac{d}{dt} \cos(5t-2) = -\sin(5t-2) \cdot \frac{d}{dt}(5t-2) = -5\sin(5t-2)$

したがって、

$\frac{d}{dt} \left( \sin(5t-2) - \cos(5t-2) \right) = 5\cos(5t-2) - (-5\sin(5t-2)) = 5\cos(5t-2) + 5\sin(5t-2)$

次に、この結果を使って元の式を書き換えます。

$\frac{d}{dt} \left[ \left( 5 + \frac{d}{dt} \right) \left( \sin(5t-2) - \cos(5t-2) \right) \right] = \frac{d}{dt} \left[ 5(\sin(5t-2) - \cos(5t-2)) + 5\cos(5t-2) + 5\sin(5t-2) \right]$

$= \frac{d}{dt} \left[ 5\sin(5t-2) - 5\cos(5t-2) + 5\cos(5t-2) + 5\sin(5t-2) \right]$

$= \frac{d}{dt} \left[ 10\sin(5t-2) \right]$

$= 10 \cdot \frac{d}{dt} \sin(5t-2)$

$= 10 \cdot \cos(5t-2) \cdot \frac{d}{dt}(5t-2)$

$= 10 \cdot \cos(5t-2) \cdot 5$

$= 50\cos(5t-2)$

3. 最終的な答え

$50\cos(5t-2)$