d/(dt)((r^3-2r^2+6t+2)3)

解

\frac{d}{d t} ((r (t)^{3} - 2 r (t)^{2} + 6 t + 2) 3)

3 (3 r^{2} \frac{d r}{d t} - 4 r \frac{d r}{d t} + 6)

解答ステップ

\frac{d}{d t} ((r^{3} - 2 r^{2} + 6 t + 2) \cdot 3)

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

= 3 \frac{d}{d t} (r^{3} - 2 r^{2} + 6 t + 2)

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

= 3 (\frac{d}{d t} (r^{3}) - \frac{d}{d t} (2 r^{2}) + \frac{d}{d t} (6 t) + \frac{d}{d t} (2))

\frac{d}{d t} (r^{3}) = 3 r^{2} \frac{d r}{d t}

\frac{d}{d t} (2 r^{2}) = 4 r \frac{d r}{d t}

\frac{d}{d t} (6 t) = 6

\frac{d}{d t} (2) = 0

= 3 (3 r^{2} \frac{d r}{d t} - 4 r \frac{d r}{d t} + 6 + 0)

簡素化

= 3 (3 r^{2} \frac{d r}{d t} - 4 r \frac{d r}{d t} + 6)