d/(dt)((x)4(x^2+1))

解

\frac{d}{d t} ((t) 4 (t^{2} + 1))

4 (3 t^{2} + 1)

解答ステップ

\frac{d}{d t} ((t) \cdot 4 (t^{2} + 1))

定数を除去:

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

= 4 \frac{d}{d t} (t (t^{2} + 1))

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = t^{2} + 1

= 4 (\frac{d t}{d t} (t^{2} + 1) + \frac{d}{d t} (t^{2} + 1) t)

\frac{d t}{d t} = 1

\frac{d}{d t} (t^{2} + 1) = 2 t

= 4 (1 \cdot (t^{2} + 1) + 2 tt)

簡素化

4 (1 \cdot (t^{2} + 1) + 2 tt) : 4 (3 t^{2} + 1)

= 4 (3 t^{2} + 1)