derivative f(t)=(1+sqrt(t))(6t^2-9)

解

導関数

f (t) = (1 + t) (6 t^{2} - 9)

\frac{30 t ^{2} + 24 t t - 9}{2 t}

解答ステップ

\frac{d}{d t} ((1 + t) (6 t^{2} - 9))

積の法則を適用:

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

f = 1 + t, g = 6 t^{2} - 9

= \frac{d}{d t} (1 + t) (6 t^{2} - 9) + \frac{d}{d t} (6 t^{2} - 9) (1 + t)

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

\frac{d}{d t} (6 t^{2} - 9) = 12 t

= \frac{1}{2 t} (6 t^{2} - 9) + 12 t (1 + t)

簡素化

\frac{1}{2 t} (6 t^{2} - 9) + 12 t (1 + t) : \frac{30 t ^{2} + 24 t t - 9}{2 t}

= \frac{30 t ^{2} + 24 t t - 9}{2 t}