derivative (1+sqrt(t))(2t^2-5)

解

導関数

(1 + t) (2 t^{2} - 5)

\frac{10 t ^{2} + 8 t t - 5}{2 t}

解答ステップ

\frac{d}{d t} ((1 + t) (2 t^{2} - 5))

積の法則を適用:

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

f = 1 + t, g = 2 t^{2} - 5

= \frac{d}{d t} (1 + t) (2 t^{2} - 5) + \frac{d}{d t} (2 t^{2} - 5) (1 + t)

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

\frac{d}{d t} (2 t^{2} - 5) = 4 t

= \frac{1}{2 t} (2 t^{2} - 5) + 4 t (1 + t)

簡素化

\frac{1}{2 t} (2 t^{2} - 5) + 4 t (1 + t) : \frac{10 t ^{2} + 8 t t - 5}{2 t}

= \frac{10 t ^{2} + 8 t t - 5}{2 t}