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

解

導関数

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

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

解答ステップ

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

積の法則を適用:

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

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

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

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

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

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

簡素化

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

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