derivative y=(5t-1)(6t-3)^{-1}

解

導関数

y = (5 t - 1) (6 t - 3)^{- 1}

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

解答ステップ

\frac{d}{d t} ((5 t - 1) (6 t - 3)^{- 1})

積の法則を適用:

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

f = 5 t - 1, g = (6 t - 3)^{- 1}

= \frac{d}{d t} (5 t - 1) (6 t - 3)^{- 1} + \frac{d}{d t} ((6 t - 3)^{- 1}) (5 t - 1)

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

\frac{d}{d t} ((6 t - 3)^{- 1}) = - \frac{2}{3 ( 2 t - 1 ) ^{2}}

= 5 (6 t - 3)^{- 1} + (- \frac{2}{3 ( 2 t - 1 ) ^{2}}) (5 t - 1)

簡素化

5 (6 t - 3)^{- 1} + (- \frac{2}{3 ( 2 t - 1 ) ^{2}}) (5 t - 1) : - \frac{1}{( 2 t - 1 ) ^{2}}

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