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

解

導関数

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

\frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}

解答ステップ

\frac{d}{d t} ((3 t - 1)^{4} (2 t + 1)^{- 3})

積の法則を適用:

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

f = (3 t - 1)^{4}, g = (2 t + 1)^{- 3}

= \frac{d}{d t} ((3 t - 1)^{4}) (2 t + 1)^{- 3} + \frac{d}{d t} ((2 t + 1)^{- 3}) (3 t - 1)^{4}

\frac{d}{d t} ((3 t - 1)^{4}) = 12 (3 t - 1)^{3}

\frac{d}{d t} ((2 t + 1)^{- 3}) = - \frac{6}{( 2 t + 1 ) ^{4}}

= 12 (3 t - 1)^{3} (2 t + 1)^{- 3} + (- \frac{6}{( 2 t + 1 ) ^{4}}) (3 t - 1)^{4}

簡素化

12 (3 t - 1)^{3} (2 t + 1)^{- 3} + (- \frac{6}{( 2 t + 1 ) ^{4}}) (3 t - 1)^{4} : \frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}

= \frac{12 ( 3 t - 1 ) ^{3}}{( 2 t + 1 ) ^{3}} - \frac{6 ( 3 t - 1 ) ^{4}}{( 2 t + 1 ) ^{4}}