derivative f(v)=(v^{-3}+3v^{-2})^3

解

導関数

f (v) = (v^{- 3} + 3 v^{- 2})^{3}

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

解答ステップ

\frac{d}{d v} ((v^{- 3} + 3 v^{- 2})^{3})

連鎖法則を適用:

3 (v^{- 3} + 3 v^{- 2})^{2} \frac{d}{d v} (v^{- 3} + 3 v^{- 2})

= 3 (v^{- 3} + 3 v^{- 2})^{2} \frac{d}{d v} (v^{- 3} + 3 v^{- 2})

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

= 3 (v^{- 3} + 3 v^{- 2})^{2} (- \frac{3}{v ^{4}} - \frac{6}{v ^{3}})

(v^{- 3} + 3 v^{- 2})^{2} = (\frac{1}{v ^{3}} + \frac{3}{v ^{2}})^{2}

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