derivative f(v)=(v^{-3}+6v^{-2})^3

Lösung

ableitung von

f (v) = (v^{- 3} + 6 v^{- 2})^{3}

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

Schritte zur Lösung

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

Wende die Kettenregel an:

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

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

\frac{d}{d v} (v^{- 3} + 6 v^{- 2}) = - \frac{3}{v ^{4}} - \frac{12}{v ^{3}}

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

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

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

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