derivative (u^{-2}+u^{-3})(u^5-4u^2)

Lösung

ableitung von

(u^{- 2} + u^{- 3}) (u^{5} - 4 u^{2})

3 u^{2} + 2 u + \frac{4}{u ^{2}}

Schritte zur Lösung

\frac{d}{d u} ((u^{- 2} + u^{- 3}) (u^{5} - 4 u^{2}))

Wende die Produktregel an:

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

f = u^{- 2} + u^{- 3}, g = u^{5} - 4 u^{2}

= \frac{d}{d u} (u^{- 2} + u^{- 3}) (u^{5} - 4 u^{2}) + \frac{d}{d u} (u^{5} - 4 u^{2}) (u^{- 2} + u^{- 3})

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

\frac{d}{d u} (u^{5} - 4 u^{2}) = 5 u^{4} - 8 u

= (- \frac{2}{u ^{3}} - \frac{3}{u ^{4}}) (u^{5} - 4 u^{2}) + (5 u^{4} - 8 u) (u^{- 2} + u^{- 3})

Vereinfache

(- \frac{2}{u ^{3}} - \frac{3}{u ^{4}}) (u^{5} - 4 u^{2}) + (5 u^{4} - 8 u) (u^{- 2} + u^{- 3}) : 3 u^{2} + 2 u + \frac{4}{u ^{2}}

= 3 u^{2} + 2 u + \frac{4}{u ^{2}}

y^{^{'}} = (\frac{y + 3}{4 x + 5})^{3}

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