derivative (6u^2)/((u^2+u)^3)

Lösung

ableitung von

\frac{6 u ^{2}}{( u ^{2} + u ) ^{3}}

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

Schritte zur Lösung

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

Entferne die Konstante:

(a \cdot f)^{'} = a \cdot f^{'}

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

Wende die Quotientenregel an:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d u} (u^{2}) = 2 u

\frac{d}{d u} ((u^{2} + u)^{3}) = 3 (u^{2} + u)^{2} (2 u + 1)

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

Vereinfache

6 \cdot \frac{2 u ( u ^{2} + u ) ^{3} - 3 ( u ^{2} + u ) ^{2} ( 2 u + 1 ) u ^{2}}{( ( u ^{2} + u ) ^{3} ) ^{2}} : \frac{6 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

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

l a pl a ce t r an s f or m [δ (t + 1)]

\int \frac{- x + 2}{x ^{2} - x + 1} d x

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in v erse l a pl a ce \frac{3 s ^{2}}{( s ^{2} + 4 ) ^{2}}

\int \frac{1}{- x ^{2} + 4 x + 5} d x