derivative ((8u^2))/((u^2+u)^3)

Lösung

ableitung von

\frac{( 8 u ^{2} )}{( u ^{2} + u ) ^{3}}

\frac{8 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

Schritte zur Lösung

\frac{d}{d u} (\frac{8 u ^{2}}{( u ^{2} + u ) ^{3}})

Entferne die Konstante:

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

= 8 \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}}

= 8 \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)

= 8 \cdot \frac{2 u ( u ^{2} + u ) ^{3} - 3 ( u ^{2} + u ) ^{2} ( 2 u + 1 ) u ^{2}}{( ( u ^{2} + u ) ^{3} ) ^{2}}

Vereinfache

8 \cdot \frac{2 u ( u ^{2} + u ) ^{3} - 3 ( u ^{2} + u ) ^{2} ( 2 u + 1 ) u ^{2}}{( ( u ^{2} + u ) ^{3} ) ^{2}} : \frac{8 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

= \frac{8 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

t an g e n t f (x) = x^{2} + x, at x = - 1

\frac{d}{d x} (\frac{2901}{1 + 4.76 e ^{- 0.3 x}})

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n = 0 \sum \infty \frac{1}{6 ^{n}}