derivative (9u^2)/((u^2+u)^3)

Lösung

ableitung von

\frac{9 u ^{2}}{( u ^{2} + u ) ^{3}}

\frac{9 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

Schritte zur Lösung

\frac{d}{d u} (\frac{9 u ^{2}}{( u ^{2} + u ) ^{3}})

Entferne die Konstante:

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

= 9 \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}}

= 9 \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)

= 9 \cdot \frac{2 u ( u ^{2} + u ) ^{3} - 3 ( u ^{2} + u ) ^{2} ( 2 u + 1 ) u ^{2}}{( ( u ^{2} + u ) ^{3} ) ^{2}}

Vereinfache

9 \cdot \frac{2 u ( u ^{2} + u ) ^{3} - 3 ( u ^{2} + u ) ^{2} ( 2 u + 1 ) u ^{2}}{( ( u ^{2} + u ) ^{3} ) ^{2}} : \frac{9 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

= \frac{9 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

\frac{d}{d x} (a^{2} + b^{2})

d er i v a t i v e f (x) = (sin^{2} (x) + 1)^{5}

\frac{d}{d x} (\frac{- 1 - x}{x - 1})

\int (x^{2} + 5)^{2} d x

t a y l or x, 25