derivative of (2x^2/((x^2+x)^3))

Lösung

\frac{d}{d x} (\frac{2 x ^{2}}{( x ^{2} + x ) ^{3}})

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

Schritte zur Lösung

\frac{d}{d x} (\frac{2 x ^{2}}{( x ^{2} + x ) ^{3}})

Entferne die Konstante:

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

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

t an g e n t f (x) = 4 x^{2} - 2 x + 4, at x = 0

t an g e n t f (x) = - 2 x^{4} + 5 x^{2} - 3, (1, 0)

\frac{d ^{2}}{d x ^{2}} (\frac{1}{x})

\frac{\partial}{\partial x} (\frac{y}{2 x + y})

\int \frac{sin ( t )}{t} d t