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

Lösung

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

- \frac{6 ( x ^{2} + 1 )}{( x ^{2} - 3 ) ^{3}}

Schritte zur Lösung

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

Entferne die Konstante:

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

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

Wende die Quotientenregel an:

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

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

\frac{d x}{d x} = 1

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

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

Vereinfache

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

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

d er i v a t i v e y = e^{- x} cos (2 x)

in v erse l a pl a ce \frac{1}{s ^{4} + s ^{2}}

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

\int - 3 tan (3 y) d y

\frac{d}{d x} (\frac{x - y}{sin ( x ) - sin ( y )})