derivative of (x^4-3x^2/((x^2-1)^2))

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (x^{4} - 3 x^{2}) = 4 x^{3} - 6 x

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

= \frac{( 4 x ^{3} - 6 x ) ( x ^{2} - 1 ) ^{2} - 4 x ( x ^{2} - 1 ) ( x ^{4} - 3 x ^{2} )}{( ( x ^{2} - 1 ) ^{2} ) ^{2}}

Vereinfache

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

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

a re a x = \frac{( y - 2 ) ^{2}}{2}, y = 6 - x

t an g e n t f (x) = \frac{x}{( 3 x - 2 ) ^{6}}, at x = 1

\int x^{5} e^{x^{3} - 7} d x

n = 1 \sum \infty \frac{3}{4 n ^{2} - 1}

d er i v a t i v e 1 + cos (2 x)