derivative of (x^2-6x+1/((x^2-1)^2))

Lösung

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

\frac{2 ( - x ^{3} + 9 x ^{2} - 3 x + 3 )}{( x ^{2} - 1 ) ^{3}}

Schritte zur Lösung

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

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

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

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

Vereinfache

\frac{( 2 x - 6 ) ( x ^{2} - 1 ) ^{2} - 4 x ( x ^{2} - 1 ) ( x ^{2} - 6 x + 1 )}{( ( x ^{2} - 1 ) ^{2} ) ^{2}} : \frac{2 ( - x ^{3} + 9 x ^{2} - 3 x + 3 )}{( x ^{2} - 1 ) ^{3}}

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

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

\int_{0}^{1} (3 1 + 7 x) d x

\int x^{2} - 9 x + 3 d x

n = 1 \sum \infty \frac{9}{6 + 9 ^{n}}

\int \frac{1}{1 - 9 x ^{2}} d x