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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = e^{2 x}

\int x \cdot 5^{x} d x

in v erse l a pl a ce \frac{e ^{- s}}{s - 1}

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

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