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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

t an g e n t \frac{2}{3} x^{3} + 4 x^{2} + 6 x - 5

\int \frac{196}{w ^{2} 49 - w ^{2}} d w

\int \frac{1}{2 x + 3} d x

\frac{\partial}{\partial y} (2 x (5 + y)^{- 1})

p a r i t y \frac{sec ( 9 x ^{2} )}{tan ( 2 x ) \cdot cos ( x )}