derivative f(x)=(-x^2-1)/((x^2-1)^2)

Lösung

ableitung von

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

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

Schritte zur Lösung

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

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

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

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

Vereinfache

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

\int tan^{3} (19 π x) d x

t a y l or \frac{1}{1 + x}

\int x - y d x

x \to \infty lim (x^{3} sin (\frac{1}{x}))

d er i v a t i v e arctan (6 y^{5} + 1)