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

Lösung

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

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

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

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

Vereinfache

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

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

\int_{0}^{1} \frac{1}{x ^{- 2}} d x

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

t^{^{'}}

\frac{\partial}{\partial x} (\frac{3 x}{4})

t an g e n t f (x) = 4 - 3 x^{2}, (- 2, - 8)