derivative of (4x+sqrt(1+x)/(x^2-1))

Lösung

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

\frac{- 8 x ^{2} - 3 x x + 1 - x + 1 - 8}{2 ( x ^{2} - 1 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

\frac{( 4 + \frac{1}{2 1 + x} ) ( x ^{2} - 1 ) - 2 x ( 4 x + 1 + x )}{( x ^{2} - 1 ) ^{2}} : \frac{- 8 x ^{2} - 3 x x + 1 - x + 1 - 8}{2 ( x ^{2} - 1 ) ^{2}}

= \frac{- 8 x ^{2} - 3 x x + 1 - x + 1 - 8}{2 ( x ^{2} - 1 ) ^{2}}

d er i v a t i v e ln (sec (5 x) + tan (5 x))

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

n = 1 \sum \infty 8 (- \frac{1}{4})^{n}

x \to 0 + lim (2 x ln (x))

x^{2} + 4, - x^{2} - 1, x = - 2, x = 3