derivative of ((2x^3+ax+a)/(x^2-1))

Lösung

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

\frac{2 x ^{4} - a x ^{2} - 6 x ^{2} - 2 a x - a}{( x ^{2} - 1 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} (2 x^{3} + a x + a) = 6 x^{2} + a

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

= \frac{( 6 x ^{2} + a ) ( x ^{2} - 1 ) - 2 x ( 2 x ^{3} + a x + a )}{( x ^{2} - 1 ) ^{2}}

Multipliziere aus

(6 x^{2} + a) (x^{2} - 1) - 2 x (2 x^{3} + a x + a) : 2 x^{4} - a x^{2} - 6 x^{2} - 2 a x - a

= \frac{2 x ^{4} - a x ^{2} - 6 x ^{2} - 2 a x - a}{( x ^{2} - 1 ) ^{2}}

\int 4 x^{2} e^{x^{3}} d x

\frac{d}{d x} (\frac{x ^{2}}{8 - 8 x})

x \to 0 lim (sin (x) + cos (x))

d er i v a t i v e y = \frac{x ^{3} + 8}{x}

f (u) = \frac{1}{u}