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

Lösung

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

\frac{x ^{4} + 2 x ^{3} - 5 x ^{2} + 8 x + 4}{x ^{2} ( x + 1 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} ((x - 1) (x^{2} + 4)) = 3 x^{2} - 2 x + 4

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

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

Vereinfache

\frac{( 3 x ^{2} - 2 x + 4 ) x ( x + 1 ) - ( 2 x + 1 ) ( x - 1 ) ( x ^{2} + 4 )}{( x ( x + 1 ) ) ^{2}} : \frac{x ^{4} + 2 x ^{3} - 5 x ^{2} + 8 x + 4}{x ^{2} ( x + 1 ) ^{2}}

= \frac{x ^{4} + 2 x ^{3} - 5 x ^{2} + 8 x + 4}{x ^{2} ( x + 1 ) ^{2}}

d er i v a t i v e \frac{100 p}{0.1 p + 1}

im pl i c i t \frac{d y}{d x}, 3 x^{3} + 9 x y^{2} = 5 x^{3}

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

in v erse l a pl a ce \frac{9}{( s - 2 ) ( s - 3 )}

d er i v a t i v e f (x) = (2 x) (4 x^{2} + 6 x^{7} + 8)