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

Lösung

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

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

Schritte zur Lösung

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

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

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

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

Multipliziere aus

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

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

\int 18 - x^{2} d x

\frac{d}{d x} (arccot (3 x))

d er i v a t i v e x^{\frac{5}{3}} - x^{\frac{2}{3}}

d er i v a t i v e x - 3

d er i v a t i v e f (t) = t^{\frac{2}{3}} - t^{\frac{1}{3}} + 4