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

Lösung

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

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

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

y^{^{'}} - \frac{9}{x} y = \frac{y ^{5}}{x ^{20}}

y^{^{''}} + 16 y = sin (4 t)

x \to 0 + lim (\frac{2}{x ^{2}})

\frac{d}{d x} ((x + 1) \cdot e^{a r c t a n (x)})

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