derivative of (4-2x-x^3/(3x-x^4))

Lösung

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

\frac{- x ^{6} - 6 x ^{4} + 10 x ^{3} - 12}{( 3 x - x ^{4} ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

(- 3 x^{2} - 2) (3 x - x^{4}) - (3 - 4 x^{3}) (4 - 2 x - x^{3}) : - x^{6} - 6 x^{4} + 10 x^{3} - 12

= \frac{- x ^{6} - 6 x ^{4} + 10 x ^{3} - 12}{( 3 x - x ^{4} ) ^{2}}

v o l u m e ab o u t y = 0, y = 9 - x^{2}, y = 0

t an g e n t f (x) = \frac{1}{x ^{4}}, (3, \frac{1}{81})

\int \frac{\frac{4}{x ^{3}} + 2}{\frac{1}{x ^{2}} - x} d x

d er i v a t i v e f (x) = 6 x arcsin (x)

\frac{\partial}{\partial x} (arcsin (4 x yz))