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

Lösung

ableitung von

\frac{- 2 x ^{2} + x + 4}{x ^{2} + 4 x - 1}

\frac{- 9 x ^{2} - 4 x - 17}{( x ^{2} + 4 x - 1 ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

(- 4 x + 1) (x^{2} + 4 x - 1) - (2 x + 4) (- 2 x^{2} + x + 4) : - 9 x^{2} - 4 x - 17

= \frac{- 9 x ^{2} - 4 x - 17}{( x ^{2} + 4 x - 1 ) ^{2}}

\int (6 - x) d x

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x \to 0 lim ((1 - x)^{\frac{5}{x}})

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