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

Lösung

ableitung von

\frac{2 x ^{2} + x - 3}{- 3 x ^{2} - 4 x + 2}

\frac{- 5 x ^{2} - 10 x - 10}{( - 3 x ^{2} - 4 x + 2 ) ^{2}}

Schritte zur Lösung

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

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

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

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

= \frac{( 4 x + 1 ) ( - 3 x ^{2} - 4 x + 2 ) - ( - 6 x - 4 ) ( 2 x ^{2} + x - 3 )}{( - 3 x ^{2} - 4 x + 2 ) ^{2}}

Multipliziere aus

(4 x + 1) (- 3 x^{2} - 4 x + 2) - (- 6 x - 4) (2 x^{2} + x - 3) : - 5 x^{2} - 10 x - 10

= \frac{- 5 x ^{2} - 10 x - 10}{( - 3 x ^{2} - 4 x + 2 ) ^{2}}

s im pl i f y (e^{- x} - e^{5}) (e^{x} + e^{- 3})

in v erse l a pl a ce \frac{400}{s ( s + 2 ) ^{3}}

\frac{\partial}{\partial x} (6 x^{\frac{y}{z}})

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

\int_{0}^{2} ∣ x ∣ d x