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

Lösung

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

\frac{- 3 x ^{2} - 8 x + 8}{( x ^{2} - 2 x ) ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Multipliziere aus

3 (x^{2} - 2 x) - (2 x - 2) (4 + 3 x) : - 3 x^{2} - 8 x + 8

= \frac{- 3 x ^{2} - 8 x + 8}{( x ^{2} - 2 x ) ^{2}}

(x + x y^{2}) d x + e^{x^{2}} y d y = 0

\int_{- 2}^{x^{2} - 3 x} e^{t^{2}} d t

d er i v a t i v e y = x^{\frac{4}{5}} (x - 3)

\frac{d}{d x} (\frac{- 2 x ^{2} + 25}{25 - x ^{2}})

\frac{d}{d θ} (cos^{2} (θ) - sin^{2} (θ))