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

Lösung

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

\frac{- 4 x ^{2} + 8 x + 8}{( 2 + x ^{2} ) ^{2}}

Schritte zur Lösung

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

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

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

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

Multipliziere aus

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

= \frac{- 4 x ^{2} + 8 x + 8}{( 2 + x ^{2} ) ^{2}}

f (x) = ln (3 x + 2)

d er i v a t i v e \frac{24}{x ^{2} + 12}

d er i v a t i v e x 3600 - x^{2}

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

\int \frac{1}{x 1 + x} d x