derivative ((2x+1)(x^3+3))/x

Lösung

ableitung von

\frac{( 2 x + 1 ) ( x ^{3} + 3 )}{x}

\frac{6 x ^{4} + 2 x ^{3} - 3}{x ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

\frac{d}{d x} ((2 x + 1) (x^{3} + 3)) = 8 x^{3} + 3 x^{2} + 6

\frac{d x}{d x} = 1

= \frac{( 8 x ^{3} + 3 x ^{2} + 6 ) x - 1 \cdot ( 2 x + 1 ) ( x ^{3} + 3 )}{x ^{2}}

Vereinfache

\frac{( 8 x ^{3} + 3 x ^{2} + 6 ) x - 1 \cdot ( 2 x + 1 ) ( x ^{3} + 3 )}{x ^{2}} : \frac{6 x ^{4} + 2 x ^{3} - 3}{x ^{2}}

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

\int - 2 e^{2 x} d x

\int \frac{e ^{4 x}}{25 - e ^{2 x}} d x

\frac{\partial}{\partial x} (ln (x^{2} + 2 y^{2}))

\frac{d y}{d x} = \frac{y}{( 1 + x ) ( 2 + x )}

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