derivative f(x)=(x^3+2x^2+3x+4)/x

Lösung

ableitung von

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

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

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

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

\frac{d}{d x} (9 - x)

d er i v a t i v e 4 π x^{2}

n = 1 \sum \infty \frac{1}{n ^{2} + 2 n}

in v erse l a pl a ce \frac{1}{s ^{2}} - \frac{1}{s} + \frac{1}{s - 2}

t an g e n t f (x) = (1 - x) (x^{2} - 3)^{2}