derivative f(x)=(2x^3-3x+1)/x

Lösung

ableitung von

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

\frac{4 x ^{3} - 1}{x ^{2}}

Schritte zur Lösung

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

Wende die Quotientenregel an:

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

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

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

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

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

Vereinfache

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

= \frac{4 x ^{3} - 1}{x ^{2}}

s l o p e y = x^{2} - 2

l a pl a ce t r an s f or m e^{- t} (t^{2})

d er i v a t i v e f (x) = x + 7 x

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

\frac{\partial}{\partial x} ((x + y)^{5} + (x - y)^{5})