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

解

導関数

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

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

解答ステップ

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

商の法則を適用:

(\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}}

簡素化

\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}}