derivative f(x)=(4x^3-2x+1)/(x^2)

解

導関数

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

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

解答ステップ

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

商の法則を適用:

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

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

\frac{d}{d x} (4 x^{3} - 2 x + 1) = 12 x^{2} - 2

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

= \frac{( 12 x ^{2} - 2 ) x ^{2} - 2 x ( 4 x ^{3} - 2 x + 1 )}{( x ^{2} ) ^{2}}

簡素化

\frac{( 12 x ^{2} - 2 ) x ^{2} - 2 x ( 4 x ^{3} - 2 x + 1 )}{( x ^{2} ) ^{2}} : \frac{2 ( 2 x ^{3} + x - 1 )}{x ^{3}}

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