derivative of x/(sqrt(1+x^2+x^4))

解

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

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

解答ステップ

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

商の法則を適用:

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

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

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

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

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

簡素化

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

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