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

解

導関数

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

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

解答ステップ

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

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

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

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

簡素化

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

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