derivative of (x^2-x+1/(x^2+x+1))

解

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

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

解答ステップ

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

商の法則を適用:

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

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

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

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

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

拡張

(2 x - 1) (x^{2} + x + 1) - (2 x + 1) (x^{2} - x + 1) : 2 x^{2} - 2

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