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

解

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

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

解答ステップ

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

指数の規則を適用する:

\frac{1}{a} = a^{- 1}

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

連鎖法則を適用:

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

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

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

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

簡素化

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

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