derivative of 1/((x^2-1(x^2+3x+3)))

解

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

- \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}

解答ステップ

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

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

\frac{d}{d x} ((x^{2} - 1) (x^{2} + 3 x + 3)) = 4 x^{3} + 9 x^{2} + 4 x - 3

= - \frac{1}{( ( x ^{2} - 1 ) ( x ^{2} + 3 x + 3 ) ) ^{2}} (4 x^{3} + 9 x^{2} + 4 x - 3)

簡素化

- \frac{1}{( ( x ^{2} - 1 ) ( x ^{2} + 3 x + 3 ) ) ^{2}} (4 x^{3} + 9 x^{2} + 4 x - 3) : - \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}

= - \frac{4 x ^{3} + 9 x ^{2} + 4 x - 3}{( x ^{2} - 1 ) ^{2} ( x ^{2} + 3 x + 3 ) ^{2}}