derivative of (x+1^2(x^2+1)^{-3})

解

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

\frac{- 4 x ^{3} - 10 x ^{2} - 4 x + 2}{( x ^{2} + 1 ) ^{4}}

解答ステップ

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

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (x + 1)^{2}, g = (x^{2} + 1)^{- 3}

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

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

\frac{d}{d x} ((x^{2} + 1)^{- 3}) = - \frac{6 x}{( x ^{2} + 1 ) ^{4}}

= 2 (x + 1) (x^{2} + 1)^{- 3} + (- \frac{6 x}{( x ^{2} + 1 ) ^{4}}) (x + 1)^{2}

簡素化

2 (x + 1) (x^{2} + 1)^{- 3} + (- \frac{6 x}{( x ^{2} + 1 ) ^{4}}) (x + 1)^{2} : \frac{- 4 x ^{3} - 10 x ^{2} - 4 x + 2}{( x ^{2} + 1 ) ^{4}}

= \frac{- 4 x ^{3} - 10 x ^{2} - 4 x + 2}{( x ^{2} + 1 ) ^{4}}