(1/(1+x^2))^'

解

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

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

解答ステップ

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

指数の規則を適用する:

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

= ((1 + x^{2})^{- 1})^{^{'}}

連鎖法則を適用:

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

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

(1 + x^{2})^{'} = 2 x

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

簡素化

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

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