derivative of 1/(4x^2+1)

解

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

- \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}}

解答ステップ

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

指数の規則を適用する:

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

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

連鎖法則を適用:

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

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

\frac{d}{d x} (4 x^{2} + 1) = 8 x

= - \frac{1}{( 4 x ^{2} + 1 ) ^{2}} \cdot 8 x

簡素化

- \frac{1}{( 4 x ^{2} + 1 ) ^{2}} \cdot 8 x : - \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}}

= - \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}}