derivative of 1/(\sqrt[3]{x^2+x+1})

解

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

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

解答ステップ

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

連鎖法則を適用:

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

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

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

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

簡素化

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

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