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

解

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

- \frac{16}{x ^{5}} - \frac{1}{x ^{2}}

解答ステップ

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

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (\frac{4}{x ^{4}}) = - \frac{16}{x ^{5}}

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

\frac{d}{d x} (2) = 0

= - \frac{16}{x ^{5}} - \frac{1}{x ^{2}} + 0

簡素化

= - \frac{16}{x ^{5}} - \frac{1}{x ^{2}}