derivative of sec(x-1/(x^3)+5)

解

\frac{d}{d x} (sec (x) - \frac{1}{x ^{3}} + 5)

\frac{3}{x ^{4}} + sec (x) tan (x)

解答ステップ

\frac{d}{d x} (sec (x) - \frac{1}{x ^{3}} + 5)

和/差の法則を適用:

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

= \frac{d}{d x} (sec (x)) - \frac{d}{d x} (\frac{1}{x ^{3}}) + \frac{d}{d x} (5)

\frac{d}{d x} (sec (x)) = sec (x) tan (x)

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

\frac{d}{d x} (5) = 0

= sec (x) tan (x) - (- \frac{3}{x ^{4}}) + 0

簡素化

= \frac{3}{x ^{4}} + sec (x) tan (x)