derivative log_{e}(sec(1/(5x)))

解

導関数

lo g_{e} (sec (\frac{1}{5 x}))

- \frac{tan ( \frac{1}{5 x} )}{5 x ^{2}}

解答ステップ

\frac{d}{d x} (lo g_{e} (sec (\frac{1}{5 x})))

簡素化

lo g_{e} (sec (\frac{1}{5 x})) : ln (sec (\frac{1}{5 x}))

= \frac{d}{d x} (ln (sec (\frac{1}{5 x})))

連鎖法則を適用:

\frac{1}{sec ( \frac{1}{5 x} )} \frac{d}{d x} (sec (\frac{1}{5 x}))

= \frac{1}{sec ( \frac{1}{5 x} )} \frac{d}{d x} (sec (\frac{1}{5 x}))

\frac{d}{d x} (sec (\frac{1}{5 x})) = - \frac{sec ( \frac{1}{5 x} ) tan ( \frac{1}{5 x} )}{5 x ^{2}}

= \frac{1}{sec ( \frac{1}{5 x} )} (- \frac{sec ( \frac{1}{5 x} ) tan ( \frac{1}{5 x} )}{5 x ^{2}})

簡素化

\frac{1}{sec ( \frac{1}{5 x} )} (- \frac{sec ( \frac{1}{5 x} ) tan ( \frac{1}{5 x} )}{5 x ^{2}}) : - \frac{tan ( \frac{1}{5 x} )}{5 x ^{2}}

= - \frac{tan ( \frac{1}{5 x} )}{5 x ^{2}}