derivative of tan^{sec(1/x}(1/x))

解答

\frac{d}{d x} (tan^{s e c (\frac{1}{x})} (\frac{1}{x}))

tan^{s e c (\frac{1}{x})} (\frac{1}{x}) (- \frac{sec ( \frac{1}{x} ) tan ( \frac{1}{x} ) ln ( tan ( \frac{1}{x} ) )}{x ^{2}} - \frac{sec ^{3} ( \frac{1}{x} )}{x ^{2} tan ( \frac{1}{x} )})

求解步骤

\frac{d}{d x} (tan^{s e c (\frac{1}{x})} (\frac{1}{x}))

使用指数法则:

a^{b} = e^{b l n (a)}

tan^{s e c (\frac{1}{x})} (\frac{1}{x}) = e^{s e c (\frac{1}{x}) l n (t a n (\frac{1}{x}))}

= \frac{d}{d x} (e^{s e c (\frac{1}{x}) l n (t a n (\frac{1}{x}))})

使用连锁律:

e^{s e c (\frac{1}{x}) l n (t a n (\frac{1}{x}))} \frac{d}{d x} (sec (\frac{1}{x}) ln (tan (\frac{1}{x})))

= e^{s e c (\frac{1}{x}) l n (t a n (\frac{1}{x}))} \frac{d}{d x} (sec (\frac{1}{x}) ln (tan (\frac{1}{x})))

\frac{d}{d x} (sec (\frac{1}{x}) ln (tan (\frac{1}{x}))) = - \frac{sec ( \frac{1}{x} ) tan ( \frac{1}{x} ) ln ( tan ( \frac{1}{x} ) )}{x ^{2}} - \frac{sec ^{3} ( \frac{1}{x} )}{x ^{2} tan ( \frac{1}{x} )}

= e^{s e c (\frac{1}{x}) l n (t a n (\frac{1}{x}))} (- \frac{sec ( \frac{1}{x} ) tan ( \frac{1}{x} ) ln ( tan ( \frac{1}{x} ) )}{x ^{2}} - \frac{sec ^{3} ( \frac{1}{x} )}{x ^{2} tan ( \frac{1}{x} )})

e^{s e c (\frac{1}{x}) l n (t a n (\frac{1}{x}))} = tan^{s e c (\frac{1}{x})} (\frac{1}{x})

= tan^{s e c (\frac{1}{x})} (\frac{1}{x}) (- \frac{sec ( \frac{1}{x} ) tan ( \frac{1}{x} ) ln ( tan ( \frac{1}{x} ) )}{x ^{2}} - \frac{sec ^{3} ( \frac{1}{x} )}{x ^{2} tan ( \frac{1}{x} )})

d er i v a t i v e f (x) = x^{4} - 6 x^{3}

\frac{d}{d x} (ln (x^{2} \cdot e^{x^{2} + 1}))

t a y l or (1 + \frac{1}{x})^{- 2}

t an g e n t f (x) = sin (x), at x = 0

\frac{d}{d x} (\frac{- 1}{x + c})