derivative of ln((ax+1/(ax-1)))

解答

\frac{d}{d x} (ln (\frac{a x + 1}{a x - 1}))

- \frac{2 a}{( a x + 1 ) ( a x - 1 )}

求解步骤

\frac{d}{d x} (ln (\frac{a x + 1}{a x - 1}))

使用连锁律:

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

= \frac{1}{\frac{a x + 1}{a x - 1}} \frac{d}{d x} (\frac{a x + 1}{a x - 1})

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

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

化简

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

= - \frac{2 a}{( a x + 1 ) ( a x - 1 )}

d er i v a t i v e f (x) = 3 x + \frac{3}{x}

\frac{d}{d x} (\frac{e ^{x}}{1 - x})

d er i v a t i v e f (x) = (x - \frac{2}{x})

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

n = 1 \sum \infty ln (n)