Question

用对数求导法求导y=(1+cosx)的1/x次y=(x-1)的三次*根号下(x-2)2次/x-3

Answer 1

1）y = (1+cosx)^(1/x),

利用对数求导法：取对数,得

ln|y| = (1/x)ln(1+cosx),

求导,得

y'/y = [x(-sinx)/(1+cosx) - ln(1+cosx)]/x^2

于是

y' = …….

2）y = [(x-1)^3]*√{[(x-2)^2]/(x-3)}

利用对数求导法：取对数,得

ln|y| = 3ln|x-1| + ln|x-2| - (1/2)ln|x-3|,

求导,得

y'/y = 3/(x-1) - 1/(x-2) - (1/2)/(x-3)),

于是

y' = …….