Question

函数求导题求解

Answer 1

y= arctan√(x^2-1) - lnx/√(x^2-1)
let
y1=arctan√(x^2-1)
tany1 = √(x^2-1)
(secy1)^2 .y1' =x/√(x^2-1)
y1' = [x/√(x^2-1)] /[1+(x^2-1) ]
= 1/[x.√(x^2-1)]
y2 = lnx/√(x^2-1)
lny2 = lnlnx - (1/2)ln(x^2-1)
(1/y2).y2' =1/(xlnx) - x/(x^2-1)
y2' = [1/(xlnx) - x/(x^2-1)] . lnx/√(x^2-1)
= 1/[x.√(x^2-1)] - xlnx/(x^2-1)^(3/2)
y'
= y1'-y2'
=1/[x.√(x^2-1)] -1/[x.√(x^2-1)] + xlnx/(x^2-1)^(3/2)
=xlnx/(x^2-1)^(3/2)