Question

求导y=arctan√(x^2-1)-(lnx/√(x^2-1)) 求详细解答过程

Answer 1

Answer 2

(arctanx)' = 1/(x² + 1)
(lnx)' = 1/x
(u/v)' = (u'v - uv')/v²
y' = [arctan√(x² - 1)]' - [lnx/√(x² - 1)]'
= [1/(x² - 1 + 1)][√(x² - 1)]' - {(1/x)√(x² - 1) - (lnx)[√(x² - 1)]'}/(x² - 1)
= (1/2)*2x/[x²√(x² - 1)] - {[√(x² - 1)/x] - (lnx)(1/2)(2x)/√(x² - 1)}/(x² - 1)
= 1/[x√(x² - 1)] - [√(x² - 1)/x]- (xlnx)/√(x² - 1)]/(x² - 1)
= 1/[x√(x² - 1)] - 1/[x√(x² - 1)] + xlnx/[(x² - 1)√(x² - 1)]
= xlnx/[(x² - 1)√(x² - 1)]