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求导y=arctan√(x^2-1)-(lnx/√(x^2-1))
人气:274 ℃ 时间:2020-05-24 15:10:03
解答
(arctanx)' = 1/(x² + 1)(lnx)' = 1/x(u/v)' = (u'v - uv')/v²y' = [arctan√(x² - 1)]' - [lnx/√(x² - 1)]'= [1/(x² - 1 + 1)][√(x² - 1)]' - {(1/x)√(x² - 1) - (lnx)[√(x...
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