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y=(1+x^2)^arctanx 对y求导
人气:464 ℃ 时间:2019-08-11 01:24:23
解答
y =(1+x^2)^arcatanx
lny = arctanx.ln(1+x^2)
(1/y)y' = 2xarctanx/(1+x^2) + ln(1+x^2)/(1+x^2)
y' = [2xarcanx+ln(1+x^2)]/(1+x^2) .(1+x^2)^arctanx
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