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设函数y=x^x+ln(arctan5x),求其导数dy / dx、微分dy
人气:440 ℃ 时间:2019-08-16 12:05:12
解答
y = e^(xlnx) + ln[arctan(5x)]
dy/dx = e^(xlnx)[lnx + 1] + 1/arctan(5x)*[1 + (5x)^2]^(-1)*5
= x^x[lnx + 1] + 5/{arctan(5x)[1 + (5x)^2]}
dy = [x^x[lnx + 1] + 5/{arctan(5x)[1 + (5x)^2]}]dx
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