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ln(lnx)的x次幂
如何求导?
人气：186 ℃　时间：2020-05-27 05:57:48
解答
{[In(In)]^x}'
=x*[In(In(x)]^(x-1)*[In(Inx)]'
=x*[In(In(x)]^(x-1)*[1/Inx]*[Inx]'
=x*[In(In(x)]^(x-1)*[1/Inx]*1/x
=[In(In(x)]^(x-1)/Inx.
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