y=x^x
=e^[ln(x^x)]
=e^(xlnx)
令u=xlnx,则y=e^u
y'=(e^u)'•u'
=(e^u)•(xlnx)'
=[e^(xlnx)]•[x'lnx+x(lnx)']
=[e^(xlnx)]•(lnx+x•1/x)
=(x^x)(1+lnx)
所以答案:
x的x次方•lnx
=(x的x次方)'•lnx+x的x次方•1/x
=(x^x)•(1+lnx)•lnx+x的(x-1)次方
y=x^x
=e^[ln(x^x)]
=e^(xlnx)
令u=xlnx,则y=e^u
y'=(e^u)'•u'
=(e^u)•(xlnx)'
=[e^(xlnx)]•[x'lnx+x(lnx)']
=[e^(xlnx)]•(lnx+x•1/x)
=(x^x)(1+lnx)
所以答案:
x的x次方•lnx
=(x的x次方)'•lnx+x的x次方•1/x
=(x^x)•(1+lnx)•lnx+x的(x-1)次方