Question

高等数学。y＝x的x次方怎么求导？

Answer 1

y = x^x, x > 0
lny = xlnx
y'/y = lnx + 1
y' = y(1+lnx) = x^x(1+lnx)

Answer 2

y'=(x^x)'
=（e^(xlnx)）'
=[e^(xlnx)]*(lnx+x/x)
=(x*x)*(1+lnx)