你求导数怎么用这么奇怪的递归函数?
x^x = e^(xlnx)
d(x^x)/dx = de^(xlnx)/dx = e^(xlnx) (lnx + 1) = x^xlnx + x^x
d((1-x)^(1-x))/dx = -(1-x)^(1-x)ln(1-x) - (1-x)^(1-x)
df(x)/dx = (1-x)^(1-x)d(x^x)/dx + x^xd(1-x)^(1-x)/dx
= (1-x)^(1-x)(x^xlnx + x^x)+x^x[-(1-x)^(1-x)ln(1-x) - (1-x)^(1-x)] =0
(1-x)^(1-x)(x^xlnx + x^x) = x^x[(1-x)^(1-x)ln(1-x) +(1-x)^(1-x)]
(1-x)^(1-x)(lnx +1) = (1-x)^(1-x)ln(1-x) +(1-x)^(1-x)
lnx = ln(1-x)
显然x=0.5时取极值