请问f(x)=x*2^(-x^2)的导数怎么求?

若复合函数F(x)=f(x)g(x)
则其导数F'(x)=f'(x)g(x)+f(x)g'(x)

题中f'(x)=2^(-x^2)+x*2^(-x^2)*ln2*(-2x)
=2^(-x^2)*[(1-2*ln2*x^2]

f(x)=x*2^(-x^2),先化简,
f(x)=x*1/2^(x^2),
令,u=x^2,
V=2^u,
y=x*1/V,
f(x)'=(x/V)'=(V-xV')/V^2=[V-x*(2^u*lnu)*u']/V^2
=[2^(x^2)-x*2^(x^2)*ln(x^2)*2x]/[2^(x^2)]^2
=[1-2x^2*ln(x^2)]/[2^(x^2]
=(1-4x^2*lnx)/[2^(x^2)].