Question

y=x^(2x)的二次导数怎么求

Answer 1

道理一样

y"=[2(lnx+1)]'*[x^(2x)]+2(lnx+1)*[x^(2x)]'

[2(lnx+1)]'=2*1/x=2/x

[x^(2x)]'=2(lnx+1)[x^(2x)]

所以y"=2*[x^(2x)]/x+2(lnx+1)*2(lnx+1)[x^(2x)]

即y"=2*x^(2x-1)+4(lnx+1)^2[x^(2x)]