帮我求导!!!

y=(cosx)^x
lny=xlncosx
(1/y)*y'=lncosx+x*1/cosx*(-sinx)
=lncosx-xtanx
所以y'=y*(lncosx-xtanx)
=(cosx)^x*(lncosx-xtanx)

y=(x^3+sinx)^(1/x)
lny=(x^3+sinx)/x
(1/y)y'=[(x^3+sinx)'*x-(x^3+sinx)*x']/x^2
=[(3x^2+cosx)x-(x^3+sinx)]/x^2
=(2x^3+xcosx-sinx)/x^2
所以y'=y*(2x^3+xcosx-sinx)/x^2
=(x^3+sinx)^(1/x)*(2x^3+xcosx-sinx)/x^2

高数最近在复习