Question

这几题的二阶导数怎么求，求帮忙

要有解题过程，谢谢！

Answer 1

(1).〔e^(-x)sinx〕'
=-e^(-x)sinx+e^(-x)cosx
=e^(-x)(cosx-sinx)
那么，〔e^(-x)sinx〕''
=-e^(-x)(cosx-sinx)+e^(-x)(-sinx-cosx)
=-2e^(-x)cosx
(2).〔f(x^2)〕'
=2xf'(x^2)
〔f(x^2)〕''
=2f'+2xf''*(2x)
=2f'+4x^2*f''
(3).〔(cos2x)^2〕'
=2cos2x*(-sin2x)*2
=-4sin2xcos2x
=-2sin4x
〔(cos2x)^2〕''
=-2cos4x*4
=-8cos4x

Answer 2

d^2/dx^2(sin(x)/e^x) = -2 e^(-x) cos(x)

d^2/dx^2(f(x^2)) = 2 (2 x^2 f''(x^2) + f'(x^2))