Question

这几题的二阶导数怎么求，求帮忙，谢谢！

要有过程，谢谢大家！

Answer 1

方法如下，
请作参考：

Answer 2

朋友，您好！详细完整清晰过程rt，希望能帮到你

Answer 3

(2)

(f(x^2))'

链式法则

=f'(x^2) . (x^2)'

=f'(x^2) . (2x)

=2x.f'(x^2)

(f(x^2))''

2阶导数=1阶导数再求导数

=((f(x^2))')'

=(2x.f'(x^2))'

链式法则

=2[ x.【f'(x^2)】' + f'(x^2). (x)']
=2[ x f''(x^2).(x^2)' + f'(x^2)]

=2[ x f''(x^2).(2x) + f'(x^2)]

=2[ 2x^2. f''(x^2) + f'(x^2)]

Answer 4

解1. f(x)=[e^(-x)]sinx，
f'(x)=[e^(-x)]'sinx+[e^(-x)](sinx)'
=-[e^(-x)]sinx+[e^(-x)]cosx=[e^(-x)](cosx-sinx)
f''(x)=[e^(-x)]'(cosx-sinx)+[e^(-x)](cosx-sinx)'=-[e^(-x)](cosx-sinx)+[e^(-x)](-sinx-cosx)
=[e^(-x)](sinx-cosx-sinx-cosx)=-2[e^(-x)]cosx