求二阶导数

Question

设f(x)可导，求y’
1）y＝f[2f(x)]
2)y=f(x^2+1)-f^2(x)
3)y=f(x^2)-x^3e^f(x)
4)y=f(e^-x)+sinf^2(x)

Answer 1

1) 解：令 u=2f(x),
则 y=f(u).
所以 du/dx=2f'(x),
dy/du=f'(u).
所以 dy/dx=(dy/du)*(du/dx)
=2f'(x)*f'(u)
=2f'(x)*f'[2f(x)].
所以 d(2)y /dx(2)
=2f''(x)*f'(u)+2f'(x)*f''(u)*u'
=2f''(x)*f'[2f(x)]+4*[f'(x)]^2 *f''[2f(x)].

= = = = = = = = =
按复合函数求导法则。
求二阶导数，只能依次求导了。

= = = = = = = = =
2)解：令 u=x^2+1, g(x)=[f(x)]^2,
则 y=f(u)+g(x).
所以 dy/dx=df/dx+dg/dx
=(df/du)*(du/dx)+(dg/df)*(df/dx)
=2x *f'(u)+2f(x) *f'(x).
=2x *f'(x^2+1)+2f(x) *f'(x).
所以 d(2)y /dx(2)
=2f'(u) +2x*f''(u) +2f'(x)*f'(x) +2f(x)*f''(x)
=2f'(x^2+1)+2x*f''(x^2+1)+2[f'(x)]^2+2f(x)*f''(x)

= = = = = = = = =
dy/dx=df/dx+dg/dx有问题.
df/dx=f'(x)
df/dx=df/du*du/dx=f'(u)*u'
f'(x)=f'(u)*u' ???
但按法则, 是f'(u)*u' .

3)解：dy/dx
=2x *f'(x^2) -3x^2 *e^f(x) -x^3 *[e^f(x)]*f'(x)
=2x *f'(x^2) -x^2 *[e^f(x)][3 -x*f'(x)].
d(2)y /dx(2)
=...

= = = = = = = = =
为避免第(2)问出现的问题，不设u了，直接按法则求导。
二次导数算不出来了。
本人计算能力不好，有可能算错了。

Answer 2

cd15363