Question

求复合函数的高阶导数

We are given a function f(x) = (x^2 - 3x + 2)^(n) * cos(x^2), where n is a constant. Answer all parts from part (a) to part (b):
(a). what is f(5)(x), the 5th-order derivative of f(x)?
(b). Does there exist a general expression for f(n)(x), the n-th order derivative of f(x)? State your reasons clearly.

Thank you in advance.

Answer 1

See picture 

The font is small, if you want the doc, I can send to you directly with bigger fonts.

追问

Thanks a lot! Just a further follow-up:
For v(x) = cos(x^2), is it possible to construct a general expression for V(n,x) = v(n)(x), the n-th order derivative of v(x)? (Please bear with my bad notations:P)?
Someone states that for u(x) = (x^2 - 3x + 2)^(n), its derivative can be written as
u(n)(x) = SUM(C(n,i) * C(n,i) * C(n, n - i) * (x - 1) ^ i * (x - 2) ^ (x - 2)^(n - i)), is it true?

追答

Not sure about cos(x^2), will take a look later.
For the other, there is, I think it's:

Please verify, where P for permutation, C for combination.

------------
yes, should be n-m+i, instead of n-m-i

追问

Thanks for the quick update! Since there are other people still attempting this question, I think I will wait for a few more days before closing this question. I really appreciate your detailed explanation. Actually I did not expect anyone would be willing to try the 5th-order derivative (indeed tedious). By the way, any thought on derivative of v(x)?

追答

there should be a general formula for v(x) also. I thought about a little, but only got partial results. I don't think I'll have much time until the weekend to think again.

Here's what I have for nth-order derivative:
1. if n is odd, it has (n+1)/2 terms, which have x, x^3, ..., x^n (along with sin(x^2) or cos(x^2)).
The term containing x^n can be expressed in (-1)^((n+1)/2)*(2x)^n*sin(x^2). The others are the sum of something related to n (you can try to find yourself).
2. if n is even, it has n/2 terms, which have x, x^3, ..., x^n (along with sin(x^2) or cos(x^2)).
The term with x^n can be expressed in (-1)^(n/2)*(2x)^n*cos(x^2). The co-efficient of the others are the sum of something related to n).

Answer 2

英语不行啊。

追问

sorry, 昨天时间有些紧，问题是(a)函数 f(x) = (x^2 - 3x + 2)^n * cos(x^2) (n是常数) 的5阶导数是什么？（可用表达式）(b)该函数的各阶导数是否有一般性的表达式。谢谢！

追答

可以用公式表示：
对于函数f(x)=g(x)*k(x)的n阶导数，有二次项展开定理公式类似的公式，即：
y'=C(n,r)[g(x)(r)]*[k(x)(n-r) ] r=0,1,,2,3....n的和。
g(x)(r),表示g(x)的r阶导数，k(x)(n-r)表示n-r阶导数。

对于本题r=5.g(x)=(x^2-3x+2)^n,k(x)=cosx^2.
y(5)=c(5,0)g(x)[(cosx^2)(5)]+c(5,1)[g(x)(1)][(cosx^2)(4)]+c(5,2)[g(x)(2)][(cosx^2)(3)]+c(5,3)[g(x)(3)][cos(x^2)(2)]+c(5,4)[g(x)(4)][(cosx^2)(1)]+c(5,5)[g(x)(5)](cosx^2).

追问

OK，如果g(x) = (x^2 - 3x + 2)^n, k(x) = cos(x^2) . 对于任意整数m （m在【0，n】上） ，是否存在g(m)(x),k(m)(x)的一般表达式(即m阶导数)？谢谢。

追答

有：
k(x)=cos(x^2)
k'(x)=-2xsin(x^2)
k''(x)=-4x^2cos(x^2)
k'''(x)=8x^3sin(x^2)
所以：k(m)(x)=(2x)^m*sin[x^2+(m+1)π/2]

g(x)=(x^2-3x+2)^n=(x-2)^n*(x-1)^n
g(m)(x)
=c(m,0)(x-2)^n*[(x-1)^n](m)]+c(m,1)[(x-2)^n(1)]*[(X-1)^n(m-2)]+......+c(m,m)[(x-2)^n(m)]*(x-1)^n

追问

那个。。。k''(x)= (k'(x))' = -4x^2 cos(x^2) - 2sin(x^2)
k(3)(x) = -8 x * cos(x^2) + 8 * x^3 * sin(x^2) - 4 x * cos(x^2)
k(4)(x) = -12 * cos(x^2) + 48* x^2 * sin(x^2) + 16 * x^4 * cos(x^2).
您觉得那个更合理呢？

g(m)(x) 的系数似乎有缺项，take n = 3 and m = 2 as an example.
因为我已经到了追问的上限，所以接下来会尽量用评论功能回复，敬请谅解。谢谢！

追答

对，关于k(x)的导数，确实你的对了，不过那样就复杂了，那要从第二项开始，应用公式了：
看成是x和sin(x^2)的两个函数的乘积的导数，则有：
k(m)(x)=-2*[c(m-1,0)*x*g(x)(m-1)+c(m-1,1)*g(x)(m-2)+c(m-1,2)*0*g(x)(m-3)+....] m>=2.
即：
k(m)(x)=-2[xg(x)(m-1)+(m-1)g(x)(m-2)]; m>=2.
k(x)'=-2xsinx^2; m=1.

g(x)=(x^2-3x+2)^n=(x-2)^n*(x-1)^n
g(m)(x)
=c(m,0)(x-2)^n*[(x-1)^n](m)]+c(m,1)[(x-2)^n(1)]*[(X-1)^n(m-1)]+c(m,2)[(x-2)^n(2)]*[(X-1)^n(m-2)]+......+c(m,m)[(x-2)^n(m)]*(x-1)^n
对，第二项中的m-2应该为m-1.