Find f'(n)(x) for n=>3
p.s. ^(n) = power of n
f'(n) = n times differentiable
更新1:
I don't know why the Q need n>=3 ...?
更新2:
Even you'v finished the Q , could u explain a little bit about it.
pure , differentiation
2009-07-20 3:03 am
Q1 : f(x) = e^(x)x^(3).
Find f'(n)(x) for n=>3
p.s. ^(n) = power of n
f'(n) = n times differentiable
Find f'(n)(x) for n=>3
p.s. ^(n) = power of n
f'(n) = n times differentiable
回答 (1)
2009-07-20 3:55 am
✔ 最佳答案
Using Leibnitz Formula for the n'th Derivative of a ProductWe have
f'(n)
=Σ(nCk)(e^x)(n-k)(3x^2)(k)
=Σ(nCk)(e^x)(3x^2)(k)
For n>=3, the expression is
(nC0)(e^x)(x^3)+(nC1)(e^x)(3x^2)+(nC2)(e^x)(6x)+(nC3)(e^x)(6)
=(e^x)(x^3)+3n(e^x)(x^2)+3n(n-1)(e^x)(x)+n(n-1)(n-2)(e^x)
2009-07-21 13:08:02 補充：
因為n>=4時﹐x^3的導數會是0
收錄日期: 2021-04-26 13:44:22
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090719000051KK01541