Find F' (x) of the following functions
f(x)= (3x 2(指數） +1) 3（指數） / (x 2(指數） -1) 4(指數）
點計?
Differentiation
2007-05-17 1:05 am
回答 (2)
2007-05-17 2:43 am
✔ 最佳答案
use quotient rule: and Chain rulethat is:
dy/dx=(v(du/dx)-u(dv/dx))/v^2
u=(3x^2+1)^3 and v=(x^2-1)^4
step1:(((x^2-1)^4)d/dx(3x^2+1)^3-((3x^2+1)^3d/dx(x^2-1)^4))
/x^2-1)^8
step2:((x^2-1)^4)(3)(3x^2+1)^2)(6x)-(3x^2+1)^3(4)(2x)(x^2-1)^3)/(x^2-1)^8
2007-05-16 18:44:40 補充：
sor no timeanswer is:[-2(3x² 1)²(x)(3x² 13)] / (x²-1)^5
參考: myself
2007-05-17 1:35 am
use quotient rule
denominator 分母 becomes [(x^2)-1]^8
numerator 分子 becomes
(x²-1)⁴(3)(6x)(3x²+1)²-(3x²+1)³(4)(2x)(x²-1)³
= -2x(3x²+13)(3x²+1)²(x-1)³(x+1)³
after simpify
f'(x) = [-2(3x²+1)²(x)(3x²+13)] / (x²-1)^5
denominator 分母 becomes [(x^2)-1]^8
numerator 分子 becomes
(x²-1)⁴(3)(6x)(3x²+1)²-(3x²+1)³(4)(2x)(x²-1)³
= -2x(3x²+13)(3x²+1)²(x-1)³(x+1)³
after simpify
f'(x) = [-2(3x²+1)²(x)(3x²+13)] / (x²-1)^5
收錄日期: 2021-04-13 13:38:20
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070516000051KK02365