Maths and stat 問題..

2006-11-06 8:01 am
a). y=cosx/1-sinx

b). y=3sin6x^3/2+cos^3x^2

c). y=x^2sinx+cos^1/2x
更新1:

d). x^2sinx+(cosx)^1/2

更新2:

find dy/dx

回答 (1)

2006-11-06 6:50 pm
✔ 最佳答案
(a):

y = cos(x) / [1 - sin(x)]
dy/dx = {[1 - sin(x)]{d[cos(x)]/dx} - {d[1-sin(x)]/dx}cos(x)} / {[1 - sin(x)] ^ 2}
={[(1 - sin(x)][-sin(x)] - [-cos(x)]cos(x)} / {[1 - sin(x)] ^ 2}
= {[cos(x) ^ 2) - sin(x)[1 - sin(x)]} / {[1 - sin(x)] ^ 2}
= {1 - [sin(x) ^ 2] - sin(x) + [sin(x) ^ 2]} / {[1 - sin(x)] ^ 2}
= [1 - sin(x)] / {[1 - sin(x)] ^ 2}
= 1 / [1 - sin(x)]

(b):

y = 3sin{6[x ^ (3 / 2)]} + cos[3(x ^ 2)]
dy/dx = 3cos{6[x ^ (3 /2 )]} d{6[x ^ (3 / 2)]}/dy + {-sin[3(x ^ 2)]}{d[3(x ^ 2)]/dx}
= 3cos{6[x ^ (3 / 2)]}{9[x ^ (1 / 2)]} + {-sin[3(x ^ 2)]}(6x)
= 27sqrt(x)cos{6[x ^ (3 / 2)]} - 6xsin[3(x ^ 2)]


(c):

y = (x ^ 2)sin(x) + cos(x / 2)
dy/dx = [d(x ^ 2)/dx]sin(x) + (x ^ 2){d[sin(x)]/dx} + [-sin(x /2)][d(x / 2)/dx]
= 2xsin(x) + (x ^ 2)cos(x) - sin(x / 2) / 2


(d):

y = (x ^ 2)sin(x) + [cos(x) ^ (1 / 2)]
dy/dx = [d(x ^ 2)/dx]sin(x) + (x ^ 2){d[sin(x)]/dx} + (1 / 2)[cos(x) ^ (-1 / 2)]{d[cos(x)]/dx}
= 2xsin(x) + (x ^ 2)cos(x) + (1 / 2)[cos(x) ^ (-1 / 2)][-sin(x)]
= 2xsin(x) + (x ^ 2)cos(x) -(1 / 2)sin(x)[cos(x) ^ (-1 / 2)]

I am not quite understand your symbols


收錄日期: 2021-04-23 15:52:11
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061106000051KK00008

檢視 Wayback Machine 備份