a). y=3xsin(x)cos(x)
b). y=[x^(1/3)][(1-x)^(2/3)]
c). y=[(2-x)^(20)][(3x+3)^(30)]
更新1:
點將3sinx cosx + 3x cosx cosx - 3x sinx sinx 轉為(3/2)sin2x + 3xcos2x
differentiation problem
2009-11-01 7:31 am
Evaluate the first derivative of the following functions.
a). y=3xsin(x)cos(x)
b). y=[x^(1/3)][(1-x)^(2/3)]
c). y=[(2-x)^(20)][(3x+3)^(30)]
a). y=3xsin(x)cos(x)
b). y=[x^(1/3)][(1-x)^(2/3)]
c). y=[(2-x)^(20)][(3x+3)^(30)]
回答 (2)
2009-11-01 8:01 am
✔ 最佳答案
a). y = 3x sinx cosxdy/dx = 3sinx cosx + 3x cosx cosx - 3x sinx sinx
= (3/2)sin2x + 3xcos2x
b). y = [x^(1/3)][(1-x)^(2/3)]
dy/dx = [(1/3)x^(-2/3)][(1 - x)^(2/3)] - [x^(1/3)][(2/3)(1 - x)^(-1/3)]
= (1/3)[(1 - x) - 2x][x^(-2/3)][(1 - x)^(-1/3)]
= (1/3)(1 - 3x)[x^(-2/3)][(1 - x)^(-1/3)]
c). y = [(2-x)^(20)][(3x+3)^(30)]
dy/dx = -20[(2 - x)^19][(3x + 3)^30] + 90[(2 - x)^20][(3x + 3)^29]
= 10[(2 - x)^19][(3x + 3)^29][-2(3x + 3) + 9(2 - x)]
= 10[(2 - x)^19][(3x + 3)^29][-6x - 6 + 18 - 9x]
= 10[(2 - x)^19][(3x + 3)^29][12 - 15x]
= 30(4 - 5x)[(2 - x)^19][(3x + 3)^29]
2009-11-01 03:18:37 補充：
sin2x = sin(x + x) = sinx cosx + cosx sinx = 2sinx cosx
cos2x = cos(x + x) = cosx cosx - sinx sinx
2009-11-01 9:09 am
點將3sinx cosx + 3x cosx cosx - 3x sinx sinx
轉為(3/2)sin2x + 3xcos2x
轉為(3/2)sin2x + 3xcos2x
收錄日期: 2021-04-23 23:19:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091031000051KK01899