y=sin²(2x+π/3)的导数要详细过程
答案:2 悬赏:80
解决时间 2021-04-09 00:11
- 提问者网友:曖昧情执
- 2021-04-08 17:15
y=sin²(2x+π/3)的导数要详细过程
最佳答案
- 二级知识专家网友:心痛成瘾
- 2021-04-08 18:37
y'=[sin²(2x+π/3)]'
=2*sin(2x+π/3)*[sin(2x+π/3)]'
=2*sin(2x+π/3)*cos(2x+π/3)*(2x+π/3)'
=2*sin(2x+π/3)*cos(2x+π/3)*2
=4*sin(2x+π/3)*cos(2x+π/3)
=2*[2sin(2x+π/3)*cos(2x+π/3)]
=2sin(4x+2π/3)
=2*sin(2x+π/3)*[sin(2x+π/3)]'
=2*sin(2x+π/3)*cos(2x+π/3)*(2x+π/3)'
=2*sin(2x+π/3)*cos(2x+π/3)*2
=4*sin(2x+π/3)*cos(2x+π/3)
=2*[2sin(2x+π/3)*cos(2x+π/3)]
=2sin(4x+2π/3)
全部回答
- 1楼网友:不羁的心
- 2021-04-08 20:04
y'=2sin(2x+π/3)*[sin(2x+π/3)]'
=2sin(2x+π/3)*cos(2x+π/3)*(2x+π/3)'
=2sin(2x+π/3)*cos(2x+π/3)*2
=4sin(2x+π/3)*cos(2x+π/3)
=2sin(4x+2π/3)
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