求导arctany/xIn√(x²+y²)(x²+y²)si
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解决时间 2021-02-06 18:55
- 提问者网友：王者佥
- 2021-02-06 11:24
求导arctany/xIn√(x²+y²)(x²+y²)si
最佳答案
- 五星知识达人网友：第幾種人
- 2021-02-06 12:49
全微分吗?z=arctan(y/x)∂z/∂x=1/(1+y²/x²)*y=x²y/(x²+y²)∂z/∂y=1/(1+y²/x²)*1/x=x/(x²+y²)dz=x²y/(x²+y²)dx+x/(x²+y²)dyz=ln√(x²+y²)∂z/∂x=1/√(x²+y²)*1/2√(x²+y²)*2x=x/(x²+y²)∂z/∂y=1/√(x²+y²)*1/2√(x²+y²)*2y=y/(x²+y²)dz=x/(x²+y²)dx+y/(x²+y²)dyz=(x²+y²)sin[3/(x²+y²)]∂z/∂x=2xsin[3/(x²+y²)]+(x²+y²)*cos[3/(x²+y²)]*[-6x/(x²+y²)²]=2xsin[3/(x²+y²)]-6xcos[3/(x²+y²)]*/(x²+y²)∂z/∂y=2ysin[3/(x²+y²)]+(x²+y²)*cos[3/(x²+y²)]*[-6y/(x²+y²)²]=2ysin[3/(x²+y²)]-6ycos[3/(x²+y²)]*/(x²+y²)dz=2xsin[3/(x²+y²)]-6xcos[3/(x²+y²)]*/(x²+y²)dx+2ysin[3/(x²+y²)]-6ycos[3/(x²+y²)]*/(x²+y²)dy======以下答案可供参考======供参考答案1:太难了。。。。。高数都忘记了
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- 1楼网友：琴狂剑也妄
- 2021-02-06 12:58
这个解释是对的
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