求二元函数s=x*y*sin(x+y)的一阶偏导数和二阶偏导数
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解决时间 2021-04-01 10:40
- 提问者网友:不爱我么
- 2021-03-31 16:24
求二元函数s=x*y*sin(x+y)的一阶偏导数和二阶偏导数
最佳答案
- 五星知识达人网友:空山清雨
- 2021-03-31 16:31
一阶偏导数
sx=y*sin(x+y)+x*y*cos(x+y)
sy=x*sin(x+y)+x*y*cos(x+y)
二阶偏导数
sxx=2y*cos(x+y)-x*y*sin(x+y)
sxy=syx
=sin(x+y)+y*cos(x+y)+x*cos(x+y)-x*y*sin(x+y)
=(1-x*y)*sin(x+y)+(x+y)*cos(x+y)
syy=2x*cos(x+y)-x*y*sin(x+y)
sx=y*sin(x+y)+x*y*cos(x+y)
sy=x*sin(x+y)+x*y*cos(x+y)
二阶偏导数
sxx=2y*cos(x+y)-x*y*sin(x+y)
sxy=syx
=sin(x+y)+y*cos(x+y)+x*cos(x+y)-x*y*sin(x+y)
=(1-x*y)*sin(x+y)+(x+y)*cos(x+y)
syy=2x*cos(x+y)-x*y*sin(x+y)
全部回答
- 1楼网友:未来江山和你
- 2021-03-31 17:00
s=xysin(x+y),
∂s/∂x=y[sin(x+y)+xcos(x+y)],
∂s/∂y=x[sin(x+y)+ycos(x+y)],
∂^2s/∂x^2=y[2cos(x+y)-xsin(x+y)],
∂^2s/∂x∂y=sin(x+y)+xcos(x+y)+y[cos(x+y)-xsin(x+y)]
=(1-xy)(sin(x+y)+(x+y)cos(x+y),
∂^2s/∂y^2=x[2cos(x+y)-ysin(x+y)].
∂s/∂x=y[sin(x+y)+xcos(x+y)],
∂s/∂y=x[sin(x+y)+ycos(x+y)],
∂^2s/∂x^2=y[2cos(x+y)-xsin(x+y)],
∂^2s/∂x∂y=sin(x+y)+xcos(x+y)+y[cos(x+y)-xsin(x+y)]
=(1-xy)(sin(x+y)+(x+y)cos(x+y),
∂^2s/∂y^2=x[2cos(x+y)-ysin(x+y)].
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