Question

求教几个高等数学题1.求f(x,y)=xsin(x+y)+ycos(x+y)的二级偏导数2.求Z=xsin(x2+y2)

Answer 1

1 Z=xsin(x+y)+ycos(x+y)

Zx=sin(x+y)+xcos(x+y)-ysin(x+y)

Zy=xcos(x+y)+cos(x+y)-ysin(x+y)

所以Zxx=cos(x+y)+cos(x+y)-xsin(x+y)-ycos(x+y)=

2cos(x+y)-xsin(x+y)-ycos(x+y)

Zxy=Zyx=cos(x+y)-xsin(x+y)-sin(x+y)-ycos(x+y)

Zyy=-xsin(x+y)-sin(x+y)-sin(x+y)-ycos(x+y)=

-2sin(x+y)-xsin(x+y)-ycos(x+y)

2 Z=xsin(x2+y2)

Zx=sin(x2+y2)+2x2cos(x2+y2)

Zy=2xycos(x2+y2)

所以全微分=[sin(x2+y2)+2x2cos(x2+y2)]dz+[2xycos(x2+y2)]dy

3 Z=2x+3y2

Zx=2

Zy=6y

△z=Z(12,8.3)-Z(10,8)=230.67-212=18.67

dz=2dx+6ydy=2*2+6*8*0.2=13.6