高数题,求(1)和(4)的过程和答案,第四题不确定,不会求对x的偏导数。谢谢
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(1)
lim [3-√(9+xy)]/(xy)=
x->0
y->0
lim [3-√(9+xy)]/(xy)=
xy->0
lim [3-√(9+xy)][3+√(9+xy)]/{(xy)*[3+√(9+xy)]}= 分子有理化
xy->0
lim [9-(9+xy)]/{(xy)*[3+√(9+xy)]}=
xy->0
lim -1/[3+√(9+xy)]=-1/(3+3)=-1/6
xy->0
答案就是-1/6
(4)u(x,y,z)=(x/y)^(1/x)=e^[1/x*ln(x/y)]
用符号"p"表示求偏导,则有
pu/px=e^[1/x*ln(x/y)]*[-1/x²*ln(x/y)+1/x*1/(x/y)*1/y]=(x/y)^(1/x)*[1-ln(x/y)]/x²
pu/py=e^[1/x*ln(x/y)]*[1/x*1/(x/y)*(-x)/y²]=(x/y)^(1/x)*[-1/(xy)]
pu/pz=0
将x=y=z=1代入,分别得到1,-1,0
故所求梯度为
1i+(-1)j+0k=i-j
lim [3-√(9+xy)]/(xy)=
x->0
y->0
lim [3-√(9+xy)]/(xy)=
xy->0
lim [3-√(9+xy)][3+√(9+xy)]/{(xy)*[3+√(9+xy)]}= 分子有理化
xy->0
lim [9-(9+xy)]/{(xy)*[3+√(9+xy)]}=
xy->0
lim -1/[3+√(9+xy)]=-1/(3+3)=-1/6
xy->0
答案就是-1/6
(4)u(x,y,z)=(x/y)^(1/x)=e^[1/x*ln(x/y)]
用符号"p"表示求偏导,则有
pu/px=e^[1/x*ln(x/y)]*[-1/x²*ln(x/y)+1/x*1/(x/y)*1/y]=(x/y)^(1/x)*[1-ln(x/y)]/x²
pu/py=e^[1/x*ln(x/y)]*[1/x*1/(x/y)*(-x)/y²]=(x/y)^(1/x)*[-1/(xy)]
pu/pz=0
将x=y=z=1代入,分别得到1,-1,0
故所求梯度为
1i+(-1)j+0k=i-j
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