求全微分，要详细过程
1个回答
展开全部
1、z=arctan [(x+y)/(1-xy)]
那么z'x=1/[1+(x+y)²/(1-xy)²] *[(x+y)/(1-xy)]' x
=(1-xy)²/(1+x²+y²+x²y²) *(1-xy+xy+y²)
=(1-xy)²(1+y²) /(1+x²+y²+x²y²)
=(1-xy)²/(1+x²)
同理z'y=(1-xy)²/(1+y²)
dz=(1-xy)²/(1+x²) dx +(1-xy)²/(1+y²)dy
2、u=x^(yx)=e^(lnx *yx)
故u'x=e^(lnx *yx) *(lnx *yx)'
=x^(yx) *(y+lnx *y)
u'y=x^(yx) *(lnx *x)
dU= x^(yx) *(y+lnx *y)dx +x^(yx) *(lnx *x) dy
那么z'x=1/[1+(x+y)²/(1-xy)²] *[(x+y)/(1-xy)]' x
=(1-xy)²/(1+x²+y²+x²y²) *(1-xy+xy+y²)
=(1-xy)²(1+y²) /(1+x²+y²+x²y²)
=(1-xy)²/(1+x²)
同理z'y=(1-xy)²/(1+y²)
dz=(1-xy)²/(1+x²) dx +(1-xy)²/(1+y²)dy
2、u=x^(yx)=e^(lnx *yx)
故u'x=e^(lnx *yx) *(lnx *yx)'
=x^(yx) *(y+lnx *y)
u'y=x^(yx) *(lnx *x)
dU= x^(yx) *(y+lnx *y)dx +x^(yx) *(lnx *x) dy
本回答被提问者和网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
|
广告 您可能关注的内容 |
