设方程e^xy+x-y=2确定了隐函数y=y(x),球y'(x)及y'(0)

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设方程e^xy+x-y=2确定了隐函数y=y(x),球y'(x)及y'(0)
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设方程e^xy+x-y=2确定了隐函数y=y(x),球y'(x)及y'(0)
设方程e^xy+x-y=2确定了隐函数y=y(x),球y'(x)及y'(0)

设方程e^xy+x-y=2确定了隐函数y=y(x),球y'(x)及y'(0)

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两边分别对x求导,
得:(y+xy'(x))e^xy+1-y'(x)=0
求得y'(x)=(ye^xy+1)/(1-xe^xy)
令x=0
求得y'(0)=0