函数y=y（x）由方程sin（x2+y2）+ex-xy2=0所确定，则dydx=2xcos(x2+y2)+ex−y22xy−2ycos(x2+y2)2xcos(x2+y2)+ex−y22xy−2ycos(x2+y2)．

Question

函数y=y（x）由方程sin（x²+y²）+e^x-xy²=0所确定，则(dy)/(dx)=

(2xcos(x2+y2)+ex−y2)/(2xy−2ycos(x2+y2))

(2xcos(x2+y2)+ex−y2)/(2xy−2ycos(x2+y2))

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Answer 1

解；  设F（x，y）=sin（x²+y²）+e^x-xy²，则
Fx＝2xcos(x2+y2)+ex−y2，Fy＝2ycos(x2+y2)−2xy
∴(dy)/(dx)＝−(Fx)/(Fy)=(2xcos(x2+y2)+ex−y2)/(2xy−2ycos(x2+y2))