maket
春芽
共回答了14个问题采纳率:100% 举报
已知xy=e^(x+y),求dy/dx.
解一:将原式写成F(x,y)=xy-e^(x+y)=0
则dy/dx=-(∂F/∂x)/(∂F/∂y)=-[y-e^(x+y)]/[x-e^(x+y)]=-(y-xy)/(x-xy)=(xy-y)/(x-xy);
解二:直接求导:y+xy′=[e^(x+y)](1+y′)=xy(1+y′)=xy+xyy′,(x-xy)y′=xy-y;故y′=(xy-y)/(x-xy);
解三:两边取对数后再求导:
lnx+lny=x+y;(1/x)+y′/y=1+y′;y+xy′=xy+xyy′;(x-xy)y′=xy-y;故y′=(xy-y)/(x-xy).
三种方法都可以.
1年前
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