2020-12-02 · 知道合伙人教育行家
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方程 sin(xy)=x²+y你确定函数y=y(x),求dy/dx;
解法(一):
两边直接对x求导:[cos(xy)]•(y+xy')=2x+y';
ycos(xy)+xy'cos(xy)=2x+y';
[xcos(xy)-1]y'=2x-ycos(xy);
故y'=[2x-ycos(xy)]/[xcos(xy)-1];
解法(二);
设F(x,y)=sin(xy)-x²-y=0;那么
dy/dx=-(∂F/∂x)/(∂F/∂y)=-[ycos(xy)-2x]/[xcos(xy)-1]=[2x-ycos(xy)]/[xcos(xy)-1];
解法(一):
两边直接对x求导:[cos(xy)]•(y+xy')=2x+y';
ycos(xy)+xy'cos(xy)=2x+y';
[xcos(xy)-1]y'=2x-ycos(xy);
故y'=[2x-ycos(xy)]/[xcos(xy)-1];
解法(二);
设F(x,y)=sin(xy)-x²-y=0;那么
dy/dx=-(∂F/∂x)/(∂F/∂y)=-[ycos(xy)-2x]/[xcos(xy)-1]=[2x-ycos(xy)]/[xcos(xy)-1];
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