设f(x,y)=xy+x/(x2+y2),求f′x(0,1),f′y(0,1).
因为 f′x(x,y)=y+[(x2+y2-x•2x)/ (x2+y2)2]=y+[(y2-x2)/(x2-y2)2], f′x(x,y)=x+[-x•2y/(x2+y2)2]= x-[2xy/(x2-y2)2], 所以f′x(0,1)=1+(1-0)/12=2, f′y(0,1)=0.
设f(x,y)=xy+x/(x2+y2),求f′x(0,1),f′y(0,1).
因为 f′x(x,y)=y+[(x2+y2-x•2x)/ (x2+y2)2]=y+[(y2-x2)/(x2-y2)2], f′x(x,y)=x+[-x•2y/(x2+y2)2]= x-[2xy/(x2-y2)2], 所以f′x(0,1)=1+(1-0)/12=2, f′y(0,1)=0.