设y=y(x)是方程xˆ2eˆy+yˆ2=1确定函数,求dy/dx│(1,0)

x^2e^y+y^2=1
两边对x求导得2xe^y+x^2e^y*y'(x)+2y*y'(x)=0
故y'(x)=-2xe^y/(x^2e^y+2y)
所以dy/dx│(1,0)=-2*1*e^0/(1^2*e^0+2*0)=-2
如果不懂,请Hi我,祝学习愉快!