Question

求由方程e^xy+x^2*y-1=0确定的隐函数,y=f（x）的导数dy/dx

Answer 1

e^xy+x^2*y-1=0,两边微分,得：

e^xy*(ydx+xdy)+2xydx+x^2dy=0,

——》dy/dx=-(y*e^xy+2xy)/(x*e^xy+x^2).