Question

求下列方程所确定的隐函数y=f（x）的导数dy/dx

x^2+y^2-xy=x

Answer 1

两边分别对x求导有：

2x+2yy'-(y+xy')=1

2x+2yy'-y-xy'=1

(2y-x)y'=y-2x+1

dy/dx=y'=(y-2x+1)/(2y-x)

Answer 2

解析
对x求导2x-y-1
对y求导2y-x
∴dy/dx=-(2x-y-1)/(2y-x)
=(y-2x+1)/(2y-x)