Question

求由下列方程确定的隐函数的导数dy/dx 1 . y＝c

求由下列方程确定的隐函数的导数dy/dx 1 . y＝cos(x+y) 2.y=x+1/2lny
3.2x^2+4y^-1=0

Answer 1

y = cos(x + y)
dy/dx = dcos(x + y)/d(x + y) · d(x + y)/dx，链式法则
dy/dx = - sin(x + y) · (1 + dy/dx)
dy/dx = - sin(x + y) - sin(x + y) · dy/dx
dy/dx · [1 + sin(x + y)] = - sin(x + y)
dy/dx = - sin(x + y)/[1 + sin(x + y)
dy/dx = - 1/[1 + csc(x + y)]