Question

求e^y+xy-e=0 对X的二阶导,

Answer 1

对函数两边微分,

d/dx【e^y+xy-e】=0

【（e^y）*dy/dx 】+【x*(dy/dx)+y】-1=0

【（e^y）*dy/dx 】+【x*(dy/dx)】=1-y

dy/dx【（e^y）+x】=1-y

再次对两边微分,

d/dx 【 dy/dx【（e^y）+x】 】=d/dx【1-y】

(d^2y/dx^2)【（e^y）+x】+(dy/dx)【（e^y)*(dy/dx)+1】=-dy/dx

(d^2y/dx^2)=-dy/dx-(dy/dx)【（e^y)*(dy/dx)+1】 / 【（e^y）+x】