设y=cosx+xe^y,则x=0时,dy/dx=?

网友回答

x=0时,y=1
dy/dx=-sinx+e^y+(xe^y)dy/dx
所以dy/dx=(-sinx+e^y)/(1-xe^y)
把x=0,y=1代入上式,得dy/dx=e
======以下答案可供参考======
供参考答案1:
y=cosx+xe^y
dy/dx = -sinx + xe^y dy/dx + e^y
dy/dx = [-sinx+e^y]/[1-xe^(y)]
x=0y = 1dy/dx|(0,1) = [0+ e]/[1-0]
= e供参考答案2:
对其求导得y'=-sinx+e^y+xe^yy',将x=0带入原等式得y=1,再将x,y值带入上等式得y=e