求下列所确定的隐函数方程y=y（x）的导数.

xy+sin(x+y)=1,两边求导数
y+xy'+cos(x+y)*(1+y')=0
xy'+cos(x+y)y'=-[cos(x+y)+y]
∴y'[cos(x+y)+x]=-[cos(x+y)+y]
∴y'=-[y+cos(x+y)]/[x+cos(x+y)]

In[4]:= s1 = D[x y[x] + Sin[x + y[x]], x]
Out[4]= y[x] + x Derivative[1][y][x] +
Cos[x + y[x]] (1 + Derivative[1][y][x])
In[7]:= Solve[s1 == 0, y'[x]]
Out[7]= {{Derivative[1][y][x] -> (-Cos[x + y[x]] - y[x])/(
x + Cos[x + y[x]])}}
楼上算的是正确的。