Question

求1.z=e^[-sin (y/x)] 2.z=arctan[(x+y)/(x-y)]的偏导数

Answer 1

求1.z=e^[-sin (y/x)] ；2.z=arctan[(x+y)/(x-y)]的偏导数
解：1。令z=e^u，u=-sinv，v=y/x；则：
∂z/∂x=(dz/du)(du/dv)(∂v/∂x)=(e^u)(-cosv)(-y/x²)=e^[-sin(y/x)][cos(y/x)](y/x²)；
∂z/∂y=(dz/du)(du/dv)(∂v/∂y)=(e^u)(-cosv)(1/x)=-e^[-sin(y/x)][cos(y/x)](1/x)；
解：2。令z=arctanu，u=(x+y)/(x-y)；则：
∂z/∂x=(dz/du)(∂u/∂x)=[1/(1+u²)][(x-y)-(x+y)]/(x-y)²= -{1/[1+arctan²(x+y)/(x-y)]}[(2y/(x-y)²]；
∂z/∂y=(dz/du)(∂u/∂y)=[1/(1+u²)][(x-y)+(x+y)]/(x-y)²={1/[1+arctan²(x+y)/(x-y)]}[(2x/(x-y)²]；

追问

可是第二题的参考答案是∂z/∂x=y/(x^2+y^2),∂z/∂y=x/(x^2+y^2)

追答

对不起，我代错了！
2。令z=arctanu，u=(x+y)/(x-y)；则：
∂z/∂x=(dz/du)(∂u/∂x)=[1/(1+u²)][(x-y)-(x+y)]/(x-y)²= -{1/[1+(x+y)²/(x-y)²]}[(2y/(x-y)²]
=-[(x-y)²/(x²+y²)][y/(x-y)²]=-y/(x²+y²)；
∂z/∂y=(dz/du)(∂u/∂y)=[1/(1+u²)][(x-y)+(x+y)]/(x-y)²={1/[1+(x+y)²/(x-y)²]}[(2x/(x-y)²]
=[(x-y)²/(x²+y²)][x/(x-y)²]=x/(x²+y²)；