求z=arctan(y⼀x)+ln根号下(x^2+y^2)的偏导数

2024-11-18 09:13:48
推荐回答(1个)
回答1:

  • z=arctan(y/x)+ln√(x²+y²)=arctan(y/x)+½ln(x²+y²)

    ∂z/∂x=(y/x)'/[1+(y/x)²]+½·2x/(x²+y²)

    =-y/(x²+y²)+x/(x²+y²)

    =(x-y)/(x²+y²)

    ∂z/∂y=(y/x)'/[1+(y/x)²]+½·2y/(x²+y²)

    =x/(x²+y²)+y/(x²+y²)

    =(x+y)/(x²+y²)