Question

求z对x和y偏导数，急急急，希望有详细步骤

Answer 1

z=(x^2+y^2).e^[-arctan(y/x)]
∂z/∂x
=e^[-arctan(y/x)] . ∂/∂x(x^2+y^2) +(x^2+y^2) .∂/∂x{ e^[-arctan(y/x)]}
=e^[-arctan(y/x)].(2x) +(x^2+y^2) . e^[-arctan(y/x)].∂/∂x[-arctan(y/x)]
=e^[-arctan(y/x)].(2x) +(x^2+y^2) . e^[-arctan(y/x)].[-1/(1+y^2/x^2)]. ∂/∂x(y/x)
=e^[-arctan(y/x)].(2x) +(x^2+y^2) . e^[-arctan(y/x)].[-1/(1+y^2/x^2)].(-y/x^2)
=e^[-arctan(y/x)].(2x) +e^[-arctan(y/x)]. y
=(2x+y).e^[-arctan(y/x)]
∂z/∂y
=e^[-arctan(y/x)].∂/∂y(x^2+y^2) +(x^2+y^2) .∂/∂y{ e^[-arctan(y/x)]}
=e^[-arctan(y/x)].(2y) +(x^2+y^2) .e^[-arctan(y/x)].∂/∂y[-arctan(y/x)]
=e^[-arctan(y/x)].(2y) +(x^2+y^2) .e^[-arctan(y/x)].[-1/(1+y^2/x^2)]∂/∂y(y/x)
=e^[-arctan(y/x)].(2y) +(x^2+y^2) .e^[-arctan(y/x)].[-1/(1+y^2/x^2)](1/x)
=e^[-arctan(y/x)].(2y) +e^[-arctan(y/x)].(-x)
=(2y-x).e^[-arctan(y/x)]