Question

求z对x和y偏导数，急急急，希望有详细步骤

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)]

Answer 2

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