微积分，求偏导，详细过程？

Question

如题所述

Answer 1

方法如下，
请作参考：

Answer 2

z=arctan(y/x)&706;z/&706;x = [1/(1 + (y/x)^2 ) ] .&706;/&706;x (y/x)= [1/(1 + (y/x)^2 ) ] . (-y/x^2)=-y/(x^2+y^2)&706;^2z/&706;x^2 =[y/(x^...

Answer 3

z=arctan(y/x)&706;z/&706;x = [1/(1 + (y/x)^2 ) ] .&706;/&706;x (y/x)= [1/(1 + (y/x)^2 ) ] . (-y/x^2)=-y/(x^2+y^2)&706;^2z/&706;x^2 =[y/(x^...

Answer 4

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

Answer 5

已知z=arctan(y/x)；求∂²z/∂x²;
解：设y/x=u；则
∂z/∂x=(∂z/∂u)(∂u/∂x)=[1/(1+u²)](-y/x²)=-[1/(1+y²/x²)](y/x²)=-y/(x²+y²);
∂²z/∂x²=2xy/(x²+y²)²