Question

z=f(x∧2+y∧2,e∧xy),求z对x,y的二阶偏导

Answer 1

z=f(x²+y²,e^(xy)),求z对x,y的二阶偏导数

设z=f(u,v)；u=x²+y²；v=e^(xy)；则：

∂z/∂x=(∂f/∂u)(∂u/∂x)+(∂f/∂v)(∂v/∂x)

=2x(∂f/∂u)+ye^(xy)(∂f/∂v)；

∂²z/∂x²=2(∂f/∂u)+2x(∂²f/∂u²)(∂u/∂x)+y²e^(xy)(∂f/∂v)+ye^(xy)(∂²f/∂v²)(∂v/∂x)

=2(∂f/∂u)+4x²(∂²f/∂u²)+y²e^(xy)(∂f/∂v)+y²e^(2xy)(∂²f/∂v²)；

∂z/∂y=(∂f/∂u)(∂u/∂y)+(∂f/∂v)(∂v/∂y)

=2y(∂f/∂u)+xe^(xy)(∂f/∂v)；

∂²z/∂y²=2(∂f/∂u)+2y(∂²f/∂u²)(∂u/∂y)+x²e^(xy)(∂f/∂v)+xe^(xy)(∂²f/∂v²)(∂v/∂y)

=2(∂f/∂u)+4y²(∂²f/∂u²)+x²e^(xy)(∂f/∂v)+x²e^(2xy)(∂²f/∂v²)；