z=(y^2-x^2)⼀(y^2+x^2)的二阶偏导数,急!!!!

还有这个也是Z=(cosy+xsiny)e ^x,过程要相信点的,谢谢!!
2025-06-22 14:07:16
推荐回答(1个)
回答1:

解:

dz=(cosy+xsiny)’e^x+(cosy+xsiny)(e^x)’

=(-sinydy+sinydx+xcosydy)e^x+(cosy+xsiny)e^xdx

=e^x(siny+xsiny+cosy)dx+e^x(xcosy-siny)dy

dz/dx= e^x(siny+xsiny+cosy)

dz/dy= e^x(xcosy-siny)

所以:

d^2z/dx^2=e^x(siny+xsiny+cosy)+e^xsiny

=e^x(2siny+xsiny+cosy).

d^2z/dy^2=e^x(-xsiny-cosy)=-e^x(xsiny+cosy).