Exercise
Find second order partial derivatives of the function
u(x,y,z)=xy+yz+zx
Final Answer
u''_{xx} (x,y,z)=u''_{yy} (x,y,z)=u''_{zz} (x,y,z)=0
u''_{xy} (x,y,z)=u''_{yx} (x,y,z)=1
u''_{yz} (x,y,z)=u''_{zy} (x,y,z)=1
u''_{xz} (x,y,z)=u''_{zx} (x,y,z)=1
Solution
Coming soon…