(1 point) Let f(x,y)=6x2y4. Then fx​(x,y)=fx​(3,y)=fx​(x,−2)=fx​(3,−2)=fy​(x,y)=fy​(3,y)=fy​(x,−2)=fy​(3,−2)=​ 1 point)

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answerhappygod
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(1 point) Let f(x,y)=6x2y4. Then fx​(x,y)=fx​(3,y)=fx​(x,−2)=fx​(3,−2)=fy​(x,y)=fy​(3,y)=fy​(x,−2)=fy​(3,−2)=​ 1 point)

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1 Point Let F X Y 6x2y4 Then Fx X Y Fx 3 Y Fx X 2 Fx 3 2 Fy X Y Fy 3 Y Fy X 2 Fy 3 2 1 Point 1
1 Point Let F X Y 6x2y4 Then Fx X Y Fx 3 Y Fx X 2 Fx 3 2 Fy X Y Fy 3 Y Fy X 2 Fy 3 2 1 Point 1 (65.83 KiB) Viewed 35 times
1 Point Let F X Y 6x2y4 Then Fx X Y Fx 3 Y Fx X 2 Fx 3 2 Fy X Y Fy 3 Y Fy X 2 Fy 3 2 1 Point 2
1 Point Let F X Y 6x2y4 Then Fx X Y Fx 3 Y Fx X 2 Fx 3 2 Fy X Y Fy 3 Y Fy X 2 Fy 3 2 1 Point 2 (12.29 KiB) Viewed 35 times
(1 point) Let f(x,y)=6x2y4. Then fx​(x,y)=fx​(3,y)=fx​(x,−2)=fx​(3,−2)=fy​(x,y)=fy​(3,y)=fy​(x,−2)=fy​(3,−2)=​
1 point) Calculate all four second-order partial derivatives and check that fxy​=fyz​ : Assume the variables are restricted to a domain on which the functon is lefined. f(x,y)=e3xy fzz​= fyy​= fzy​= fyz​=
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