数学,高等数学多元函数问题,谢谢
2个回答
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不可以。因 z = z(x, y) , 不是 z = f(x, y)
z = z(x, y), x = f(u, v), y= g(u, v),
z'<u> = z'<x>x'<u>+z'<y>y'<u>
= z'<x>f'<u>+z'<y>g'<u>
z'<v> = z'<x>f'<v>+z'<y>g'<v>
z''<uu> = z''<xx>[f'<u>]^2+z''<xy>f'<u>g'<u>
+z''<yx>g'<u>f'<u>+z''<yy>[g'(u)]^2
= z''<xx>[f'<u>]^2+2z''<xy>f'<u>g'<u>+z''<yy>[g'(u)]^2
= z''<xx>[f'<u>]^2-2z''<xy>f'<u>f'<v>+z''<yy>[f'(v)]^2
z''<vv> = z''<xx>[f'<v>]^2+2z''<xy>f'<v>g'<v>+z''<yy>[g'(v)]^2
= z''<xx>[f'<v>]^2+2z''<xy>f'<v>f'<u>+z''<yy>[f'(u)]^2
得 z''<uu> + z''<vv>
= (z''<xx>+z''<yy>)[(f'<u>)^2+(f'<v>)^2]
z = z(x, y), x = f(u, v), y= g(u, v),
z'<u> = z'<x>x'<u>+z'<y>y'<u>
= z'<x>f'<u>+z'<y>g'<u>
z'<v> = z'<x>f'<v>+z'<y>g'<v>
z''<uu> = z''<xx>[f'<u>]^2+z''<xy>f'<u>g'<u>
+z''<yx>g'<u>f'<u>+z''<yy>[g'(u)]^2
= z''<xx>[f'<u>]^2+2z''<xy>f'<u>g'<u>+z''<yy>[g'(u)]^2
= z''<xx>[f'<u>]^2-2z''<xy>f'<u>f'<v>+z''<yy>[f'(v)]^2
z''<vv> = z''<xx>[f'<v>]^2+2z''<xy>f'<v>g'<v>+z''<yy>[g'(v)]^2
= z''<xx>[f'<v>]^2+2z''<xy>f'<v>f'<u>+z''<yy>[f'(u)]^2
得 z''<uu> + z''<vv>
= (z''<xx>+z''<yy>)[(f'<u>)^2+(f'<v>)^2]
追问
不好意思,好乱啊。。看不懂
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