Question

一道高数题，求大佬。

Answer 1

由定义,f(x,y)在原点可微即f(x,y)-f(0,0)-x·∂f(0,0)/∂x-y·∂f(0,0)/∂y = o(√(x²+y²)).
代入f(0,0) = 0,∂f(0,0)/∂x = ∂f(0,0)/∂y = 0即f(x,y) = o(√(x²+y²)).
然而对x = y,有f(x,y) = f(x,x) = x³/(2x²) = x/2 = √(x²+y²)/(2√2).
因此(x,y) → (0,0)时,f(x,y)并不是√(x²+y²)的高阶无穷小,也即f(x,y)在原点不可微.
注:也可以用另一种说法:f(x,y)在原点对向量(1,1)的方向导数 = 1/2,
并不等于(1,1)·(∂f(0,0)/∂x,∂f(0,0)/∂y) = ∂f(0,0)/∂x+∂f(0,0)/∂y = 0,因此在原点不可微.