Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 3 / 3B (Analysis II)

A function is a differentiable function at $a_0=(a,b)$ if there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ withThe map $L=D{f}_{a_0}$ is its derivative; equivalently $L(h_1,h_2)=A{h}_1+B{h}_2$ for fixed constants. Since every such linear map is bounded, $|f(a_0+h)-f(a_0)|\le \Vert L\Vert \Vert h\Vert +|r(h)|\to 0$. This proves differentiability implies continuity, not merely continuity along each coordinate axis.

Put $r=(x^2+y^2{)}^{1/2}$. Since $x^2{y}^2\le (x^2+y^2{)}^2/4$, the nonzero-point expression satisfiesThus the defining remainder condition for differentiability holds with the zero linear map:The estimate controls every approach direction at once, which is stronger than merely finding the two zero partial derivatives.

On the nonzero $x$ axis the function equals one, whereas its assigned value at the origin is zero. Therefore $f(x,0)$ does not tend to $f(0,0)$ as $x\to 0$. It is not a continuous function there, and differentiability implies continuity shows