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For the differentiability from continuous partial derivatives argument, choose a small box around zero inside the given open set. For $h=(h_1,\dots ,h_p)$ in that box, telescope $f(h)-f(0)$ by changing one coordinate at a time. The one-dimensional mean value theorem on the $j$th coordinate segment giveswhere ${\xi }_j$ lies on that segment, and therefore $\Vert {\xi }_j{\Vert }_2\le \Vert h{\Vert }_2$. A zero-length segment contributes zero and needs no choice of intermediate point. Set $L(h)={\sum }_j{f}_j(0)h_j$. For every $\epsilon >0$, continuity of the finitely many partial derivatives at zero gives $|f_j({\xi }_j)-f_j(0)|<\epsilon /\sqrt{p}$ when $h$ is sufficiently small. Then the Cauchy-Schwarz inequality givesThus $f$ is Fréchet differentiable at zero, with derivative $L$. No continuity of the partial derivatives away from zero was used.

For the supplied one-dimensional function,The derivative at zero follows from $g(s)/s=s\sin (1/s)\to 0$. For nonzero $s$ the derivative is continuous. At zero, the sequences $s_k=1/(2\pi k)$ and $t_k=1/((2k+1)\pi )$ give derivative values $-1$ and $1$. Since $f_x(x,y)=g^{\prime }(x)$, the partial derivative is continuous exactly where $x\ne 0$, irrespective of $y$.

Nevertheless $f$ is differentiable at every point of the plane. The one-dimensional function $g$ is differentiable at every real number. At a fixed $(x_0,y_0)$,Their sum has remainder $o(\sqrt{h^2+k^2})$, which proves Fréchet differentiability with derivative $(h,k)\mapsto g^{\prime }(x_0)h+g^{\prime }(y_0)k$. In particular, at the origin $|f(h,k)|\le h^2+k^2$, an immediate remainder bound. Continuity of all partial derivatives is sufficient, not necessary, for differentiability.

The function is Fréchet differentiable at $(a,b)$ when there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatIf it exists, $L(h,k)=f_x(a,b)h+f_y(a,b)k$, as is seen by taking increments along each coordinate axis.

To prove differentiability from continuous partial derivatives, take a small rectangle in the given neighbourhood and split the increment along its sides. The one-variable mean value theorem givesfor some $0<\theta ,\eta <1$ when the corresponding increment is nonzero. If an increment is zero, its term is simply omitted. The required one-variable continuity follows from existence of the corresponding partial derivative on the segment. Subtracting $L(h,k)$ gives a remainder whose absolute value, by continuity of both partial derivatives at $(a,b)$, is at mostfor sufficiently small increments. Since $ϵ$ is arbitrary, the normalized remainder tends to zero. Thus the function is differentiable at $(a,b)$, with the displayed derivative.