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If all partial derivatives exist near a point and are continuous there, successive coordinate increments and the one-variable mean value theorem express $f(a+h)-f(a)$ as ${\sum }_j{h}_j{\partial }_{j}f(a+{\eta }_j(h))$ with ${\eta }_j(h)\to 0$. Subtracting the displayed linear map leaves $o(\Vert h\Vert )$, proving Fréchet differentiability. In two dimensions the absolute remainder is at most $ϵ(|h_1|+|h_2|)\le \sqrt{2}ϵ\Vert h\Vert$ for small enough increments.

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.

Differentiability at $(a,b)$ means that there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatNecessarily $L(h,k)=f_x(a,b)h+f_y(a,b)k$ when those partial derivatives exist. Under the stated assumptions, apply the one-variable mean value theorem separately to the two coordinate increments:for some intermediate $\theta ,\eta \in (0,1)$, omitting a term if its increment is zero. Existence of the partial derivatives in a neighborhood supplies the differentiability along each segment required for that theorem. Their continuity at $(a,b)$ makes the difference from $f_x(a,b)h+f_y(a,b)k$ bounded by $(|h|+|k|)\epsilon (h,k)$, where $\epsilon (h,k)\to 0$. Since $|h|+|k|\le \sqrt{2}\sqrt{h^2+k^2}$, this is the required little-o remainder and proves Fréchet differentiability.

For the particular function, $f(h,0)=f(0,k)=0$, so the defining difference quotients give $f_x(0,0)=f_y(0,0)=0$. However $f(t,t)=1/2$ for every $t\ne 0$. It is not continuous at the origin, and therefore not differentiable there. This illustrates why existence of partial derivatives alone is insufficient.

Set $r=\sqrt{x^2+y^2}$. The bound $|xy|\le r^2/2$ and $|x^2-y^2|\le r^2$ give $|f(x,y)|\le r^2/2$. Hence $f\to 0$ at the origin, proving continuity, and $|f(x,y)|/r\to 0$, proving Fréchet differentiability there with derivative equal to the zero linear map.

Along the coordinate axes, direct difference quotients giveThese formulas include the origin. Therefore the two mixed partial derivatives differ:The explicit operator order removes any ambiguity about the notation for mixed derivatives. Also $f_{xx}(0,0)=f_{yy}(0,0)=0$, since the first derivatives restricted to their own axes vanish identically.

All four second-order partial derivatives are discontinuous at the origin. Away from it,Along $x=y\ne 0$ these are one and minus one, rather than their zero values at the origin. Meanwhile ${\partial }_y{\partial }_{x}f(x,0)=1$ for $x\ne 0$, whereas its value at the origin is minus one; and ${\partial }_x{\partial }_{y}f(0,y)=-1$ for $y\ne 0$, whereas its origin value is one. This gives a concrete unequal mixed partial derivatives example despite differentiability of the original function.

Fix $(x,y)\in U$ and a sufficiently small rectangle inside $U$. Its rectangular incrementcan be evaluated twice using the fundamental theorem of calculus, givingDivide by $hk$ and let $h,k\to 0$. Continuity of the mixed partial derivatives makes the two limits $D_2{D}_{1}f(x,y)$ and $D_1{D}_{2}f(x,y)$. They are equal everywhere on $U$.

Repeated interchange of adjacent partial derivatives reduces every order-$m$ derivative of a smooth function to $D_1^j{D}_2^{m-j}f$, $0\le j\le m$. Hence there are at most $m+1$ distinct functions. This maximum is attained by $f(x,y)=e^{x+2y}$, whose listed partial derivatives are the distinct functions $2^{m-j}{e}^{x+2y}$. The answer is $m+1$, rather than the $2^m$ possible written orders.

For the supplied $f$, $f(t,t)=1/2$ for $t\ne 0$, whereas $f(0,0)=0$. It is not even continuous at the origin, so $f$ is neither differentiable nor infinitely differentiable there.

For $g$, $|g(x,y)|\le |xy|\le (x^2+y^2)/2$. Thus $g(x,y)=o(\sqrt{x^2+y^2})$, proving Fréchet differentiability at the origin with derivative zero. However,including zero at the origin. Therefore $D_2{D}_{1}g(0,0)=-1$ and $D_1{D}_{2}g(0,0)=1$. $g$ is differentiable but not infinitely differentiable at the origin: it cannot have continuous second partial derivatives near that point.

For the operator norm, submultiplicativity bounds the remainder of the matrix exponential byTherefore $\Vert \exp A-\exp 0-A\Vert /\Vert A\Vert \to 0$. This is precisely Fréchet differentiability at zero, with

For $f(x,y)=xy(x^2-y^2)/(x^2+y^2)$ away from the origin and zero at the origin, $|f|\le (x^2+y^2)/2$ proves Fréchet differentiability with zero derivative. Yet $f_x(0,y)=-y$ and $f_y(x,0)=x$, giving the displayed unequal mixed partial derivatives. They exist but are not continuous at the origin, so Clairaut's theorem does not apply.