Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / F (Multivariable Calculus)

The $i$th partial derivative is the limit ${\partial }_{i}f(x_0)={\lim }_{t\to 0}[f(x_0+t{e}_i)-f(x_0)]/t$. For a nonzero vector $v$, use the directional convention ${\partial }_{v}f(x_0)={\lim }_{t\to 0}[f(x_0+tv)-f(x_0)]/t$, without normalizing $v$. Differentiability means $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$ for a linear map $L=df(x_0)$.

For the given example take $v=(v_1,v_2)$. If $v_1\ne 0$, then $f(tv)=t^2{v}_2^3/v_1$, so its directional difference quotient tends to zero. If $v_1=0$, the function vanishes along that line. Thus $\boxed{{\partial }_{v}f(0)=0\text{ for every }v\ne 0}$. However along $(x,y)=(t^3,t)$ with $t\ne 0$, one has $f(x,y)=1$, while $f(0,0)=0$. Hence the function is neither continuous nor differentiable at zero. Existence of all directional derivatives, even with a linear dependence on the direction, does not by itself imply total differentiability.

Define the scalar function $H(s)=⟨f(x+s(y-x)),z⟩$ on $[0,1]$. It is continuous there and differentiable inside, with derivative $⟨df(x+s(y-x))(y-x),z⟩$ by the chain rule. The one-dimensional mean value theorem gives $H(1)-H(0)=H^{\prime }(\theta )$ for some $0<\theta <1$, proving the displayed vector-paired version. No claim of a single intermediate point giving a vector equality without pairing is required.

For the final assertion, use the scalar mean value theorem directly on $s\mapsto g(sh)$: it is continuous at $s=0$ and differentiable for $s>0$. Thusfor some $0<{\theta }_h<1$. The assumed limits of all partial derivatives imply $\Vert \nabla g(x)\Vert \to 0$ as $x\to 0$, $x\ne 0$. Consequently $|g(h)-g(0)|/\Vert h\Vert \le \Vert \nabla g({\theta }_{h}h)\Vert \to 0$. Hence $\boxed{dg(0)=0}$ and $g$ is differentiable there. This argument uses only differentiability on the interior of the line segment, so it does not assume the conclusion at its endpoint.