Past exam of the options II course of the University of Oxford / 2009 / F / F2

The $i$th partial derivative is ${\partial }_{i}f(x_0)={\lim }_{t\to 0}[f(x_0+t{e}_i)-f(x_0)]/t$. The directional derivative in direction $v$ is the same limit with $v$ in place of $e_i$, using the specified vector rather than normalizing it. Differentiability means that there is a linear map $L:{\mathbb{R}}^n\to \mathbb{R}$ withFor $v=(v_1,v_2)$ with $v_1\ne 0$, the difference quotient at the origin is $t{v}_2^3/v_1$, tending to zero. If $v_1=0$, the function is identically zero on the directional line, again giving derivative zero. Thus every directional derivative exists and is zero.

Along the curved path $(x,y)=(t^3,t)$, however, $f(t^3,t)=1$ for $t\ne 0$, while $f(0,0)=0$. Hence the function is not continuous at the origin. Differentiability implies continuity, so it is not differentiable there either. Direction-by-direction limits alone do not control general approaches to a point.

For $x\ne y$, apply the scalar mean value theorem to $h(t)=⟨f(x+t(y-x)),z⟩$ on $[0,1]$. It is continuous on the closed interval and differentiable inside by the stated segment and regularity assumptions. The chain rule givesThus $h(1)-h(0)=h^{\prime }(\theta )$ for some $\theta \in (0,1)$, precisely the asserted identity. If $x=y$, both sides are zero when the zero-direction derivative is naturally taken as $Df(x)0$.

For the final claim, all partial derivatives tending to zero imply $\Vert \nabla g(x)\Vert \to 0$ away from the origin. For a nonzero small vector $h$, the one-variable function $t\mapsto g(th)$ is continuous on $[0,1]$ by continuity at zero and differentiable on $(0,1)$. Thereforefor some $\theta \in (0,1)$. For any $\epsilon >0$, if $\Vert h\Vert$ is sufficiently small, the entire punctured segment has gradient norm below $\epsilon$. Hence $|g(h)-g(0)|\le \epsilon \Vert h\Vert$, provingThis argument does not presuppose differentiability at the origin.