Past exam of the options I course of the University of Oxford / 2012 / F (Multivariable Calculus)

The function is continuously differentiable when all its first partial derivatives exist on $U$ and are continuous there. Equivalently, $h$ has a Fréchet derivative at every point and the map $x\mapsto Dh(x)$ is continuous. For a scalar-valued function,On the open domain, continuous first partial derivatives ensure that this linear expression is the total derivative.

For differentiable maps with compatible domains, the multivariable chain rule isIn Jacobian matrix notation this is matrix multiplication. In particular, if both maps are continuously differentiable, the composition is continuously differentiable. Componentwise,

Let $L(x,y,z)=(x-y,y-z,z-x)$. This is a linear map, hence smooth; composing it with the given $C^1$ function proves that $g=f\circ L$ is continuously differentiable by the chain rule. Write $f_1,f_2,f_3$ for the first partial derivatives, evaluated at $L(p)$. ThenAdding these identities givesGeometrically, $L(p+t(1,1,1))=L(p)$, so $g$ is constant along every line in the diagonal direction; its derivative in that direction must vanish.

An embedded $k$-dimensional submanifold $M\subseteq {\mathbb{R}}^n$ is locally a coordinate $k$-plane: for every $p\in M$, there are open neighbourhoods $U\ni p$, $V\subseteq {\mathbb{R}}^n$ and a diffeomorphism $\Phi :U\to V$ withFor the present $C^1$ setting, $\Phi$ and its inverse are continuously differentiable; in the smooth convention both are smooth.

A tangent vector to $M$ at $p$ is a velocity ${\gamma }^{\prime }(0)$ of a continuously differentiable curve $\gamma :(-ϵ,ϵ)\to M$ with $\gamma (0)=p$. The tangent space is the collection of all these vectors:In the local coordinates above, it is $D\Phi (p{)}^{-1}({\mathbb{R}}^k\times \{0{\}}^{n-k})$, so it is a $k$-dimensional linear subspace of ${\mathbb{R}}^n$.

The regular level set theorem says that for a $C^1$ map $g:U\subseteq {\mathbb{R}}^n\to {\mathbb{R}}^r$, if $Dg$ has rank $r$ on $g^{-1}(c)$, then $g^{-1}(c)$ is an $(n-r)$-dimensional submanifold andHere $r=1$ and the given rank condition makes the zero set a two-dimensional submanifold. It is nonempty because $g(0)=f(0)=0$. Part (c) givesConsequentlyThere is also a direct curve proof: $g(p+t(1,1,1))=g(p)=0$, so the entire line through $p$ in this direction lies in $M$. Its velocity at zero is exactly the required tangent vector.