Let $f: U \rightarrow \mathbf{R}^{n}$ be a map on an open subset $U \subset \mathbf{R}^{m}$. Explain what it means for $f$ to be differentiable on $U$. If $g: V \rightarrow \mathbf{R}^{m}$ is a differentiable map on an open subset $V \subset \mathbf{R}^{p}$ with $g(V) \subset U$, state and prove the Chain Rule for the derivative of the composite $f g$.

Suppose now $F: \mathbf{R}^{n} \rightarrow \mathbf{R}$ is a differentiable function for which the partial derivatives $D_{1} F(\mathbf{x})=D_{2} F(\mathbf{x})=\ldots=D_{n} F(\mathbf{x})$ for all $\mathbf{x} \in \mathbf{R}^{n}$. By considering the function $G: \mathbf{R}^{n} \rightarrow \mathbf{R}$ given by

$$G\left(y_{1}, \ldots, y_{n}\right)=F\left(y_{1}, \ldots, y_{n-1}, y_{n}-\sum_{i=1}^{n-1} y_{i}\right)$$

or otherwise, show that there exists a differentiable function $h: \mathbf{R} \rightarrow \mathbf{R}$ with $F\left(x_{1}, \ldots, x_{n}\right)=$ $h\left(x_{1}+\cdots+x_{n}\right)$ at all points of $\mathbf{R}^{n} .$