mean value property of derivatives in high dimensions Let $E$ be a path-connected subset of $\mathbb{R}^n$ and $f$ a differentiable function on $E$. Prove or disprove:
for any $x,y\in E$, there exists $z\in E$ such that $f(x)-f(y)=\nabla f(z)\cdot (x-y)$. 
How to solve this? Thanks.
I have tried but failed: 
(1). Suppose $E$ is convex. Let $g(t)=f(x+t(y-x))$. Then $g'(t)=\nabla f(x+t(y-x))\cdot (y-x)$. By mean value theorem, there exists $t_0\in (0,1)$ such that  $g'(t_0)=g(1)-g(0)=f(y)-f(x)$. We choose $z=x+t_0(y-x)$. 
(2). In general, $E$ is only assumed path-connected. If I let $\gamma$ be a curve such that $\gamma(0)=x$ and $\gamma(1)=y$, and $g(t)=f(\gamma(t))$, then $g'(t)=\nabla f(\gamma(t))\cdot \gamma'(t)$. However,  $\gamma'(t)\neq y-x$. Hence I cannot solve...

Hint: consider $t\to (\cos t, \sin t)$.
