Mean Value Theorem for Higher Dimensions

Let f be differentiable on a connected set E\subseteq \mathbb{R}^n, then for any x,y\in E, there exists z\in E such that f(x)-f(y)=\nabla f(z)\cdot (x-y).

Proof: The trick is to use the Mean Value Theorem for 1 dimension via the following construction:

Define g:[0,1]\to\mathbb{R}, g(t)=f(tx+(1-t)y). By the Mean Value Theorem for one variable, there exists c\in (0,1) such that g'(c)=\frac{g(1)-g(0)}{1-0}, i.e.

\nabla f(cx+(1-c)y)\cdot (x-y)=f(x)-f(y). Here we are using the chain rule for multivariable calculus to get: g'(c)=\nabla f(cx+(1-c)y)\cdot (x-y).

Let z=cx+(1-c)y, then \nabla f(z)\cdot (x-y)=f(x)-f(y) as required.

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