## 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.

## The Scientific (Mathematical) Way to Cut a Cake

Ever wondered if there is an alternative way to cutting cake so that it can stay fresh and softer in the refrigerator?

This is how!

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# Proving Quotient Rule using Product Rule

This is how we can prove Quotient Rule using the Product Rule.

First, we need the Product Rule for differentiation: $\displaystyle\boxed{\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}}$

Now, we can write $\displaystyle\frac{d}{dx}(\frac{u}{v})=\frac{d}{dx}(uv^{-1})$

Using Product Rule, $\displaystyle \frac{d}{dx}(uv^{-1})=u(-v^{-2}\cdot\frac{dv}{dx})+v^{-1}\cdot(\frac{du}{dx})$

Simplifying the above will give the Quotient Rule! :

$\displaystyle\boxed{\frac{d}{dx}(\frac{u}{v})=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}}$

You can also try proving Product Rule using Quotient Rule!