01 - Differentiability Criterion

Differentiability Criterion

Suppose $f:{\mathbb{R}}^2\to \mathbb{R}$ has continuous partial derivatives then

$f$ is differentiable in ${\mathbb{R}}^2$ with derivative at $\underline{a}=(a,b)$ by

$$df(\underline{a})=(f_x(a,b),f_y(a,b))$$

Proof

Let $(a,b)\in \Omega$ and $(h,k)\in {\mathbb{R}}^2$
Define

$$L:=(f_x(a,b),f_y(a,b))$$

Need to show that

$$\underset{(h,k)\to \underline{0}}{\lim }\frac{f(a+h,b+k)-f(a,b)-L(h,k)}{|(h,k)|}=0$$

Using the telescoping sum

$$\begin{array}{rl}& f(a+h,b+k)-f(a,b)\\ & =(f(a+h,b+k)-f(a+h,b))+(f(a+h,b)-f(a,b))\end{array}$$

By Mean Value Theorem in one variable, there exists ${\theta }_h,{\theta }_k\in (0,1)$ s.t.

$$\begin{array}{rl}f(a+h,b+k)-f(a+h,b)& =f_y(a+h,b+{\theta }_{k}k)k\\ f(a+h,b)-f(a,b)& =f_x(a+{\omega }_{h}h,b)h\end{array}$$

(This works as the partial derivatives exist so we can use MVT)

Hence

$$f(a+h,b+k)-f(a,b)=f_y(a+h,b+{\theta }_{k}k)k+f_x(a+{\theta }_{h}h,b)h$$

Treating $(h,k)$ as a column vector for matrix $L$ then

$$L(h,k)=f_x(a,b)h+f_y(a,b)k$$

So

$$\begin{array}{rl}& f(a+h,b+k)-f(a,b)-L(h,k)\\ & =(f_x(a+{\theta }_{h}h,b)-f_x(a,b))h+(f_y(a+h,b+{\theta }_{k}k)-f_y(a,b))k\\ & =(f_x(a+{\theta }_{h}h,b)-f_x(a,b),f_y(a+h,b+{\theta }_{k}k)-f_y(a,b))\cdot (h,k)\end{array}$$

As it is a dot product of two vector, then by Cauchy Schwarz Inequality
The absolute value is at most

$$\sqrt{(f_x(a+{\theta }_{h}h,b)-f_x(a,b){)}^2+(f_y(a+h,b+{\theta }_{k}k)-f_y(a,b){)}^2}\cdot \sqrt{h^2+k^2}$$

For the limit to equal $0$ we need

$$\sqrt{(f_x(a+{\theta }_{h}h,b)-f_x(a,b){)}^2+(f_y(a+h,b+{\theta }_{k}k)-f_y(a,b){)}^2}\to 0$$

as $(h,k)\to \underline{0}$

Therefore

$$f_x(a+{\theta }_{h}h,b)\to f_x(a,b)\text{ and }{f}_y(a+h,b+{\theta }_{k}k)\to f_y(a,b)$$

Hence it tends to $0$ thus the limit equals $0$

Proof - Uniqueness of Total Derivatives from Partial Derivatives

For notational convenience and simplicity suppose

$$n=2\text{ and }m=1$$

Let $\Omega ={\mathbb{R}}^2$ and function $f:{\mathbb{R}}^2\to \mathbb{R}$
Suppose $\underline{a}=(a,b)$ and $\underline{h}=(h,k)$, and $f(a,b)\text{ represent }f((a,b))$ for notation

Consider the two directions $(0,1)$ and $(1,0)$ as the basis for ${\mathbb{R}}^2$

Let the two directional derivatives be $f_x(a,b)$ and $f_y(a,b)$ (partial derivatives)
Note that they can also be written as $\frac{\partial f}{\partial x}(a,b)$ and $\frac{\partial f}{\partial y}(a,b)$ but it takes longer :)

Assume that $f:{\mathbb{R}}^2\to \mathbb{R}$ is differentiable at point $\underline{a}=(a,b)$
With linear map $L=df(\underline{a}):{\mathbb{R}}^2\to \mathbb{R}$ being the derivative, then

$$df(\underline{a}):(1,0)\to f_x(a,b),df(\underline{a}):(0,1)\to f_y(a,b)$$

Hence $df(\underline{a})$ is unique, as partial derivatives are uniquely defined