Proving Function Polynomial in Coordinates is Differentiable Everywhere

[i_a_n]

The question is:
Using the chain rule to prove that a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$ which is polynomial in the coordinates is differentiable everywhere.
(The chain rule is for the use under function composition circumstances, how to apply it here to prove that the function $f$ which is polynomial in the coordinates is differentiable everywhere?)

 

[Fernando Revilla]

(The chain rule is for the use under function composition circumstances, how to apply it here to prove that the function $f$ which is polynomial in the coordinates is differentiable everywhere?)

We can write:

$f:\mathbb{R}^n\to \mathbb{R},\quad f(x_1,\ldots,x_n)=\displaystyle\sum_{i_1\geq 0,\ldots,i_n\geq 0}a_{i_1,\ldots,i_n}x_1^{i_1}\ldots x_n^{i_n}$

with finitely many nonnull real coefficientes $a_{i_1,\ldots,i_n}$. Now, use the following properties:$(1)\;$ The projections $\pi_i:\mathbb{R}^n\to\mathbb{R},\quad \pi_i(x_1,\ldots,x_n)=x_i$ are differentable on $\mathbb{R}^n$.

$(2)\;$ Every constant funcion is differentiable on $\mathbb{R}^n$.

$(3)\;$ The product of two differentiable functions on $\mathbb{R}^n$ is differentiable on $\mathbb{R}^n$.

$(4)\;$ The sum of two differentiable functions on $\mathbb{R}^n$ is differentiable on $\mathbb{R}^n$.

 

[i_a_n]

We can write:

$f:\mathbb{R}^n\to \mathbb{R},\quad f(x_1,\ldots,x_n)=\displaystyle\sum_{i_1\geq 0,\ldots,i_n\geq 0}a_{i_1,\ldots,i_n}x_1^{i_1}\ldots x_n^{i_n}$

with finitely many nonnull real coefficientes $a_{i_1,\ldots,i_n}$. Now, use the following properties:$(1)\;$ The projections $\pi_i:\mathbb{R}^n\to\mathbb{R},\quad \pi_i(x_1,\ldots,x_n)=x_i$ are differentable on $\mathbb{R}^n$.

$(2)\;$ Every constant funcion is differentiable on $\mathbb{R}^n$.

$(3)\;$ The product of two differentiable functions on $\mathbb{R}^n$ is differentiable on $\mathbb{R}^n$.

$(4)\;$ The sum of two differentiable functions on $\mathbb{R}^n$ is differentiable on $\mathbb{R}^n$.

But where you used the chain rule? I mean, how to use the chain rule to prove that $f$ is differentiable everywhere?

 

[Fernando Revilla]

But where you used the chain rule? I mean, how to use the chain rule to prove that $f$ is differentiable everywhere?

Perhaps an example will provide you the adequate outline. Consider $\pi_7(x_1,\ldots,x_n)=x_7$ and $g:\mathbb{R}\to \mathbb{R}$ given by $g(t)=t^3$. As $\pi_7$ and $g$ are differentiable, $(g\circ \pi_7)[(x_1,\ldots,x_n)]=x_7^3$ is also differentiable, etc, etc,...

 

[blue_raver22]

The chain rule is a powerful tool in calculus that allows us to calculate the derivative of a composite function. In this case, we are interested in proving that a function $f:\mathbb{R}^n\rightarrow \mathbb{R}$, which is polynomial in the coordinates, is differentiable everywhere.

To begin, let us consider the definition of differentiability. A function $f$ is said to be differentiable at a point $a$ if the following limit exists:

$$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

Using the chain rule, we can rewrite this limit as:

$$\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\cdot \frac{h}{h}=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\cdot \lim_{h\to 0}\frac{h}{h}$$

Since we are dealing with a composite function, we can express $f(a+h)$ as $f(g(h))$, where $g(h)$ is another function. Applying the chain rule, we can rewrite the limit as:

$$\lim_{h\to 0}\frac{f(g(h))-f(g(0))}{g(h)-g(0)}\cdot \lim_{h\to 0}\frac{g(h)-g(0)}{h}$$

Now, since $g$ is a polynomial in the coordinates, we can express it as $g(h)=a_n h^n + a_{n-1}h^{n-1} + ... + a_1 h + a_0$. Thus, we can rewrite the limit as:

$$\lim_{h\to 0}\frac{f(a_n h^n + a_{n-1}h^{n-1} + ... + a_1 h + a_0)-f(a_0)}{a_n h^n + a_{n-1}h^{n-1} + ... + a_1 h + a_0-a_0}\cdot \lim_{h\to 0}\frac{a_n h^n + a_{n-1}h^{n-1} + ... + a_1