Prove $f(x) = \frac{1}{x}$ is smooth (infinitely differentiable).

I have never proved that a function is smooth (infinitely differentiable) before. The only function that comes to mind which is smooth is $g(x) = e^{x}$, because it is defined on all of $\Bbb R$, continuous everywhere, and once you prove that $g'(x) = e^{x}$, you are done in showing that it is infinitely differentiable, i.e., smooth.

How would I prove that the function $f(x) = \frac{1}{x}$ is smooth everywhere except at $0$? It's not hard to show that the derivative at each $x \neq 0$ of $f$ is $-\frac{1}{x^{2}}$. So, I have that the first derivative exists. How do I go about showing that derivatives of all orders exist?

• Find a closed form for $f^{(n)}(x)$. – Alex R. Feb 10 '15 at 23:07
• @AlexR. Are you saying this because you know a closed form will exist? – layman Feb 10 '15 at 23:08
• yes, it exists. The main takeaway is that the closed form looks like $c/x^{n+1}$, where $c$ is a constant that only depends on $n$. – Alex R. Feb 10 '15 at 23:09

$f'(x)=\frac{-1}{x^2}$ $f''(x)=\frac{2}{x^3}$ $f'''(x)=\frac{-6}{x^4}$ so

$f^{(n)}(x)=n!\frac{(-1)^{n}}{x^{n+1}}$

• I guess what I really wanted was to prove these using the limit definition of derivative. But now that I think about it, there is no need. Once we have an established identity, why work with the harder thing when we can work with the easier thing? – layman Feb 10 '15 at 23:29

First fact: every rational function is differentiable and its derivative is a rational function.

This is clear since

$$\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x) g(x) - f(x) g'(x)}{g^2(x)}$$

Iterating the argument, you can show that every rational functions is infinitely differentiable (formally you have to use induction).

• I think it's because rational functions have numerator and denominator polynomials, which are infinitely differentiable, that this ultimately holds. Maybe we can take it even further and say the quotient of infinitely differentiable functions is infinitely differentiable when the denominator is non-zero. – layman Feb 10 '15 at 23:43

If you can differentiate $f^n(x) = c_n x^{-n}$ and show it is continuous except at $x = 0$, then you can prove smoothness by induction.