$g(x)\approx0\implies g'(x)\approx0?$

Question

I was answering another question, and a step required that I use the statement in the title. I did prove it in the answer, but I still want to make sure because @maxmilgram says that it implies that the derivative of every function is zero. I don't really understand how this follows, since the statement in the question title talks about taking a function to be approximately zero in the first place. Surely there's a difference between the exact value of the function and the approximation $0.$ Here is the proof. \begin{eqnarray} g(x)&\approx&0\\ g'(x)&=&\lim_{h\to0}\frac{g(x+h)-g(x)}h\\ &\approx&\lim_{h\to0}\frac{0-0}h\\ &=&0 \end{eqnarray} I didn't write it formally because it's so trivial in my opinion.

Answer 1

You've been given an example to show that you're wrong. I'd like to explain what's wrong with your 'reasoning'. In the below, "small" is used informally.

Not writing things formally because they're trivial is something people often do, but do be careful... $g(x)\approx 0$ so $g(x+h)-g(x)\approx 0$ is fine to say. However, $\frac{g(x+h)-g(x)}{h}\approx\frac{0}{h}$ is not fine. Imagine saying, "well, $0.00001\approx0$ so of course $0.00001\times n\approx 0$ for any $n$, right?" Clearly not; take $n$ to be a googolplex. There is a similar issue here; $g(x+h)-g(x)$ is "small" but is it small relative to $h$? Is it still small after dividing by $h$? This is not necessarily true.

For example, $\sqrt{h}$ is small, but it is not small relative to $h$; $\frac{\sqrt{h}}{h}=\frac{1}{\sqrt{h}}\to\infty,\,h\to0^+$ and $\infty$ is most definitely not small. That's why counterexamples like Eric's can exist.

However, if $g\approx0$ is so small that you can actually say, $g\in o(x)$ as $x\to0$, then we can precisely argue that $g'(0)=0$. We need precise bounds on the "smallness" to argue like that. Note: $g'(x)$ doesn't need to be small, just $g'(0)=0$ is forced in this case.

Answer 2

Any small function can have huge derivatives. Take for example the functions $f_n(x) = \frac{1}{n} \sin(n^2x)$. They have values in $[-\frac1n,\frac1n]$. The derivatives have huge values, since $f'_n(x) = n \cos(n^2x)$ takes on all the values in $[-n,n]$.