Krdytkyu

Suppose $f(x)$ is a continuously differentiable function on $mathbb{R}$ such that its derivative is unbounded.

Claim: $f(x)$ is not uniformly continuous.

Assume f(x) is uniformly continuous, then given $epsilon$ we have a suitable $delta$. Since $f'(x)$ is not bounded, their exists an element $c$ such that $f'(c) > k frac{epsilon}{delta}$, $k gg 0$. Since $f'(x)$ is continuous, if we choose a $delta$ neighbourhood of $c$, we would then by mean value theorem have
$$f(b) - f(a) = f'(c')(b-a), c' in Big(c- frac{delta}{2},; c + frac{delta}{2}Big).$$ or

$$ f'(c') = frac{epsilon}{delta}$$ - a contradiction since $f'(c')$ has value in around some small neighborhood of $k frac{epsilon}{delta}$.

So finally I have this conclusion:

If $f(x)$ is continuously differential function, then $f(x)$ is uniformly continuous iff $f'(x)$ is bounded?

Is my conclusion right?

New details:

I realize that that my argument $f'(c) to f'(c')$ is slippery and might not yield the desired conclusion every time, but suppose if I have a specific $f(x)$ and I can somehow show that $m leq mid f'(x) mid leq M$ for some interval $I$ where $f'(c) = kfrac{epsilon}{delta}$ and $m, M$ are suitable close enough (say $epsilon^{1000000})$, then does my argument holds?

Take any unbounded continuous integrable function $g$ and define $f(x)=int_0^{x} g(y), dy$. Then $f$ is uniformly continuous but its derivative is not bounded. To construct such a function $g$ draw triangles with bases $(n-frac 1 {n^{3}}, n+frac 1 {n^{3}})$ and height $n$. Think of a function $g$ whose graph is made up of these triangles. $f$ is uniformly continuous because it is continuous and it has finite limits at $pm infty$.

The error in OP's argument is that they fix $epsilon, delta$ beforehand for the uniformity claim, but later use the fact that the derivative $f'$ is above a certain value in an interval of length $delta$, which is not guaranteed, as Kavi explained.

Intuitively, if the interval where $f'$ is high is very small, and becomes smaller with higher $f'$, for a given $epsilon$ the 'spikes' in $f'$ (as described by Kavi) are not able to increase $f$ by much (more than $epsilon$) except for a finite number at the beginning.

No, you are not right. Consider for example the $C^{infty}$ function
$$f(x)=frac{cos(e^x)}{1+x^2}.$$
$f$ is uniformly continuous in $mathbb{R}$ because it is continuous and its limit at $pm infty$ is zero.

On the other hand, its derivative is unbounded in $mathbb{R}$:
$$f'(x)=-frac{e^xsin(e^x)}{1+x^2}-frac{2xcos(e^x)}{(1+x^2)^2}$$
and $lim_{nto +infty} f'(x_n)=+infty$ where $x_n=ln(3pi/2+2pi n)$.