Is there a uniformly continuous function values in $f: \mathbb{R}{\to}\mathbb{R}$ such that its first derivative is not bounded and is defined on a non-compact set?
And if $f: X_{1}{\to}\mathbb{R}$? (let $(X_{1},d{1})$ metric space ).
And if $f: X_{1}{\to}X_{2}$? (let $(X_{2},d{2})$ metric space ).
I know that the first derivative is bounded f:I-->R (let I interval) iff f is a lipschitz function; If X1 is a compact thanks to the Heine-Cantor theorem f is uniformly continuous.
On
The function $f(x)=\frac{\sin(e^x)}{1+x^2}$ has derivative $\frac{e^x\cos(e^x)(1+x^2)-2x\sin(e^x)}{(1+x^2)^2}$, which is clearly unbounded. Yet, $f$ is uniformly continuous as for $f(x)-f(y)$ to be $\ge \epsilon$, we clearly need at least one of $|f(x)|, |f(y)|$ to be $\ge \frac\epsilon2$, hence $|x|$ or $|y|$ must be $\le\sqrt{ \frac2\epsilon-1}$, so we may essentially consider $f$ on a compact interval.
EDIT: To elaborate, I'll give an explicit proof of uniform continuity - the above short argument seems to have been ambiguous.
Let $\epsilon>0$ be given. We are looking for $\delta>0$ such that for all $x,y$ with $|x-y|<\delta$ we have $|f(x)-f(y)|<\epsilon$. Let $\epsilon_1=\min\{\epsilon,1\}$ and $L=1+\sqrt{\frac2{\epsilon_1}-1}$. Then $L\ge2$. The restriction $g\colon [-L,L]\to \mathbb R$, $x\mapsto f(x)$ of $f$ to the compact interval $[-L,L]$ is continuous and hence uniformly continuous. Therefore, there exists $\delta_0>0$ such that for all $x,y\in[-L,L]$ with $|x-y|<\delta_0$, we have $|g(x)-g(y)|<\epsilon$. Let $\delta=\min\{\delta_0,1\}$. Now if $x,y\in\mathbb R$ with $|x-y|<\delta$, I claim that $|f(x)-f(y)|<\epsilon$. Indeed, if both $x,y$ are $\in[-L,L]$, then $|f(x)-f(y)|=|g(x)-g(y)|<\epsilon$ because $|x-y|<\delta\le\delta_0$. If on the other hand one of $x,y$ has absolute value $>L$, then the other has absolute value $>L-\delta\ge L-1$, hence both $|x|,|y|$ are $>L-1=\sqrt{\frac2{\epsilon_1}-1}$. This implies $x^2+1>\frac2{\epsilon_1}$, hence $|f(x)|<\frac{\epsilon_1}2$ because the sine is bounded by $1$. Since similarly $|f(y)|<\frac{\epsilon_1}2$, we find $|f(x)-f(y)|<|f(x)|+|f(y)|<\epsilon_1\le\epsilon$. Thus in all possible cases $|f(x)-f(y)|<\epsilon$, as was to be shown.$_\square$
Finally, an explicit argument why $\frac{e^x\cos(e^x)(1+x^2)-2x\sin(e^x)}{(1+x^2)^2}$ is unbounded.
The inequality $e^t\ge1+t$ for all $t\in\mathbb R$ should be well-known.
It implies $e^t=(e^{t/3})^3\ge(1+\frac t3)^3$ for all $t\ge-3$.
Therefore, if $x\ge-3$ then
$$ \frac{e^x}{1+x^2}\ge \frac{1+x+\frac13x^2+\frac1{27}x^3}{1+x^2}=\frac{x}27+\frac13+\frac{26x+18}{27(x^2+1)}.$$
The last summand is bounded,hence the total expression is
$$ \frac{e^x}{1+x^2}\ge\frac{x}{27}+C$$
for some constant $C$.
Given $M\in\mathbb R$ (where without loss of generality, $M>C$), select an integer $k>\frac{e^{27(M-C)}}{2\pi}>0$ and let $x=\ln(2\pi k)\ge\ln(2\pi)>0$.
Then
$$f'(x)=\frac{e^x\cos(e^x)(1+x^2)-2x\sin(e^x)}{(1+x^2)^2} = \frac{e^x\cos(2k\pi)(1+x^2)-2x\sin(2k\pi)}{(1+x^2)^2} =\frac{e^x}{1+x^2} \ge \frac x{27}+C.$$
But $x=\ln(2\pi k)>27(M-C)$ then implies
$$ f'(x)>M.$$
Similarly, by letting $x=\ln((2k+1)\pi)$ we find $x$ with $f'(x)<-M$.
Thus $f'$ is neither bounded from above nor from below.$_\square$
There is a simpler function than $g(x)= \frac{sin(e^{x})}{1+x^2}$ uniformly continuous such that its first derivative is not bounded and is defined on a non-compact set.
The function $f:R^{+}\to R^{+}$ $f(x):=\sqrt{x}$ is a 0.5-holder function and then $f(x)$ is uniformly continuous. $f^{'}(x)= \frac{1}{2\sqrt{x}}$ which is unbounded in $\mathbb{R^{+}}$. Obviously the set $\mathbb{R^{+}}$ is non-compact with euclidean metric because is unbounded.