Let $f :\left ( a,b \right ) → \mathbb{R}$ be an unbounded differentiable function.
Prove that the derivative $f{}':\left ( a,b \right )\rightarrow \mathbb{R}$ is also unbounded.
I've looked at some solutions and got some ideas from them. But, I'd like to get some feedback on my approach as well because I'm not quite sure about my understanding of concepts of derivatives in the Real-Analysis. So, I will try to write in detail to show my level of understanding.
Proof: Suppose that $\left | {f}'\left ( x \right ) \right |$ is bounded for any $x\in \left ( a,b \right )$.
$\Rightarrow \exists M>0$ s.t $\left | {f}'\left ( x \right ) \right |< M$.
Let $c\in \left ( a,b \right )$.
Since $f$ is differentiable on $\left ( a,b \right )$, it is continuous on $\left ( a,b \right )$. $\Rightarrow f\left ( c \right )$ is well-defined. $\Rightarrow \left |f\left ( c \right ) \right |$ has some fixed value.
Let $\left ( x_{n} \right )$ be any sequence that satisfies $x_{n}\neq c$ $\forall n$, but $\lim_{n\rightarrow \infty }x_{n}= c.$ Also assume that $\forall \left (x_{n} \right )$, we have $x_{n}\in \left ( a,b \right )$ $\forall n$.
Then $\forall \left (x_{n} \right )$, $f$ is continuous on $\left [ \min \left ( x_{n},c \right ),\max \left ( x_{n},c \right ) \right ]$, and $f$ is differentiable on $\left ( \min \left ( x_{n},c \right ),\max \left ( x_{n},c \right ) \right )$.
Then by Mean Value Theorem, $\exists k\in \left ( \min \left ( x_{n},c \right ),\max \left ( x_{n},c \right ) \right ) $ s.t $${f}'\left ( k \right )=\frac{f\left ( \max \left ( x_{n},c \right ) \right )-f\left ( \min \left ( x_{n},c \right ) \right )}{\max\left ( x_{n},c \right )-\min \left ( x_{n},c \right )}$$
$$\Rightarrow \left | {f}'\left ( k \right ) \right |=\frac{\left | f\left ( x_{n} \right )-f\left ( c \right ) \right |}{\left | x_{n}-c \right |}$$
$$\Rightarrow \left | {f}'\left ( k \right ) \right |\left | x_{n}-c \right |=\left | f\left ( x_{n} \right )-f\left ( c \right ) \right |$$
Now, let's consider $\left | f\left ( x_{n} \right ) \right |$ $\forall \left ( x_{n} \right )$.
$$\left | f\left ( x_{n} \right ) \right |=\left | f\left ( x_{n} \right )-f\left ( c \right )+f\left ( c \right ) \right |\leq \left | f\left ( x_{n} \right )-f\left ( c \right ) \right |+\left | f\left ( c \right ) \right |=\left | x_{n} -c\right |\cdot \left | {f}'\left ( k \right ) \right |+\left | f\left ( c \right ) \right |$$ $$< M\left | x_{n}-c \right |+\left | f\left ( c \right ) \right |< M\left ( b-a \right )+\left | f(c) \right |$$
But since $\left | f\left ( x \right ) \right |$ is unbounded, $\exists \left ( x_{n} \right )$ that makes $\left | f\left ( x_{n} \right ) \right |$ unbounded.
For such $\left ( x_{n} \right )$, $\exists N$ s.t $\left | f\left ( x_{N} \right ) \right |>M(b-a)+|f(c)|$.
For such $N$, $M(b-a)+|f(c)|<\left | f\left ( x_{N} \right ) \right |<M(b-a)+|f(c)|$.
This is a contradiction derived from our supposition "$\left | {f}'\left ( x \right ) \right |$ is bounded for any $x\in \left ( a,b \right )$."
Hence, $\left |{f}'\left ( x \right ) \right |$ is unbounded. $\blacksquare$
We need to assume that $$ -\infty<a<b<\infty, $$ for otherwise, $f(x)=x$ is unbounded in $(-\infty,\infty)$ and its derivative is bounded.
Assume that $|f'(x)|\le M$ and pick a arbitrary $x_0\in (a,b)$. Then $$ f(x)-f(x_0)=f'(\xi)(x-x_0) $$ and hence $$ |f(x)|=|f(x_0)+f'(\xi)(x-x_0)|\le |f(x_0)|+M|x-x_0|\le |f(x_0)|+M|b-a|=M_1, $$ for every $x\in (a,b)$.