Show $f^\prime : (a,b) \rightarrow R$ is unbounded. Question: Let $f:(a,b) \rightarrow R$ be an unbounded differentiable function. Show $f^\prime : (a,b) \rightarrow R$ is unbounded.  
My attempt: We know from the question that $f(x) > K$ for $K \in R$, and $\lim_{x\to c} \frac {f(x)-f(c)}{x-c} =L$ for $c \in (a,b)$. 
Then, $f^\prime (c) =\lim_{x\to c} \frac {f(x)-f(c)}{x-c}\ge \lim_{x\to c} \frac {K-f(c)}{x-c}\ge \lim_{x\to c} \frac {K-f(c)}{x} = \frac {K-f(c)}{c} \forall c\in(a,b)$. Therefore, $f^\prime (x) \forall x\in(a,b)$ is also unbounded.
Could you tell me if it is correct? and please give me some hint if wrong. 
Thank you in advance. 
 A: That is not true. What does it mean by $f(x)>K$ for $K\in{\bf{R}}$? Unbounded here means, for each $M>0$, one can find a corresponding $x\in(a,b)$ such that $|f(x)|>M$.
One can prove the assertion by this way: Assume the contrary that $f'$ is bounded, then there exists some $M>0$ such that $|f'(x)|\leq M$ for all $x\in(a,b)$.
We pick a $c\in(a,b)$ and then $|f(x)|\leq|f(x)-f(c)|+|f(c)|=|f'(\xi)||x-c|+|f(c)|\leq(b-a)M+|f(c)|$ for all $x\in(a,b)$, this is a contradiction.
A: Your answer makes no sense. For instance, what does it mean to assert that “$f'(x)\forall x\in(a,b)$ is also unbounded”? Besides, you never use the essential fact that you're working on a bounded interval $(a,b)$.
If $f'$ was bounded, let $M\in(0,+\infty)$ be such that $(\forall x\in(a,b)):\bigl|f'(x)\bigr|<M$. Pick $c\in(a,b)$. Then, for each $x\in(a,b)$,$$\bigl|f(x)\bigr|\leqslant\bigl|f(c)\bigr|+\bigl|f(x)-f(c)\bigr|\leqslant\bigl|f(c)\bigr|+M.|x-c|,$$by the mean value theorem and the definition of $M$. Therefore$$(\forall x\in(a,b)):\bigl|f(x)\bigr|\leqslant\bigl|f(c)\bigr|+M.\max\{b-c,c-a\}.$$But this is impossible, since we are assuming that $f$ is unbounded.
