Let $f:(a,b) \to \mathbb{R}$ to be differentiable and unbounded. Prove: $f'$ is not bounded on $(a,b)$ Let $f:(a,b) \to \mathbb{R}$ to be unbounded and a differentiable function. 
Prove: $f'$ isn't bounded on $(a,b)$ 
My Attempt:
suppose towards contradiction that f' is bounded in $(a,b)$.
then there exist some $M >0$ such that $f$ is a M-Lipschitz function. 
Therefore, $f$ is a uniformly continuous functions such that for every $\epsilon > 0$ there exist some $ \delta > 0$ such that for every $x , y \in (a, b): \ |f(x) - f(y)|< \epsilon$
But, given that $f$ isn't bounded and continues, than because (a,b) is an open interval - WLOG: 
$lim_{x \to b^-} = \infty$ (WLOG - doesn't matter whether its $+\infty$ or $-\infty$ - only that the limit is infinite). 
this is a contradiction to the assumption that $f'$ is bounded - Thus $f'$ isn't bounded. 
is this correct? I'm confused about the fact the argument about a M-Lipschitz argument isn't an "if only and only if" argument. 
 A: The last part of your argument is not clear for me, You can remark that since $f$ is uniformly continuous on $(a,b)$, you can extend it by continuity to $[a,b]$  and a continuous function on a closed interval is bounded,
another approach
Suppose that there exists $M>0$ such that $|f'(x)|<M$, for every integer $n$, there exists $x_n,y_n$ such that $|f(x_n)-f(y_n)|>n$, we have $|f(x_n)-f(y_n)|=|f'(c_n)||x_n-y_n|\leq M|b-a|$, this implies that every integer $n$ verifies $n\leq M|b-a|$. Contradicition.
A: Attempt:
Assume $f'$ is bounded on $(a,b)$, i.e.
$|f'(x)| < B (>0)$ , real.
Since $f$ is not bounded on $(a,b)$ there is 
a sequence $x_n \in (a,b)$ such that 
$y_n:= f(x_n)$ is not bounded.
Let $c \in (a,b)$.
$|f(x_n)| \le |f(x_n)-f(c)| +|f(c)| =$
$|f'(t)(x_n -c)| + |f(c)| \lt$ 
$B|(b-a)| +|f(c)|$, where 
$t \in (\min (x_n, c), \max (x_n, c)) $.
Since $|f(x_n)|$ is unbounded :
For every $M >0$, real, there is a 
$n_M$ such that $|f(x_M)| >M.$
Choose $M > B|b-a|+|f(c)|$, a contradiction.
Used: MVT
