Derivates being bounded in some subinterval Suppose $f(x)$ is differentiable on $(a,b)$. Prove that there exists a subinterval $(c,d)\subset(a,b)$ such that $f'(x)$ is bounded on $(c,d)$.
I have no idea then.
 A: Modifying the Carother's proof mentioned in my comment above:
For each positive integer $n$, define $f_n(x)={f(x+1/n)-f(x)\over 1/n}$. Then $(f_n)$ converges to $f'$ pointwise on $(a,b)$. 
Now, for $n$ a positive integer, define the set $E_n=\cap_{i,j\ge n}\{ x : |f_i(x)-f_j(x)|\le 1\}$. By the pointwise convergence of $(f_n)$, it follows that $\cup_{i=1}^\infty E_i=(a,b)$. Also, since each $f_i$ is continuous, each $E_n$ is closed. 
From the Baire Category Theorem (or rather, the corollary that an open interval is not a countable union of nowhere dense sets)  it follows that there is an $n$ and an open interval $I $ which is a proper subset of $(a,b)$ such that $I\subset E_n$. From the definition of $E_n$, it follows that for all $x\in I$ and all $i,j\ge n$, we have $|f_i(x)-f_j(x)|\le 1$. Consequently, for all $x\in I$, we have $|f'(x)-f_n(x)|\le 1$.   
Now, as $f_n$ is continuous on $I$, it follows that there is an $M$ so that $|f_n(x)|\le M$ for all $x\in I$.  But then for all $x\in I$, we have $|f'(x)|\le M+1$, which proves the result.
