Prove that if $f$ is differentiable on $[a,b]$ and $f$ is Lipschitz, then $f$ has a bounded derivative. Prove that if $f$ is differentiable on $[a,b]$ and $f$ is Lipschitz continuous (LC), then $f$ has a bounded derivative.
My proof: 
$f$ is LC $\Rightarrow$ f has bounded a derivative: there exists $M\gt 0$ such that $$|f(x)-f(y)|\le M|x-y|, \forall x,y\in [a,b]$$ 
then $${|f(x)-f(y)|\over {|x-y|}}\le M$$(if $ x\neq y$) 
If $y\lt x$ (without loss of generality $x\lt y$) then $[y,x]\subseteq [a,b]$ and by hypothesis $f$ is differentiable on $[a,b]$ hence $f$ is differentiable on $[y,x]$. Then, using Mean Value Theorem there must exist
$c\in (y,x)$ such that $$f´(c)={f(x)-f(y)\over x-y}$$ hence $$|f´(c)|\le M$$
As $x,y$ are arbitrary elements of $[a,b]$, then $c$ is also an arbitrary element but of $(a,b)$ therefore $$|f´(c)|\le M$$ $$\forall  c\in (a,b)$$ 
but does this imply that $$|f´(c)|\le M, \forall  c\in [a,b]?$$
I would really appreciate your help
 A: Your "proof" could be used for the converse: When $|f'(x)|\leq M$ for all $x\in[a,b]$ then $f$ is Lipschitz continuous on $[a,b]$, and $M$ can serve as a Lipschitz constant.
In fact we are asked to prove that there is an $M>0$ with $|f'(x)|\leq M$ for all $x\in[a,b]$, given that $f$ is Lipschitz continuous on $[a,b]$.
Now when $|f(y)-f(x)|\leq M\>|y-x|$ for all $x$, $y\in[a,b]$ then
$$\left|{f(y)-f(x)\over y-x}\right|\leq M\qquad(y\ne x)\ .$$
It follows that for each fixed $x\in[a,b]$ we have
$$\left|\lim_{y\to x}{f(y)-f(x)\over y-x}\right|\leq M$$
and therefore $|f'(x)|\leq M$.
A: Fix a point $x_0 \in (a,b)$ and pick an arbitrary point $x \in (a,b)$ 
Then 
$$|f'(x) - f'(x_0)| = \lim_{w \rightarrow x,z \rightarrow x_0} |\frac{f(x)-f(w)}{w-x} -
\frac{f(x_0)-f(z)}{x_0-z}| \leq 2K$$
where $K$ is the Lipschitz constant. 
But then $\forall x \in (a,b)$
$$|f'(x)| \leq |f'(x) - f'(x_0)| + |f'(x_0)| \leq 2K + f'(x_0)$$
For the endpoints a similar argument will work, using monotonic limits in the above calculations. 
