Using the Constant Function Theorem to prove the Increasing Function Theorem I quote Thomas W.Tucker: 

... By the way, I view the Constant Function Theorem as even more basic than the IFT. It would be nice to use it as our theoretical cornerstone, but I know of no way to use it to get the IFT. ...

$\,$ from $\,$ Rethinking Rigor in Calculus: The Role of the Mean Value Theorem - The American Mathematical Monthly, Vol. 104, No. 3 (Mar., 1997), page 233 $ \,$ where IFT means Increasing Function Theorem. 
For convenience:  


*

*Increasing function $f$ means that $\,$if $c \lt d$, then $f(c) \le f(d)$;  

*IFT: if $f'(x) \ge 0$ on $[a,b]$, then $f$ is increasing on $[a,b]$;  

*CFT: if $f'(x)=0$ on $[a,b]$, then $f$ is constant on $[a,b]$.


Is it impossible to get the IFT from the CFT ?
 A: It would be nice to have a clear statement of what it means to "get from CFT to IFT".  In the most standard sense of logical implication, we have two true statements $A$ and $B$, so the statement "$A \implies B$" is certainly true.  But surely something other than this trivial observation is meant.
One nice way of construing statements like this is developed in Jim Propp's article Real Analysis in Reverse.  Namely, both assertions are meaningful when applied to an arbitrary ordered field, so one can explore the class of ordered fields in which they hold.  It turns out that many, but not all, of the interesting theorems in calculus imply the Dedekind completeness of the ordered field, so only hold in the real numbers.  That is the case here, as is treated in Propp's article: a gap in an ordered field leads to a locally constant function which is not constant, so CFT implies Dedekind completeness (hence so does IFT).
Does the above argument -- i.e., showing that CFT implies Dedekind completeness and then giving, say, the usual proof of the Mean Value Theorem and then deducing IFT -- count as "getting from CFT to IFT"?  I would guess not.  But then, as I pointed out above, it's less than clear what this "getting from" business really means.
Note that exactly this issue comes up in $\S 5.1$ of this note of mine on real induction, in which I suggest a rather inadequate meaning for equivalence of theorems like this by saying that each one "immediately implies" the other...whatever that means!
A: IFT to CFT: consider $f(x)$ and $g(x) := -f(x)$ s.t. $f(x)$ is differentiable... (the standard set-up). Now, $f'(x) \ge 0$, $g'(x) \le 0$, so $f$ is increasing, and $g$ is decreasing.
(All that jazz^ to show that we get decreasing functions if $g'(x) \ge 0$)
Now, if $h(x)$ also differentiable and so on, $h'(x) = 0$, we have
$h'(x) = 0 \implies h'(x) \ge 0 \implies h \text{ is increasing}$, i.e, $h(x) \ge h(y)$ if $x>y$
$h'(x) = 0 \implies h'(x) \le 0 \implies h \text{ is decreasing}$, i.e., $h(x)\le h(y)$ if $x>y$
so if $x>y$, putting both inequalities together: $h(x) = h(y)$ (note this is for all $x$ and $y$), so $h$ must be constant as it equals a constant value $h(x)$ for all values of $x$.
I know this isn't the most rigorous answer, but you can fill in the details.
