lower bounding first derivative of polynomial f is uni-variate polynomial of degree d.
I am interested in lower bounding modulus first derivative of f (i.e. |f'|)in interval [a,b] given the promise that in interval [a,b] f' don't have any root.
Like in Markoff theorem..to upper bound first derivative of polynomial in open interval (a,b)
 is
       |f'(x)|<=(2*M*d*d)/(b-a)
where M is upper bound of f in (a,b),but I don't know proof of this theorem.
I know polynomial f i.e. its coefficient,degree and upper bound as well as lower bound of f in interval [a,b], Now I want to lower bound f' in [a,b] in terms of these terms..say like Markoff theorem.
Pls let me know whether this time I able to put my question clearly :)
Thanks
Ram 
 A: I think you need to be a bit more precise. For example, choose any polynomial $g(x)$ that does not vanish on $[a,b]$, let $f'(x)=(x-a+\epsilon)g(x)$, let $F(x)$ be an antiderivative of $f'(x)$, and let $f(x)=F(x)+C$ for some constant $C$. Choosing $\epsilon$ small, you can make $f'(a)=\epsilon g(a)$ small, while choosing an appropriate $C$, you can make $\sup |f(x)|$ or $\inf |f(x)|$ on the interval $[a,b]$ to be anything that you want. So it's really not clear what sort of lower bound you have in mind. 
A: This is basically repeating what Joe Silverman wrote, in a concrete example.
(I do so, as a misunderstanding between asker and answerers seems to persist.)
It might be difficult (or impossible) to get a lower bound similar to the upper bound.
Let us fix the interval $[0,1]$.


*

*For $x^2$ the derivative is $0$ in $0$, and this polynomial is thus not admissible by your question.

*For $(x+0.001)^2$ the derivative is $0.002$ in $0$, so very small, and this polynomial is admissible by your question.

*For $(x+1)^2$ the derivative is $2$ in $0$, so not too small. 
So, one would need to distinguish 2 and 3. With this example one could still guess this might be due to the fact that in 2 the polynomial is small at 0.
But replacing $(x+0.001)^2$ by 2'. $(x+0.001)^2 + 1$ the minimum is not too small anymore and just looking at rough parameters like maximum and minimum on the interval the polynomials 2' and 3 are not that different. While they are very different with respect to what you are looking for.
Thus, if a bound of the form you are looking for exists, it seems it definitely has to take into account other/finer quantities than maximum/minimum and degree.
(I know you do not restrict exclusively to this situation, so this is not a full answer;
but is meant to provide some 'bound' on what one can hope for.) 
A: As already discussed the answer to the question, as stated, is NO. However, one can get close if one is willing to go to a measure theoretic statement. For example, the statement
$$
 |\{x \in [a,b]:\quad |f'(x)| < \varepsilon\}| \to 0
$$
as $\varepsilon \to 0$, is an easy consequence of continuity. However, one can quantify this. If I remember correctly, it is relatively easy to prove
$$
 |\{x \in [a,b]:\quad |f'(x)| < \varepsilon\}| \leq C \varepsilon^{1/d}
$$
where $C$ is an (universal) constant and $d$ the degree of the polynomial $f$. I think one can give a proof by just rescaling Polya's inequality.
A: Okay, here goes:

1.  Calculate the coefficients $\langle c_j : j\in d\rangle$ of the degree d-1 polynomial f'.

2.  Run quantifier elimination over real closed fields on the formulas

$((\displaystyle\bigwedge_{j\in d} ((a \leq x_j) \land (x_j \leq b))) \implies$
$(((((\displaystyle\sum_{j\in n} 1)\cdot (\displaystyle\sum_{j\in d} (c_j \cdot (\displaystyle\prod_{j\in d} x_j))))+1) \leq 0) \lor (1 \leq ((\displaystyle\sum_{j\in n} 1)\cdot (\displaystyle\sum_{j\in d} (c_j \cdot (\displaystyle\prod_{j\in d} x_j))))))$

until the result is $ \;  (0 = 0) \;  $.

3.  $\frac1n$ is now a lower bound for $|f'|$ in the interval $[a,b]$.
A: We have the following inegalities :
$c_1 n (k+1) \leq \sup_{0 \neq f \in \mathcal P_{k,n}} \frac{ \| f' \|_{[-1,1]} }{ \| f \|_{[-1,1]} } \leq c_2 n (k+1)$
Where $c_1 > 0$, $c_2 > 0$ are absolute constants and $\mathcal P_{k,n}$ is the set of all
polynomials of degree at most $n$ with real coecients and with at most $k$ ($0 \leq k \leq n$)
zeros in the open unit disk.
