For a $C^1$ function, the difference $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |$ is small when $|d-c|$ is small 
Suppose $g\in C^1 [a,b]$. Prove that for all $\epsilon > 0$, there is $\delta > 0$ such that $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{< \epsilon }$ for all points $c,d \in [a,b]$ with $0 <|d-c|< \delta$

First, I don't know what $C^1 [a,b]$ means.
Some ideas:
By Mean value theorem, $g'(c) =  {{g(b)-g(a)} \over {b-a}} $ since $c\in[a,b]$. 
To show $|{g'(c)} - {{g(d)-g(c)} \over {d-c}} |{< \epsilon }$ whenever $0 <|d-c|< \delta$ for $ c,d \in [a,b]$. I guess I have to use the definition of limit of continuous function. But, I don't know how to connect all of these ideas. 

This problem has nothing to do with the mean value theorem. This problem is designed to test your understanding of uniform continuity and/or compactness. It seems likely to me that you have just learned (in the context of whatever book or class this is from) the compactness argument that a continuous function on a compact interval is uniformly continuous.
Once you've parsed this and your question, the answer amounts to writing down a definition of uniform continuity.
