Showing that two functions are different on some interval with differentiation. Suppose I have two functions $f(x)$ and $g(x)$. Both functions are differentiable at some constant c. I want to show that the two functions are different on at least one interval with a positive lebesgue measure. 
I can show that $f’(c)\ne g’(c)$. My guess is it is enough to show that $f(x) \ne g(x)$ on a small neighbourhood of $c$.
Am I correct? If so, how should I argue this properly?
 A: WLOG assume $f'(c) > g'(c)$. Let $h(x) = f(x + c) - g(x + c)$. Then $h$ is differentiable at $x=0$, with $h'(0) = f'(c) - g'(c) > 0$.
There are two possibilities:
Csae 1: $h(0) \neq 0$
In this case, since $h$ is continuous at zero, we have $h(x) \neq 0$ in some $\epsilon$-neighborhood $(-\epsilon, \epsilon)$, hence $f(x+c) \neq g(x+c)$ in that neighborhood, or equivalently $f(x) \neq g(x)$ for $x \in (c-\epsilon, c+\epsilon)$. 
Case 1: $h(0) = 0$
Working from the definition of the derivative, we have
$$\lim_{x \to 0}\frac{h(x)}{x} = h'(0) > 0$$
Hence there is some $\epsilon > 0$ such that for all $x$ with $0 < |x| < \epsilon$, we have
$$\left|\frac{h(x)}{x}\right| = \frac{h(x)}{x} > \frac{h'(0)}{2} > 0$$
This means that
$$|h(x)| > \frac{h'(0)}{2}|x| > 0$$
and in particular $h(x) \neq 0$. Referring to the definition of $h$, this means that for $0 < |x| < \epsilon$, we have
$$f(x+c) - g(x+c) \neq 0$$
so $f(x+c) \neq g(x+c)$. Equivalently, $f(x) \neq g(x)$ for all $x \in (c-\epsilon, c+\epsilon) \setminus \{c\}$.
