Assume that real function $f$ is differentiable at $x_0$ with $f'(x_0)>0$. How would one show that there exists a $\delta>0$ such that $$ f(x)>f(x_0) $$ for all $x$ in between $x_0$ and $ x_0 + \delta $.
Is it always true that $f$ is increasing on an interval $(x_0, x_0+\epsilon)$ for some $\epsilon>0$?
Since $f'(x_0) > 0$, you know that $$ \lim_{h\to 0} \frac{f(x_0+h) - f(x_0)}{h} > 0 $$ Hence (why?), $\exists \delta > 0$ such that $$ 0 < h < \delta \Rightarrow f(x_0+h) - f(x_0) > 0 $$ Hence $$ f(x) > f(x_0) \quad\forall x\in (x_0, x_0+\delta) $$
Hint: write the definition of differentiability as $$ f(x) = f(x_0) + f'(x_0)(x-x_0)+ (x-x_0)\epsilon(x-x_0) $$with $\lim_0\epsilon = 0$ and write an $\epsilon-\delta$ thing with the function $\epsilon$.
details:
when $(x-x_0)$ is small enough and positive, $|\epsilon(x-x_0)| < f'(x_0)/2$ because $f'(x_0)>0$. For such an $x$ we have $$ f(x) - f(x_0) = (f'(x_0)+\epsilon(x-x_0))(x-x_0) > (f'(x_0)-f'(x_0)/2)(x-x_0) = f'(x_0)(x-x_0)/2>0 $$
Hint: Find $\delta$ such that if $|x - x_0| < \delta$, then $\frac12 f'(x_0) < \frac{f(x) - f(x_0)}{x - x_0} < \frac32 f'(x_0)$.
