Let $\;f : \; \stackrel{\circ}{D}\; \subset \mathbb{R} \to \mathbb{R}$ differentiable in $x_0 \in \; \stackrel{\circ}{D}\;$ and $f\;'(x_0) > 0$.
Does exists a neighborhood $A \subset \; \stackrel{\circ}{D}\;$ of $x_0$ where $f$ is crescent?
Let $\;f : \; \stackrel{\circ}{D}\; \subset \mathbb{R} \to \mathbb{R}$ differentiable in $x_0 \in \; \stackrel{\circ}{D}\;$ and $f\;'(x_0) > 0$. Does exists a neighborhood $A \subset \; \stackrel{\circ}{D}\;$ of $x_0$ where $f$ is crescent? 


No. The function $$f(x)= x+2x^2\sin\frac{1}{x},\quad x\in\mathbb R,$$ is not monotonic in any neighborhood of $x=0$ yet $f'(0)=1$. 

