Given that $\displaystyle y=2x^3-6x^2+18x$, find $\displaystyle \frac{dy}{dx}$ Hence show that $\displaystyle \frac{dy}{dx}> 0$ for all x.

I know how to find the derivative but what does it mean by:

Hence show that $\displaystyle \frac{dy}{dx}> 0$ for all x, above? By finding the derivative how do I show this? If the derivative is positive do I assume its greater than 0? Is that how I show it?