1. ## Derivatives

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?

2. Originally Posted by Mr Rayon
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?
$\displaystyle y=2x^3-6x^2+18x$

$\displaystyle \frac{dy}{dx}=6x^2-12x+18$

$\displaystyle \frac{dy}{dx}=6(x^2-2x+3)$

$\displaystyle \frac{dy}{dx}=6[(x-1)^2+2]$

Hence $\displaystyle \frac{dy}{dx}>0$ for all x.