Help with a simple derivative I am trying to solve $\dfrac {6} {\sqrt {x^3+6}}$ 
and so far I made it to $6(x^3+6)^{-\frac 1 2}$ 
then I continued and now I have $(x^3+6)^{- \frac 3 2} * 3x^2$
and I cannot figure out what how to find the constant that should be before the parenthesis.

$\dfrac{d}{dx} \left(\dfrac{6}{\sqrt{x^3+6}}\right) = 6 \dfrac{d}{dx}\left(\dfrac{1}{\sqrt{x^3+6}}\right) = 6\cdot -\dfrac{1}{2}\left(x^3+6\right)^{\frac{-3}{2}}\cdot 3x^2 = -9x^2\left(x^3+6\right)^{\frac{-3}{2}}$ via chain rule. 
