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.

$$\frac{d}{dx}(\frac{6}{\sqrt{x^3+6}})= 6\frac{d}{dx}(\frac{1}{\sqrt{x^3+6}})\\
\implies 6\frac{d}{dx}(\frac{1}{\sqrt{x^3+6}})=6\frac{d}{d(x^2+6)}(\frac{1}{\sqrt{x^3+6}})\frac{d}{dx}(x^3+6)=\frac{-3}{(x^3+6)^{3/2}}\cdot 3x^2=-9\frac{x^2}{(x^3+6)^{3/2}}$$
