<s>[INST] How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$  x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$  over $({x^2+3)^6}$
I'm not sure if this is right. 
 [/INST] $$\frac {d}{dx} \frac 2{(x^2+3)^3}=\frac {d}{dx} (2(x^2+3)^{-3})=2\frac {d}{dx} (x^2+3)^{-3}=2\times(-3)(x^2+3)^{-4}2x=-\frac {12x}{(x^2+3)^{4}}$$
</s>