<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] You can use the quotient rule or the product rule. Let's use the quotient rule for this: 
$$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'g-g'f}{g^2}$$
In your case we have: $f(x)=2 \implies f'(x)=0$ and $g(x)=x^2+3^3 \implies g'(x)=2x$
It follows:
$$\implies \frac{f'g-g'f}{g^2}=\frac{0\cdot (x^2+3^3)-2x\cdot2}{(x^2+3^3)^2}=\frac{-4x}{(x^2+3^3)^2}$$
The other approach would be to use the product rule:
$$\frac{d}{dx}(f(x)\cdot g(x))=f'g+g'f$$
In your case we have $$\frac{2}{x^2+3^3}=2(x^2+3^3)^{-1}$$
The corresponding functions and derivatives are: $f(x)=2 \implies f'(x)=0$ and $g(x)=(x^2+3^3)^{-1} \implies g'(x)=-2x(x^2+3^3)^{-2}$ (I used the chain rule here)
$$\implies f'g+g'f=0 \cdot (x^2+3^3)^{-1}-2x(x^2+3^3)^{-2}\cdot 2=\frac{-4x}{(x^2+3^3)^2}$$
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