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. 
 A: Hint. One may apply
$$
\left( \frac1{f}\right)'=-\frac{f'}{f^2} \tag1
$$ with
$$
f(x)=x^2+3^3.
$$
Find $f'(x)$ then use $(1)$ and conclude with $\dfrac2{x^2+3^3}=2 \times \dfrac1{x^2+3^3}=2 \times \dfrac1{f}$.
A: 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)
$$\implies f'g+g'f=0 \cdot (x^2+3^3)^{-1}-2x(x^2+3^3)^{-2}\cdot 2=\frac{-4x}{(x^2+3^3)^2}$$
A: Start with 3^3 = 27 so the fraction changes to 2/(x^2+27). Then you get u = 2; u' = 0; v = x^2+27; v' = 2x; The put each into: (vu'-uv')/v^2
A: Classical approach:
$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ whereas $u$ and $v$ are functions in $x$. 
So when we have 
$$\frac{2}{x^2+3^3}$$
We have 
$$u(x) = 2$$
$$v(x) = x^2+ 3^3$$
And from that it follows
$$u'(x) = 0$$
$$v'(x) = 2x$$
Using the above formula
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{2}{x^2+3^3}\right) = \frac{0 -2\cdot2x}{(x^2+3^3)^2} = \frac{-4x}{x^4+2\cdot 3^3 x^2 + 3^6} = \frac{-4x}{x^4+54x^2+729}$$
A: $$\frac {d}{dx} \frac 2{(x^2+3)^3}=\frac {d}{dx} (2(x^2+3)^{-3})=2\frac {d}{dx} (x^2+3)^{-3}=2\times(-3)(x^2+3)^{-4}2x=-\frac {12x}{(x^2+3)^{4}}$$
