derive using the chain rule Given the polinomyal $f(x)=\frac{x^3}{(4-x^2)^3}$ find $f'(x)$
So, If I try to derive this, first I must to apply the chain rule in the denominator and then derive of the division (...)
$$f'(x)=\frac{x^3}{3(4-x^2)^2(-2)} = \frac{x^3}{-6(4-x^2)^2}$$
(...)?
Or there is another way to do this?
 A: You're doing wrong. A good way to do this when still not expert in derivatives, is to write
$$
f(x)=\frac{g(x)}{h(x)}
$$
where
$$
g(x)=x^3,\qquad h(x)=(4-x^2)^3.
$$
Then, first of all, you have to differentiate the quotient:
$$
f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2}
$$
So you need to compute $g'(x)$ and $h'(x)$. The first one is easy:
$$
g'(x)=3x^2.
$$
The second one needs the chain rule:
$$
h'(x)=3(4-x^2)^2(-2x)=-6x(4-x^2)^2.
$$
Now plug in the formula above:
\begin{align}
f'(x)&=
\frac{3x^2\cdot(4-x^2)^3 - x^3\cdot(-6x(4-x^2)^2)}{(4-x^2)^6}\\[2ex]
&=\frac{3x^2(4-x^2)^3+6x^4(4-x^2)^2}{(4-x^2)^6}\\[2ex]
&=\frac{3x^2(4-x^2)+6x^4}{(4-x^2)^4}\\[2ex]
&=\frac{12x^2+3x^4}{(4-x^2)^4}=\frac{3x^2(4+x^2)}{(4-x^2)^4}
\end{align}
A different strategy may be noting that
$$
f(x)=\left(\frac{x}{4-x^2}\right)^3
$$
so, letting
$$g(x)=\frac{x}{4-x^2}$$
you have, by the chain rule,
$$
f'(x)=3g(x)^2g'(x)
$$
The derivative of $g(x)$ can be computed again with the quotient rule:
$$
g'(x)=\frac{1\cdot(4-x^2)-x\cdot(-2x)}{(4-x^2)^2}=
\frac{4-x^2+2x^2}{(4-x^2)^2}=\frac{4+x^2}{(4-x^2)^2}
$$
and so
$$
f'(x)=3\left(\frac{x}{(4-x^2)}\right)^2\frac{4+x^2}{(4-x^2)^2}
$$
which gives the same result as before.
A: You can probably use quotient rule,
$$h(x)=\dfrac{f(x)}{g(x)}$$
$$h'(x)=\dfrac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}$$
A: You want to use the quotient rule then the chain rule.
$$\frac{d}{dx}\frac{f(x)}{g(h(x))}=\frac{g(h(x))f'(x)-f(x)g'(h(x))h'(x)}{g(h(x))^2} $$
so:
$$\frac{d}{dx}\frac{x^3}{(4-x^2)^3} = \frac{(4-x^2)^3(3x^2)-x^3(3(4-x^2)^3)(-2x)}{(4-x^2)^6} $$
