Very confused? Derivative or chain rule? I have a bunch of questions like this 

I have been learning about the chain rule, derivatives, the extended power rule, product rule, exponent rule, etc. But so far my book has not had any similar examples like this so I am confused as to where I begin. Do I use the chain rule? If so where do I start? Is there a particular "name" for these or just "finding the derivative?" I really appreciate your help!
Thanks!
 A: You may use any method that works.  A convenient feature of mathematics is that it works no matter which (valid) method you use.
For this kind of problem, it is easiest to ask yourself, "how many 'places' is there an $x$?"  By "places", I mean "numerator and denominator".  If there are $x$s in both places, then you (probably) will prefer to use the quotient rule.  If there are $x$s only in the numerator, then there is no quotient and the result is straightforward.  If there are $x$s only in the denominator, then the quotient rule will be very easy; the lack of $x$s in the numerator makes the derivative of the numerator especially simple.  Similarly, the chain rule makes it easy in the case that there are only $x$s in the denominator since this is "some function of $x$" raised to "some negative power".
Consider $g(x) = \frac{1}{f(x)}$.  We may apply the quotient rule: $$g'(x) = \frac{f(x) \cdot 0 - 1 \cdot f'(x)}{f(x)^2} = \frac{-f'(x)}{f(x)^2}$$ or we may apply the chain rule: $$g'(x) = \left( f(x)^{-1} \right)' = (-1)f(x)^{-2}f'(x) = \frac{-f'(x)}{f(x)^2}$$.  We get the same answer either way, so it doesn't matter which method we use.
Some people find it easier to go one way or the other.  It doesn't matter which way is chosen as long as the method is applied correctly.
A: If you differentiate $\frac{1}{x}$, you get $\frac{-1}{x^2}$.
Thus by the chain rule, if you differentiate $\frac{1}{g(x)}$, you get $\frac{-1}{g(x)^2}\cdot g'(x)$.
Let $g(x) =4x^2+4$, and substitute.
A: Remember: the chain rule is for finding derivatives of functions that look like $g(h(x))$, where $g(u)$ and $h(x)$ are both differentiable functions.
In this case, note that
$$
\frac{1}{4x^2+4}=g(h(x)),
$$
where
$$
g(u)=\frac{1}{u}\qquad\text{and}\qquad h(x)=4x^2+4.
$$
A: Therefore, the derivative is $$ \frac{d}{dx} \, \left( \frac{1}{4 \, x^2 + 4} \right) = \frac{8 x}{(4 \, x^2 + 4)^2}. $$
