Strategy to calculate $ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right) $. I am asked to calculate the following: $$ \frac{d}{dx} \left(\frac{x^2-6x-9}{2x^2(x+3)^2}\right). $$
I simplify this a little bit, by moving the constant multiplicator out of the derivative:
$$ \left(\frac{1}{2}\right) \frac{d}{dx} \left(\frac{x^2-6x-9}{x^2(x+3)^2}\right) $$
But, using the quotient-rule, the resulting expressions really get unwieldy:
$$ \frac{1}{2} \frac{(2x-6)(x^2(x+3)^2) -(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} $$
I came up with two approaches (3 maybe):


*

*Split the terms up like this: $$ \frac{1}{2}\left( \frac{(2x-6)(x^2(x+3)^2)}{(x^2(x+3)^2)^2} - \frac{(x^2-6x-9)(2x(2x^2+9x+9))}{(x^2(x+3)^2)^2} \right) $$
so that I can simplify the left term to $$ \frac{2x-6}{x^2(x+3)^2}. $$
Taking this approach the right term still doesn't simplify nicely, and I struggle to combine the two terms into one fraction at the end.

*The brute-force-method. Just expand all the expressions in numerator and denominator, and add/subtract monomials of the same order. This definitely works, but i feel like a stupid robot doing this.

*The unofficial third-method. Grab a calculator, or computer-algebra-program and let it do the hard work.
Is there any strategy apart from my mentioned ones? Am I missing something in my first approach which would make the process go more smoothly?
I am looking for general tips to tackle polynomial fractions such as this one, not a plain answer to this specific problem.

Note that $x^2-6x-9 = (x-3)^2 - 18$. So after pulling out the factor of $\frac 12$, it suffices to compute 
$$\frac{d}{dx} \left(\frac{x-3}{x(x+3)}\right)^2$$
and
$$\frac{d}{dx} \left(\frac{1}{x(x+3)}\right)^2.$$
These obviously only require finding the derivative of what's inside, since the derivative of $(f(x))^2$ is $2f(x)f'(x)$.
For a final simplification, note that
$$\frac{1}{x(x+3)} = \frac{1}{3} \left(\frac 1x - \frac{1}{x+3}\right),$$
so you'll only ever need to take derivatives of $\frac 1x$ and $\frac {1}{x+3}$ to finish, since the $x-3$ in the numerator of the first fraction will simplify with these to give an integer plus multiples of these terms.
As a general rule, partial fractions will greatly simplify the work required in similar problems.
