let $g(x):= \frac{f(x)}{h(x)}$ and : $f(x) =a_nx^n+a_{n-1}x^{n-1}+...+a_0$ let : $$g(x):= \frac{f(x)}{h(x)}$$ and :
$$f(x) =a_nx^n+a_{n-1}x^{n-1}+...+a_0 \ \ \ n\in \mathbb{N}$$
$$h(x) =b_mx^m+b_{m-1}x^{m-1}+...+b_0\ \ \ m\in \mathbb{N}$$
Now :
$$g'(x)=?$$
My Try :
$$g'(x)=\frac{(n a_{n}x^{n-1}+(n-1)a_{n-1}x^{n-2}+...+a_1)h(x)-( mb_mx^{m-1}+(m-1)b_{m-1}x^{m-2}+...+b_1)f(x)}{(h(x))^2}$$
Now :
if $n=m$
What is its simplest form?
 A: It is usually considered that $n < m$ or you can make it to that level. Then, you can obtain the zeros of $g(x)$, and apply partial fraction expansion to break $g(x)$ down to simple terms, which are easily differentiable.
Edit:
If $m = n$,
$$g(x) = \frac{a_n}{b_n} + \frac{(a_{n-1} - \frac{b_{n}a_{n-1}}{a_m})x^{m-1}+...+(a_{0} - \frac{a_{0}b_n}{a_m})}{b_mx^m+b_{m-1}x^{m-1}+...+b_0} = k +  h(x)$$
Where, $k = \frac{a_n}{b_n}$ is a constant, and hence vanishes upon differentiation. You can apply partial fraction on $h(x)$, and then differentiate.
A: EDIT-1: I've just realized in the question $m = n$, so I redo the example, but the idea is still the same.
EDIT-2: I've fixed the definition of simplest form.
I don't know if there is a standard definition of simplest form. But as a working definition, it seems appropriate to define the simplest form of a rational function to be $c \frac {f(x)} {g(x)}$ such that $c \in \mathbb{R}$, $f(x)$, $g(x)$ are monic (i.e. with leading coefficient $1$) polynomial with no common polynomial divisors (i.e. $\gcd(f, g) = 1$ as polynomial).
Now we claim for a general rational function $\frac {f(x)} {g(x)}$, it can happen that $\frac {g(x)f'(x) - f(x)g'(x)} {(g(x))^2}$ is already in the simplest form after expanding the numerator and denominator.
Consider
$$\frac {f(x)} {g(x)} = \frac {2 x^2 + x + 1} {x^2 + x + 1}$$
We have
$$g(x)f'(x) - f(x)g'(x) = x^2 + 2 x$$
and
$$(g(x))^2 = x^4 + 2 x^3 + 3 x^2 + 2 x + 1$$
Now  $x^2 + 2x$ factors completely into $x (x + 2)$. By direct checking, we see that $0$ and $-2$ are not roots of $x^4 + 2 x^3 + 3 x^2 + 2 x + 1$, so we must have $x^2 + 2x$ and $x^4 + 2 x^3 + 3 x^2 + 2 x + 1$ having no common factors. Hence $\frac {x^2 + 2x} {x^4 + 2 x^3 + 3 x^2 + 2 x + 1}$ is already in the simplest form by our definition.
So in general, we cannot hope to find a formula that is simpler than $\frac {g(x)f'(x) - f(x)g'(x)} {(g(x))^2}$. Of course, if $\deg f = \deg g = n$ is given, one can do the expansion and collect-like-terms steps explicitly. But in general, the numerator and denominator may not have cancellation.
