Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}
$$
Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?

$$\eqalign{x+xa^2-a-ax^2&= x-a+xa^2-ax^2 \\ & = x-a+x(a^2-ax) \\ &= x-a+x(a(a-x)) \\ &= x-a+x(-a(x-a)) \\ &=\color{blue}{x-a}-ax\color{blue}{(x-a)} \\ &=(x-a)(1-ax).\;\checkmark
}$$
Therefore you can conclude that: $$\eqalign{\require{cancel}\dfrac{f(x)-f(a)}{x-a}&=\dfrac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)} \\ &=\dfrac{\color{red}{\cancel{\color{black}{(x-a)}}}(1-ax)}{(1+x^2)(1+a^2)\color{red}{\cancel{\color{black}{(x-a)}}}} \\ &= \dfrac{1-ax}{(1+x^2)(1+a^2)}.
}\tag{$x\neq a$}$$
