I have come across an interesting result,
$$ \frac{d}{dx}\left( \frac{1}{1-x}\right) = \frac{d}{dx}\left( \frac{x}{1-x}\right) \quad\text{for}\ |x| \leq 1 $$
since,
$$ \frac{1}{1-x} = 1 + x + x^2 + \mathcal{O}(x^3) \quad\text{for}\ |x| \leq 1\\ \frac{x}{1-x} = x + x^2 + x^3 + \mathcal{O}(x^4) \quad\text{for}\ |x| \leq 1\\ $$ so in the limit to $\infty$, neglecting the 1, $$ \frac{d}{dx} \left( \frac{1}{1-x} \right) = \frac{d}{dx} \left( \frac{x}{1-x} \right) $$
Is the correct way to explain this as: the rate at which the functions increase with $x$ is the same save a constant?
It is a way, but you should use the power series (which converge only for $\lvert x\rvert<1$). However, writing the second rational function in canonical form is simpler: $$\frac{x}{1-x}=-1+\frac1{1-x},$$ hence the derivatives are equal.