Why are the derivatives of $\frac{x^2+1}{x^2+x+1}$ and $\frac{-x}{x^2+x+1}$ the same? 
Why are the derivatives of these functions the same?
$$\frac{x^2+1}{x^2+x+1} \qquad\qquad \frac{-x}{x^2+x+1}$$

original exercise text (See part (e).)
I have tried to answer this question and consulted the answer booklet but this did not make much sense. Would someone be able to simplify or elaborate further? Thank you :)
The answer guide said the derivates are the same as differ at most by an additive constant.

$$
{d\over dx} (f(x)+1)={df\over dx}+{d1\over dx}={df\over dx}
$$
so we can add $1$ to a function and the derivative of the sum is the same as the derivative of the function. So
$$
\begin{align}
\frac{-x}{x^2+x+1}+1
&=\frac{-x}{x^2+x+1}+\frac{x^2+x+1}{x^2+x+1}\\
&={x^2+x+1-x\over x^2+x+1}\\
&={x^2+1\over x^2+x+1}\\
\end{align}
$$
