Differentiate $f(x) = \ln (x^2 + x + 1)$. Check your answer if its reasonable by comparing the graphs of $f$ and $f'$.

$ \begin{equation} \begin{aligned} \text{if } f(x) =& \ln (x^2 + x + 1), \text{ then} \\ \\ f'(x) =& \frac{\displaystyle \frac{d}{dx} (x^2 + x + 1) }{x^2 + x + 1} \\ \\ f'(x) =& \frac{2x + 1}{x^2 + x + 1} \end{aligned} \end{equation} $

We see from the graph that $f$ is increasing when $f'(x)$ is positive. On the other hand, $f$ is decreasing whenever $f'(x)$ is negative. We can say that our answer is reasonable.