If $f(x) = \left(x + 1\right) \left(x + 2\right) \left(x + 3\right)$, what is $f'\left(0\right)$?

Explanation

1. 01

   $\log y = \log(x + 1) + \log(x + 2) + \log(x + 3)$, so $\frac{y'}{y} = \frac{1}{x + 1} + \frac{1}{x + 2} + \frac{1}{x + 3}$.

2. 02

   At 0: $y = 6$ and $\frac{y'}{y} = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$, so $y' = 11$.