If $f(x) = \sqrt{1 + \sqrt{x}}$, what is $f'\left(9\right)$?

Explanation

1. 01

   $f'(x) = \frac{1}{2\sqrt{1 + \sqrt{x}}} \cdot \frac{1}{2\sqrt{x}}$, one factor per layer.

2. 02

   At 9: $\sqrt{x} = 3$ and $\sqrt{1 + 3} = 2$, so $\frac{1}{4} \cdot \frac{1}{6} = \frac{1}{24}$.