Derivative of an accumulation function via FTC Part 1

By the Fundamental Theorem of Calculus Part 1, since $f(t) = \frac{1}{\sqrt{t}}$ is continuous near $t = 9$, we have $F'(x) = f(x) = \frac{1}{\sqrt{x}}$. This holds for all $x$ in the domain where $f$ is continuous, in particular near $x = 9$.

Substituting $x = 9$ gives $F'(9) = \frac{1}{\sqrt{9}} = \frac{1}{3}$.