1. **State the problem:** We are given the function $$g(x) = \int_0^x \sqrt{1 + 6t} \, dt$$ and asked to find its derivative $$g'(x)$$ using the Fundamental Theorem of Calculus. 2. **Recall the Fundamental Theorem of Calculus:** If $$G(x) = \int_a^x f(t) \, dt$$, then $$G'(x) = f(x)$$ provided $$f$$ is continuous. 3. **Apply the theorem:** Here, $$f(t) = \sqrt{1 + 6t}$$, so $$g'(x) = \sqrt{1 + 6x}$$ 4. **Final answer:** $$\boxed{g'(x) = \sqrt{1 + 6x}}$$