Derivative of an accumulation function at a point

By the Fundamental Theorem of Calculus Part 1, since $e^{-t^2}$ is continuous everywhere, the accumulation function $F(x) = \int_3^x e^{-t^2}\, dt$ satisfies $F'(x) = e^{-x^2}$. Notice the lower limit being $3$ (a constant) plays no role in this formula — it only affects the value of $F$, not its derivative.

Substituting $x = 1$ into $F'(x) = e^{-x^2}$ gives $F'(1) = e^{-1^2} = e^{-1}$. So the final answer is $F'(1) = e^{-1} = \dfrac{1}{e}$.

No, the derivative formula would not change. By the Fundamental Theorem of Calculus Part 1, the derivative of the accumulation function only depends on the integrand evaluated at x, giving F prime of x equals e to the negative x squared. The lower limit is just a constant that affects the value of F itself, not its rate of change. So changing three to seven would still give the same derivative formula.