# How do you use part 1 of the fundamental theorem of calculus to find f ' (4) for f(x) = int sqrt ((t^2)+3) dt from 1 to x?

FTC 1 tells us that $f ' \left(x\right) = \sqrt{{x}^{2} + 2}$
So $f ' \left(4\right) = \sqrt{{4}^{2} + 2} = \sqrt{18} = 3 \sqrt{2}$