Question

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?

Answer 1

FTC 1 tells us that `f'(x) = sqrt(x^2+2)`

So `f'(4) = sqrt(4^2+2) = sqrt18 = 3sqrt2`