Let's say that we have the function capital F of x, which we're going to define as the definite integral from one to sine of x, so that's an interesting upper bound right over there, of two t minus one, and of course, dt. And what we are curious about is trying to figure out what is F prime of x going to be equal to? So pause this video and see if you can figure that out. All right, so some of you might have been a little bit challenged by this notion of, hey, instead of an x on this upper bound, I now have a sine of x. If it was just an x, I could have used the fundamental theorem of calculus. Just to review that, if I had a function, let me call it h of x, if I have h of x that was defined as the definite integral from one to x of two t minus one dt, we know from the fundamental theorem of calculus that h prime of x would be simply this inner function with the t replaced by the x. It would just be two x minus one.