2011 Calculus BC free response #6c

Part C. Find the value of the sixth derivative of f evaluated at 0. So you could imagine if you just tried to find the sixth derivative of f, that would take you forever. And then to evaluate it at 0, because this is x squared here. And you'd have to keep doing the product rule over and over again, and the chain rule and all the rest. It would become very, very, very messy. But we have a big clue here. The fact that they made us find the first four terms of the Taylor series of f about x equals 0 tells us that there might be a simpler way to do this, as opposed to just taking the sixth derivative of this and evaluating at 0. The simplest way to do this is to just go back. In the last problem, we were able to come up with the first four non-zero terms of the Taylor series of f. And if you look at your definition of the Taylor series right here-- and we go into depth on this in another Khan Academy video where we talk about why this makes sense-- you see that each degree term of the Taylor series, its coefficient is that derivative. And this Taylor Series is centered around 0, and that's what we care about in terms of this problem. We see the coefficient is that derivative divided by that degrees, that derivative evaluated at 0 divided by that degrees factorials. So the second degree term, it's the second derivative of f evaluated at 0 divided by 2 factorial. The fourth degree term is the fourth derivative of f evaluated at 0 divided by 4 factorial. So the sixth degree term-- let's remind ourselves what we're even trying to figure out-- so they want us to figure out the sixth derivative of f evaluated at 0. That's what they want us to figure out. Well, if you think about the Taylor series centered at 0, or at 0, or approximated around 0, the sixth degree term in the Taylor series approximation of f is going to be f prime of the sixth derivative of f evaluated at 0 times x to the sixth over 6 factorial. This is going to be the sixth degree term in Taylor approximation, in Taylor series. And we have that term sitting right over here. This is the sixth degree term. We figured it out in the last problem. This right here is the sixth degree term. So you have x to the sixth over here, x to the sixth over here, you have 6 factorial over here, 6 factorial over here. So this negative 121 must be the sixth derivative of f evaluated at 0. So that's our answer. This is equal to negative 121. And we're done.