Calculus, all content (2017 edition)
- 2015 AP Calculus BC 2a
- 2015 AP Calculus BC 2b
- 2015 AP Calculus BC 2c
- 2015 AP Calculus BC 2d
- 2015 AP Calculus BC 5a
- 2015 AP Calculus BC 5b
- 2015 AP Calculus BC 5c
- 2015 AP Calculus BC 5d
- 2015 AP Calculus BC 6a
- 2015 AP Calculus BC 6b
- 2015 AP Calculus BC 6c
- AP Calculus BC exams: 2008 1 a
- AP Calculus BC exams: 2008 1 b&c
- AP Calculus BC exams: 2008 1 c&d
- AP Calculus BC exams: 2008 1 d
- Calculus BC 2008 2 a
- Calculus BC 2008 2 b &c
- Calculus BC 2008 2d
- 2011 Calculus BC free response #1a
- 2011 Calculus BC free response #1 (b & c)
- 2011 Calculus BC free response #1d
- 2011 Calculus BC free response #3a
- 2011 Calculus BC free response #3 (b & c)
- 2011 Calculus BC free response #6a
- 2011 Calculus BC free response #6b
- 2011 Calculus BC free response #6c
- 2011 Calculus BC free response #6d
2011 Calculus BC free response #6c
Calculating the 6th derivative at 0 from the Taylor Series approximation. Created by Sal Khan.
Want to join the conversation?
- Is there a difference between the words 'order' and 'degree' when looking at these types of equations?(4 votes)
- I'm not sure why you can't just take the 6th derivative of f(x) at 0 by hand especially because all the terms with x just go away. Or am I missing something?(1 vote)
- You could, but you would need to contend with multiple applications of the product rule, things can get messy and that is where errors can be made. Doing it that way is often called the "Brute Force" method. Nothing wrong with that.
In this case, the question, and its several sections, are leading you, step by step, to an analytic solution, which is yours for the taking. It is one thing to be able to do the "cookbook" calculations, that is, relying on the application of the rules of differentiation and integration that apply to a given situation, and quite another to start to get a feel for the meaning of what you are doing and the implications of those meanings which will help guide you to analytic solutions.
Don't get me wrong, I am not belittling "Brute Force". You need to be an expert at it and do it many times to start to get a feel for the concepts and implications I am talking about. I am also not belittling you either because you may not yet see the implications, or, if you do, chose not to use that knowledge. Everyone progresses at their own rate, the key is to keep working at it.
Many mathematicians like to joke that they are lazy, but that laziness is an admirable trait because it is actually an optimization of effort, meaning that you can quickly dispense with one problem so as to move on to another.
Keep Studying!(5 votes)
- What is the difference between a taylor polynomial and the taylor approximation and the taylor series approximating an insert-the-degree term? Or is there a difference at all?(2 votes)
- If instead the question asked what the f(5)(x) derivative value was, would it be equal to 0? As there is no fifth degree term so we can presume it is zero?(1 vote)
- Yes, the fifth derivative of f(x) evaluated at 0 would be equal to 0 because there is no fifth degree term.
If you keep taking derivatives of each term in the series, by the time you reach the fifth derivative, the only terms left would be the terms of a higher degree than -121x (the sixth degree term). Because these terms would all equal 0 when x = 0, we know that the fifth derivative is also equal to 0.(1 vote)
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.