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## AP®︎/College Calculus BC

### Course: AP®︎/College Calculus BC > Unit 11

Lesson 2: AP Calculus BC 2015- 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

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# 2015 AP Calculus BC 6b

Maclaurin series for derivative.

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- When taking the derivative of the series, doesn't the index shift up one?(0 votes)

## Video transcript

- [Voiceover] Part b,
write the first four, write the first nonzero terms of the Maclaurin series for
f prime, the derivative of f. Express f prime as a rational function for the absolute value
of x being less than R, our radius of convergence. So, if we wanna find f prime, we could just take each of, we could just take the derivative of each of these terms with respect to x. And so, we could just say, if this is f, then f prime, and I'm not gonna scroll down just so I can see this up here. We could just say f prime, f prime of x is going to be, or the Maclaurin series for f prime of x, maybe I should write it that way, so let me write it this way. Maclaurin, Maclaurin, Maclaurin, that a looked funny, Maclaurin series for f prime, f prime of x, well, it's going to be the sum from n equals one to infinity, and we would just take the derivative of this right over here with respect to x. And so, this is just
application of the power rule. Take the exponent right over here, multiply it by the coefficient, so if you take n times this, it cancels out with this n, so it's going to be negative three to the n minus one, and then decrement your exponent, times x to the n minus one. And so, they want us
to write the first four nonzero terms of the Maclaurin series, so that is going to be equal to, so, I'll write approximately equal to, because we're one going
to write the first four of terms of this infinite series, and just to be clear what I did, I just did the power rule here, I looked at this exponent, which is n, multiplied by this coefficient, which had a n in the denominator, so that n and this n cancel out, so I'm just left with negative
three to the n minus one, and then I decrement that exponent. That's straight out of the power rule, it's one of the first things you'll learn about taking derivatives. And so, if we want the
first four nonzero terms, when n equals one this is going to be negative three to the one minus one power, let me just write it, negative three to the one minus one times x to the one minus one, that's when n equals one, plus negative three, negative three to the two minus one, two minus one, times x to the two minus one, and then, and actually I could've reassured this is negative three
x to the n minus one, actually let me me do that just for, so this is, this I could just write, because they have the same exponent, this will simplify a little bit, as negative three x to the n minus one, and so this going to be approximately, when n is equal to one, this is gonna be equal to zero, so negative three x to the zero
power is just gonna be one, when n is equal to two, this is going to be two minus one, so it's gonna be the first power, so negative three x to the first power, so I could just write
this as negative three x, and then when n is three, well, this is going to be
negative three x squared, so negative three x
squared is going to be, negative three squared is nine x squared, and then the fourth term is gonna be negative three x
to the four minus one power, so to the third power, so negative three to the
third power is negative 27 times x to the third power. So there you have it, that's the first four nonzero
terms of the Maclaurin series. You could've also just looked over here, and said okay, the derivative of x, with respect to x is one, derivative of negative
three have x squared, with respect to x is negative
three, you could've said, the derivative of this is nine
x squared, right over there, and then you'd have to
write out the fourth term and take out the
derivative in the same way, and you'd have gotten
this right over here. So we did the first part, we wrote the first four nonzero terms of the Maclaurin series for f prime, the derivative of f, and then they said, express f prime as a rational function for the absolute value
of x being less than R. So this sum, if we assume it converges, and we know the radius of
convergence already, so, or assuming that we're dealing with x that are withing
the radius of convergence. So this right over here, you might recognize this, so I could write it like that, or I could also write it, if I start at n equals zero, so I could also write this as from n equals one to infinity of, or actually, from n
equals zero to infinity of negative three x to the n minus one, either, I'm sorry, to the n, because now the first term
was to the zero power, that first term was to zero power, so whether you do one minus one, is where you start, or you just at zero, these two things are equivalent. You might recognize these
as a geometrix series with common ration of negative, of negative three x. And so, what's the sum
of a geometric series with common ration, with
a certain common ration? Well, it's going to be
equal to the first term, and regardless of how you view this, the first term is going to be one divided by one minus the common ratio. So our common ratio is negative three x, one minus negative three x, well, that's just going
to be one plus three x. If what I just did here
looks unfamiliar to you, I encourage you to watch the sum of infinite geometric series, and not only do we show you this formula and how to apply it, but we show how you
can prove this formula, it's actually a pretty fun prove. But anyway, regardless of how you view
this Maclaurin series, it is an infinite geometric series, and this assuming that our x is in our radius of convergence. This is what our sum is, this is what we are going to converge to.