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### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 13: Taylor & Maclaurin polynomials intro- Taylor & Maclaurin polynomials intro (part 1)
- Taylor & Maclaurin polynomials intro (part 2)
- Worked example: Maclaurin polynomial
- Worked example: coefficient in Maclaurin polynomial
- Worked example: coefficient in Taylor polynomial
- Taylor & Maclaurin polynomials
- Taylor polynomial remainder (part 1)
- Taylor polynomial remainder (part 2)
- Worked example: estimating sin(0.4) using Lagrange error bound
- Worked example: estimating eˣ using Lagrange error bound
- Lagrange error bound
- Visualizing Taylor polynomial approximations
- Worked example: Taylor polynomial of derivative function

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# Worked example: Taylor polynomial of derivative function

Rather than approximating a function, this time we are asked to approximate the derivative of a function.

## Want to join the conversation?

- Here's the step I don't understand: g(x) = g(2) + g'(2)(x - 2) + g''(2)(x - 2)^2 + ... When taking the derivative of g(x), why aren't the (x - 2) terms differentiated?(10 votes)
- He never takes the derivative, he just replaces the function f(x) with g'(x). It is sort of like a substitution.(23 votes)

- Anyone else think this question was ambiguous? I determined the second degree Taylor polynomial of g(x), then took the first derivative of that and evaluated at x=1. In other words, instead of not using the data g(2)=3 like Sal, I didn't use g'''(2)=2. Admittedly, what I did would probably give a less accurate approximation, but I think it still satisfies the wording of the question. If anyone wants to set me straight on this though, please do.(6 votes)
- Hi!

Thank you for your clarification. Yes, you could also go about it the way you did, and, as you correctly stated, your answer would be slightly less accurate. But you can always increase your accuracy by using the third derivative. So, just throw in an extra term g^(3)(x)*(x-2)^3/3!.

This extra term will further increase the accuracy of your Taylor polynomial and you should end up with the same answer that they have gotten in the video above.(4 votes)

- It's only an approximation, so the answer will vary, but can't you also solve this by finding the Taylor series for g(x)? g(x) will approximately equal 3 + (x - 2) - 1/2(x-2)^2. Now, what does g'(1) mean? It means finding the slope of the tangent line at g(1). Therefore, if we take the derivative of our approximate function, we get 1 - (x-2) or 3 - x. Substituting 1 in for x, the approximation of the slope at g(1) becomes 2, or g'(1) approximately equals 2. It's not the exact answer Sal got, but since these both are approximations of the real answer, is this one equally valid?(6 votes)
- how come in none of the taylor/maclaurin vids has he gone into convergence and divergence and radius of convergence??(5 votes)
- check out his videos on convergence tests, just google it.(1 vote)

- Where does the (x-2) term come from in the above example?(2 votes)
- We have an
`(x-2)`

term because this particular Taylor polynomial is centered at`x=2`

. Remember that in general, the formula for the nth order term of a Taylor polynomial is`( f^(n)[c] * (x-c)^n ) / n!`

where`c`

is the center of our Taylor polynomial.

Importantly,`c`

is also the number at which the derivatives are evaluated to find the coefficients.

Hope that helps.(5 votes)

- i did the problem a slightly different way by calculating taylor series for g(x) to get the polynomial (-x^2+5x+1) then differentiated that one and solved for x=1 and got the same answer! is that okay?(3 votes)
- Yes this is a valid, though somewhat more laborious, way to solve this problem. However, when I do the calculations this way, I get the same intermediate step as An Duy ...

Note also that**not**simplifying the expression and just taking the derivatives of each term (using the power rule) is less painful!(2 votes)

- So even though it says "centered at x=2", we actually make c=2 instead of x, and then treat the x as being whatever we want later? Is that correct?(2 votes)
- Yes, it's approximating the general function g(x), around a given x = c. c is the x-value that is the center of the approximation. Since this gives us the approximation of the general function, we can then plug in whatever x-value we want, to get an approximation of the function that that point.(3 votes)

- What does second degree mean? does it mean that instead of starting with g(t) we start with the g'(t)? thanks(1 vote)
- It means that you'll get the Taylor polynomial up to the term where you use the second derivative and elevate (x-c) to the second power.

For example if instead of the second degree polynomial he used the third degree it would add: (f'''(2)(x-2)^3)/3! to the Taylor Polynomial.(2 votes)

- A function has an n th derivative evaluated at x = a that is proportional to n. What is the radius of convergence of the Taylor series of this function expanded about the point x = a?(1 vote)
- How is it that Sal made g(x) = g'(2) ... and so forth?

Shouldn't the equation look like g(s) = g'(2)... since every equation after it, the x is subsituted with 2?(1 vote)

## Video transcript

- So, let's say we've been given all this information about the function g and it's derivative
evaluated at x equals two. We know g of two is equal to three. G prime of two is equal to one. The second derivative of g
evaluated two is negative one. The third derivative of g
evaluated at two is two. Given that, what we're
being tasked with is we want to use the second
degree Taylor polynomial centered at x equals two to
approximate g prime of one. Not g of one, g prime of
one and so I encourage you to pause this video and try
to think about it on your own. I'm assuming you've had a go at it. Let's just remind ourselves what a second degree Taylor polynomial
centered at x equals two would look like for a
general function f of x. F of x would approximately be equal to, it would be f of two plus f prime of two times x minus two plus
f prime prime of two times x minus two squared, all
of that over two factorial. That would get us to a second degree place because it's x minus two squared. This is gonna give us a
second degree polynomial. This is the general case. If we want to find the
approximation for f centered at x equals but we're
gonna do it for g prime. Let me write this down. Alright, so I'll do it in blue. So, we have g prime of
x is what we're going to try and approximate and then we're going to evaluate it at x equals 1. G prime of x is going to
be approximately equal to, well same thing, it's
going to be the function that I'm going to try to approximate evaluated at two so g prime of two. Notice, so I'm approximate f of x, it's that function evaluated at two. If I'm approximating g prime of x, it's that function evaluated at two. Then, plus the first
derivative of this thing which is the second derivative of g. G prime prime of two times x minus two and then plus the second
derivative of the function that I'm trying to approximate
but the second derivative of g prime is going to be
the third derivative of g. It's going to be g prime prime
prime of two times x minus two squared, all of
that over two factorial. Now, they tell us what these things are. Let me use some new colors here. They tell us g of two is equal to three. This right here, oh, actually no, that's not what we're gonna wanna use. They tell us we're using g of two. They're telling us g prime
of two is equal to one. G prime of two, so this right
over here is equal to one. G prime prime of two is
equal to negative one. This is negative one right over here and then finally the third derivative of g evaluated at two is two. Two over two factorial, two factorials is two times one or it's just two. So that and that cancel out. What are we left with
for our approximation our second degree approximation of g prime of x centered at x equals two? We are left with g prime
of x is approximately equal to one minus x minus two. One minus x minus two,
I guess I could write that as plus two minus
x so plus two minus x. The negative of x minus
two is two minus x. Plus x minus two squared and obviously I could simplify this even more. This is three minus x plus x minus two squared and now I can evaluate it. If I wanna approximate g prime of one, I could say g prime of one
is going to be approximately equal to, wherever I see the x is here I put in a one there, so it's going to be three minus one plus
one minus two squared. Well, this is going to be two and then one minus two is negative one but then if I square it I get a positive one. So this whole thing is one. Two plus one is equal to three. Once again, this is an
approximation for g prime of one. What did we do here? We found the Taylor series. The second degree Taylor
series approximation for g prime of x centered
around x equals two and then we evaluated that approximation at x equals one to
approximate g prime of one. Anyway, hopefully, you found that fun.