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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.

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• 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?
• He never takes the derivative, he just replaces the function f(x) with g'(x). It is sort of like a substitution.
• 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.
• 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.
• 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?
• how come in none of the taylor/maclaurin vids has he gone into convergence and divergence and radius of convergence??
• check out his videos on convergence tests, just google it.
(1 vote)
• Where does the (x-2) term come from in the above example?
• 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.
• 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?
• 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!
• 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?
• 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.
• 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.