If f(x) = sin(x^3) find f^(15)(0)

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Homework Help Overview

The problem involves finding the 15th derivative of the function f(x) = sin(x^3) at the point x = 0, which relates to the study of power series and derivatives in calculus.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster inquires about the necessity of calculating all derivatives or if a more efficient method exists. Some participants suggest using the power series expansion for sin to derive the necessary information about the derivatives.

Discussion Status

Participants are exploring the use of the Maclaurin series for sin(x) with a substitution for x^3. There is a suggestion that the 15th derivative can be inferred from the series expansion, and one participant expresses confidence in their calculation of the 15th derivative based on this approach.

Contextual Notes

There is an implicit assumption that the participants are familiar with Taylor series and Maclaurin series expansions, as well as the properties of derivatives.

Calcotron
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Short and sweet, If f(x) = sin(x^3) find f^(15)(0) by which I mean find the 15th derivative of f. Do I have to write out all the derivatives and look for a pattern somewhere or is there an easier way?
 
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Do you know the power series expansion for sin? If so, you can read off the derivatives from the Taylor series expansion
[tex] f(x) = \sum_{n=0}^\infty\frac{x^n}{n!}f^{(n)}(0).[/tex]
 
Ok, so from the maclaurin series with x^3 substituted I get f(x) = x^3 - (x^9/ 3!) + (x^15/ 5!) and just from this I can see the 15th derivative of the function is going to be 15!/ 5!. Is that all correctÉ
 
yes!
 

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