Compute 6th Derivative of Sin(x^2) at x=0 | Maclaurin Polynomial Hint

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Homework Help Overview

The problem involves computing the sixth derivative of the function f(x) = sin(x^2) at x = 0, with a suggestion that the Maclaurin polynomial may be useful in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the Maclaurin series for sin(x) and how substituting x^2 affects the expansion. There is inquiry about the specific alterations to the series due to this substitution.

Discussion Status

Some participants have provided hints regarding the use of the Maclaurin series and its application to the function sin(x^2). There is an ongoing exploration of the series' terms and their coefficients, with acknowledgment of the presence of zero coefficients for certain powers.

Contextual Notes

Participants note that the problem may involve several terms in the series expansion, some of which have zero coefficients, which could influence the computation of the derivatives.

me9900
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Homework Statement



Compute the sixth derivative of f(x)=sin(x^2) at x=0. Hint: Maclaurin polynomial may be helpful to you.

Homework Equations


The taylor expansion/maclaurin expansion for sin.


The Attempt at a Solution


Help.
 
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Hint: Maclaurin polynomial may be helpful to you.
Do you know the Maclaurin series for sin(x)?
 
yes but how does the x^2 alter the expansion?
 
Replace each occurrence of x in the Maclaurin series expansion for sin(x) with x^2.
 
So the expansion would be:

x^2-x^6/3!+x^10/5! ?
 
me9900 said:
So the expansion would be:

x^2-x^6/3!+x^10/5! ?

That would be the first three nonzero terms, yes. Keep in mind there are quite a few terms whose coefficients are zero, so sin(x^2) = 0 + 0x + x^2 + 0x^3 + 0x^4 + 0x^5 - x^6/3! + ...

Regarding your original problem, keep in mind the general formula for the Maclaurin series, and how you get the coefficients of the various powers of x.
 
Thanks
 

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