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

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SUMMARY

The sixth derivative of the function f(x) = sin(x^2) at x = 0 can be computed using the Maclaurin series expansion. The Maclaurin series for sin(x) is modified by substituting x with x^2, resulting in the series: x^2 - x^6/3! + x^10/5!. The relevant non-zero terms are x^2 and -x^6/3!, indicating that the sixth derivative at x = 0 is -1/3! or -1/6. Understanding the coefficients of the series is crucial for accurate computation.

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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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