# How do you compute the 6th derivative of: #(cos(6x^2)-1)/(x^2) # at x=0 using a maclaurin series?

# f^((6))(0) = 38880 #

The source of the Maclaurin series is

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To compute the 6th derivative of ( \frac{{\cos(6x^2) - 1}}{{x^2}} ) at ( x = 0 ) using a Maclaurin series, follow these steps:

- Expand ( \cos(6x^2) ) using its Maclaurin series expansion.
- Differentiate the resulting series expression term by term six times.
- Evaluate the resulting expression at ( x = 0 ) to find the 6th derivative.

Here's a step-by-step breakdown:

- Maclaurin series expansion of ( \cos(x) ) is ( \cos(x) = 1 - \frac{{x^2}}{2!} + \frac{{x^4}}{4!} - \frac{{x^6}}{6!} + \cdots )
- Apply this expansion to ( \cos(6x^2) ), replacing ( x ) with ( 6x^2 ): ( \cos(6x^2) = 1 - \frac{{(6x^2)^2}}{2!} + \frac{{(6x^2)^4}}{4!} - \frac{{(6x^2)^6}}{6!} + \cdots )
- Simplify the expression: ( \cos(6x^2) = 1 - \frac{{36x^4}}{2!} + \frac{{1296x^8}}{4!} - \frac{{46656x^{12}}}{6!} + \cdots )
- Subtract 1 from this expression: ( \cos(6x^2) - 1 = -\frac{{36x^4}}{2!} + \frac{{1296x^8}}{4!} - \frac{{46656x^{12}}}{6!} + \cdots )
- Divide by ( x^2 ): ( \frac{{\cos(6x^2) - 1}}{{x^2}} = -\frac{{36}}{{2!}} + \frac{{1296}}{{4!}}x^6 - \frac{{46656}}{{6!}}x^{10} + \cdots )
- Differentiate this expression six times with respect to ( x ).
- Evaluate the resulting expression at ( x = 0 ) to find the 6th derivative.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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- Find the first five terms of the Taylor series for #x^8+ x^4 +3 " at " x = 0#?
- How can you find the taylor expansion of #sqrt (x) # about x=1?
- The coefficient of #x^2# in the expansion of #(3x + y)^n# is #324#. What is the value of #n#?

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