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

Answer 1

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

We seek the value of #f^((6))(0)# where:
# f(x) = (cos(6x^2)-1)/x^2 #

The source of the Maclaurin series is

# f(x) = f(0) + (f'(0))/(1!)x + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ... (f^((n))(0))/(n!)x^n + ...#
Starting with the well known Maclaurin Series for #cosx#
# cosx = 1 - x^2/(2!) + x^4/(4!) - x^6/(6!) + ... \ \ \ \ \ \ AA x in RR#
Then, the #n^(th)# of #cos(x)# is given by:
# u_n{cosx} = ((-1)^n)/((2n)!)x^(2n) #
Thus, for #f(x)#, we can write:
# u_n{f(x)} = #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ = {((-1)^n)/((2n)!)(6x^2)^(2n) - 1}/x^2#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ((-1)^n 6^(2n))/((2n)!)x^(4n-2) - 1/x^2#
In order to rapidly compute #f^((6))(0)# we can substitute #n=6# into the general term of definition of the Maclaurin Series:
# u_6{f(x)} = (f^((6))(0))/(6!) x^6#
And we can therefore compute #f^((6))(0)# by comparing coefficient in the derived Maclaurin series for #f(x)# with the #x^6# term, for which we require that:
# 4n-2 = 6 => n=2 #
So substituting #n=4# into the derived Maclaurin series general term, we get:
# u_2{f(x)} = ((-1)^2 6^(4))/((4)!)x^6 - 1/x^2#
Then, by comparing coefficients of #x^6#, we have:
# (f^((6))(0))/(6!) = ((-1)^2 6^(4))/((4)!)#
# :. f^((6))(0) = ((-1)^2 6^(4) \ 6!)/((4)!)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( 6^(4) \ 6*5*4!)/((4)!)#
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 6^(4) * 6*5 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1296 * 30 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 38880 #
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Answer 2

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

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

Here's a step-by-step breakdown:

  1. Maclaurin series expansion of ( \cos(x) ) is ( \cos(x) = 1 - \frac{{x^2}}{2!} + \frac{{x^4}}{4!} - \frac{{x^6}}{6!} + \cdots )
  2. 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 )
  3. Simplify the expression: ( \cos(6x^2) = 1 - \frac{{36x^4}}{2!} + \frac{{1296x^8}}{4!} - \frac{{46656x^{12}}}{6!} + \cdots )
  4. Subtract 1 from this expression: ( \cos(6x^2) - 1 = -\frac{{36x^4}}{2!} + \frac{{1296x^8}}{4!} - \frac{{46656x^{12}}}{6!} + \cdots )
  5. 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 )
  6. Differentiate this expression six times with respect to ( x ).
  7. Evaluate the resulting expression at ( x = 0 ) to find the 6th derivative.
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Answer from HIX Tutor

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.

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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