Question

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`

Explanation:

We seek the value of `f^((6))(0)` where:

`f(x) = (cos(6x^2)-1)/x^2`

The Maclaurin series is given by

`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`