Question

How do you compute the 9th derivative of: `arctan((x^3)/2)` at x=0 using a maclaurin series?

Answer 1

`f^(9^(th))(0)=-1/24*9!`

It should be
`f^((9))(0)/(9!) = -1/24 => f^((9))(0) = -(9!)/24 = -362880/24 = -15120`

Explanation:

The Mac Laurin series of `arctan(x)=x-x^3/3+x^5/5-x^7/7+o(x^7)`.

If we replace the variable `x` with `x^3/2`,
it becomes `arctan(x^3/2)=x^3/2-(x^3/2)^3/3+(x^3/2)^5/5+o(x^15)`

that rewritten assumes the form
`arctan(x^3/2)=x^3/2-x^9/24+x^15/160+o(x^15)`

By recalling that in the Maclaurin series the coefficient of the each term of n degree is just the n-th derivative of `f(x)` evaluated in 0, it is immediate to deduce that `f^(9^(th))(0)=-1/24*9!`.
Enjoy