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

1 Answer

#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