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

Nov 4, 2016

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 $\arctan \left(x\right) = x - {x}^{3} / 3 + {x}^{5} / 5 - {x}^{7} / 7 + o \left({x}^{7}\right)$.

If we replace the variable $x$ with ${x}^{3} / 2$,
it becomes $\arctan \left({x}^{3} / 2\right) = {x}^{3} / 2 - {\left({x}^{3} / 2\right)}^{3} / 3 + {\left({x}^{3} / 2\right)}^{5} / 5 + o \left({x}^{15}\right)$

that rewritten assumes the form
$\arctan \left({x}^{3} / 2\right) = {x}^{3} / 2 - {x}^{9} / 24 + {x}^{15} / 160 + o \left({x}^{15}\right)$

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