How do you find #f^6(0)# where #f(x)=arctanx/x#?

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Answer 1 · 2017-03-02T18:01:22

#f^((6))(0) = -720/7#

We start by constructing the MacLaurin series for #arctan x#.
Consider the function:

#f(x) = 1/(1+x^2)#

This is the sum of a geometric series of ratio: #-x^2#, so that we have:

#1/(1+x^2) =sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^nx^(2n) #

with radius of convergence #R=1#.

Within the interval #x in (-1,1)# we can therefore integrate term by term:

#int_0^x dt/(1+t^2) =sum_(n=0)^oo (-1)^n int t^(2n)dt #

#arctan x=sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1) #

and dividing by #x# both sides:

#arctanx/x = sum_(n=0)^oo (-1)^n x^(2n)/(2n+1) #

Now consider the standard expression of the MacLaurin series of #f(x)#

#f(x) = sum_(n=0)^oo f^((n))(0)/(n!)x^n#

The two series are equal only if the coefficients of the same degree in #x# are equal, so that for #x^6# we have:

#f^((6))(0)/(6!) = -1/7#

and:

#f^((6))(0) = -720/7#