Compute the 6th derivative of f(x) = arctan((x^2)/4) at x = 0. Hint: Use the Maclaurin series for f(x).
The maclaurin series of arctanx which is ((-1)^n)*x^(2n+1)/2n+1
The Attempt at a Solution
I subbed in x^2/4 for x into the maclaurin series and got *summation* ((-1)^n)*x^(4n+2)/(4^2n+1)(2n+1). After that, i'm not sure what to do. I know that the 6th derivative in a general maclaurin series would be represented by *the 6th derivative of f at 0*/n! and x is to the power of 6. My prof said to find the value where the exponent of x in the series is 6.
Any help would be greatly appreciated. Thanks!