# Finding Derivatives Using Taylor/Maclaurin Polynomials

1. Mar 12, 2010

### HypeBeast23

1. The problem statement, all variables and given/known data

Compute the 6th derivative of f(x) = arctan((x^2)/4) at x = 0. Hint: Use the Maclaurin series for f(x).

2. Relevant equations

The maclaurin series of arctanx which is ((-1)^n)*x^(2n+1)/2n+1

3. 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!

2. Mar 12, 2010

### HallsofIvy

Okay, so you know that the general term is $((-1)^n)*x^{4n+2}/(4^{2n+1})(2n+1)$.
And you know that the coefficient is the nth derivative over n!. Okay. For what n is
4n+2= 6? What is the coefficient of $x^6$?

3. Mar 12, 2010

### HypeBeast23

Yes, I did as such and found n = 1 to get x^6. Thus, the coefficient of x would be -(1/192), no? Apparently, the answer was wrong.