Error Estimation for Simpsons Method

[3.141592654]

Homework Statement

Find the error estimation for the integral \int(1+x^2)^\frac{1}{4} with limits [0, 2] for n=8.

Homework Equations

\epsilon\leq \frac{|f^{(4)}_{max}(x)|(b-a)^5}{180n^4}

The Attempt at a Solution

The tricky part about this problem is finding f^{(4)}_{max}(x):

f(x)=(1+x^2)^\frac{1}{4}

f&#039;(x)=\frac{1}{4}(1+x^2)^\frac{-3}{4}*2x <br /> <br /> =\frac{1}{2}x(1+x^2)^\frac{-3}{4}

f&#039;&#039;(x)=(\frac{1}{2}x*\frac{-3}{4}(1+x^2)^\frac{-7}{4}*2x)+((1+x^2)^\frac{-3}{4}*\frac{1}{2})

=\frac{1}{2} [\frac{-6}{4}x^2(1+x^2)^\frac{-7}{4}+(1+x^2)^\frac{-3}{4}]

=\frac{1}{2} [\frac{-3}{2}x^2(1+x^2)^\frac{-7}{4}+(1+x^2)^\frac{-3}{4}]

f&#039;&#039;&#039;(x)=\frac{1}{2} [(\frac{-3}{2}x^2*\frac{-7}{4}(1+x^2)^\frac{-11}{4}*2x)+((1+x^2)^\frac{-7}{4}*\frac{-6}{2}x)+(\frac{-3}{4}(1+x^2)^\frac{-7}{4}*2x)]

=\frac{1}{2} [\frac{42}{8}x^3(1+x^2)^\frac{-11}{4}-\frac{6}{2}x(1+x^2)^\frac{-7}{4}-\frac{6}{4}x(1+x^2)^\frac{-7}{4}]

=\frac{1}{2} [\frac{21}{4}x^3(1+x^2)^\frac{-11}{4}-3x(1+x^2)^\frac{-7}{4}-\frac{3}{2}x(1+x^2)^\frac{-7}{4}]

=\frac{3}{2} [\frac{7}{4}x^3(1+x^2)^\frac{-11}{4}-x(1+x^2)^\frac{-7}{4}-\frac{1}{2}x(1+x^2)^\frac{-7}{4}

\frac{3}{2} [\frac{7}{2}x^3(1+x^2)^\frac{-11}{4}-3x(1+x^2)^\frac{-7}{4}]

I entered the derivatives I calculated along with the derivatives taken from an online derivative calculator into my graphing calculator to make sure I was coming out with the right answer. It appears I've made a mistake with the 3rd derivative but I can't find the error. Thanks for any help.

 

[3.141592654]

Looks like I made a mistake, the error estimation formula for Simpson's Method is:

\epsilon\leq\frac{|f^{(4)}_{max}(x)|(b-a)^5}{180n^4}

So I'm trying to find the maximum of the fourth derivative. However, I still can't find the error in my third deriv.