Science:Math Exam Resources/Courses/MATH103/April 2013/Question 06 (b)/Solution 1

Taking the derivative of the cdf found in part a we have

${\begin{aligned}p(x)&=F'(x)=\left({\frac {2}{\pi }}\left(\arctan \left({\frac {x}{2}}\right)+{\frac {\pi }{4}}\right)\right)'\\&={\frac {2}{\pi }}\cdot \left({\frac {1}{1+\left({\tfrac {x}{2}}\right)^{2}}}\cdot {\frac {1}{2}}+0\right)\\&={\frac {1}{\pi }}\cdot {\frac {4}{4+x^{2}}}={\frac {4}{\pi (4+x^{2})}}\end{aligned}}$

completing the question.