Answer
$g'(\theta)=\dfrac{\sec^3{\dfrac{\theta}{2}}+\sec{\dfrac{\theta}{2}}\tan^2{\dfrac{\theta}{2}}}{2}$.
Work Step by Step
Using both the Product Rule and the Chain Rule:
Product Rule $g'(\theta)=((u(\theta)(v(\theta))’=u’(\theta)v(\theta)+u(\theta)v’(\theta))$
$u(\theta)=\sec{\dfrac{\theta}{2}} ;u’(\theta)=\dfrac{1}{2}\sec{\dfrac{\theta}{2}}\tan{\dfrac{\theta}{2}}$.
$v(\theta)=\tan{\dfrac{\theta}{2}} ;v’(\theta)=\dfrac{1}{2}\sec^2{\dfrac{\theta}{2}} $.
$g'(\theta)=(\dfrac{1}{2}\sec{\dfrac{\theta}{2}}\tan{\dfrac{\theta}{2}}(\tan{\dfrac{\theta}{2}})+(\dfrac{1}{2}\sec^2{\dfrac{\theta}{2}})(\sec{\dfrac{\theta}{2}})$
$=\dfrac{\sec^3{\dfrac{\theta}{2}}+\sec{\dfrac{\theta}{2}}\tan^2{\dfrac{\theta}{2}}}{2}$.
