Answer
$g'(\theta)=e^{\theta}(\tan^{2}\theta+\tan\theta-\theta)$
Work Step by Step
$g(\theta)=e^{\theta}(\tan\theta-\theta)$
Differentiate using the product rule:
$g'(\theta)=(e^{\theta})(\tan\theta-\theta)'+(\tan\theta-\theta)(e^{\theta})'=...$
$...=(e^{\theta})(\sec^{2}\theta-1)+(\tan\theta-\theta)(e^{\theta})=...$
We can use the following identity: $\sec^{2}\theta-1=\tan^{2}\theta$:
$...=e^{\theta}\tan^{2}\theta+(\tan\theta-\theta)e^{\theta}=...$
Take out common factor $e^{\theta}$ for a better looking answer:
$...=e^{\theta}(\tan^{2}\theta+\tan\theta-\theta)$
