## Calculus: Early Transcendentals 8th Edition

$g'(\theta)=e^{\theta}(\tan^{2}\theta+\tan\theta-\theta)$
$g(\theta)=e^{\theta}(\tan\theta-\theta)$ 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: $...=e^{\theta}(\tan^{2}\theta+\tan\theta-\theta)$