#### Answer

$g'(θ)=e^{θ}(tan(θ)+sec^{2}(θ)-θ-1) $

#### Work Step by Step

Using the product rule, $\frac{d}{dx} ((e^{θ})(tan(θ)-θ)$
$=(\frac{d}{x}e^{θ})(tan(θ)-θ)+e^{θ}(\frac{d}{dx}((tan(θ)-θ))$
$= e^{θ}(tan(θ)-θ)+e^{θ}(sec^{2}(θ)-1)$
Simplifying then gives $e^{θ}(tan(θ)+sec^{2}(θ)-θ-1)$