# Chapter 2 - Derivatives - 2.4 Derivatives of Trigonometric Functions - 2.4 Exercises: 6

$g'(θ)=e^{θ}(tan(θ)+sec^{2}(θ)-θ-1)$

#### 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 $e^{θ}(tan(θ)+sec^{2}(θ)-θ-1)$

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