Answer
$$y'=\sec\theta$$
Work Step by Step
$$y=\ln(\sec\theta+\tan\theta)$$
Recall the following Derivative Rules: $$\frac{d}{d\theta}(\ln u)=\frac{1}{u}\frac{du}{d\theta}$$ $$\frac{d}{d\theta}(u+v)=\frac{du}{d\theta}+\frac{dv}{d\theta}$$
We have $$y'=\Big(\ln(\sec\theta+\tan\theta)\Big)'$$
$$y'=\frac{1}{\sec\theta+\tan\theta}(\sec\theta+\tan\theta)'=\frac{\sec\theta\tan\theta+\sec^2\theta}{\sec\theta+\tan\theta}$$
$$y'=\frac{\sec\theta(\sec\theta+\tan\theta)}{\sec\theta+\tan\theta}=\sec\theta$$
