If, indeed, it is that \(\displaystyle \frac{d}{d\theta} sec(\theta)\)that is bothering you, remember that \(\displaystyle sec(\theta)= \frac{1}{cos(\theta)}\)
Let \(\displaystyle u= cos(\theta)\)so that \(\displaystyle sec(\theta)=\frac{1}{u}\) and then \(\displaystyle \frac{d\frac{1}{u}}{du}= -\frac{1}{u^2}\) and \(\displaystyle \frac{du}{d\theta}= -sin(\theta)\) so that \(\displaystyle \frac{d}{d\theta}sec(\theta)= \frac{d \frac{1}{u}}{du}\frac{du}{d\theta}\)\(\displaystyle = \left(frac{1}{u^2}\right)\left(sin(\theta)\right)\)\(\displaystyle =\frac{sin(\theta)}{cos^2(\theta)}\)\(\displaystyle = \left(\frac{sin(\theta)}{cos(\theta)}\right)\left(\frac{1}{cos(\theta)}\right)= tan(\theta)sec(\theta)\)