Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 11

Differentiate $\displaystyle f(\theta) \frac{\sec \theta}{1 + \sec \theta} $


$
\begin{equation}
\begin{aligned}

f'(\theta) =& \frac{(1 + \sec \theta) \frac{d}{d\theta} (\sec \theta) - \left[ (\sec \theta) \frac{d}{d\theta} (1 + \sec \theta) \right]}{(1 + \sec \theta)^2}
&& \text{Using Quotient Rule}
\\
\\
f'(\theta) =& \frac{(1 + \sec \theta) (\sec \theta \tan \theta) - (\sec \theta) (\sec \theta \tan \theta) }{(1 + \sec \theta)^2}
&& \text{Simplify the equation}
\\
\\
f'(\theta) =& \frac{\sec \theta \tan \theta + \cancel{\sec^2 \theta \tan \theta} -\cancel{\sec^2 \theta \tan \theta} }{(1 + \sec \theta)^2}
&& \text{Combine like terms}
\\
\\
f'(\theta) =& \frac{\sec \theta \tan \theta}{(1 + \sec \theta)^2}
&& \text{}
\\
\\

\end{aligned}
\end{equation}
$