Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 31

a.) Differentiate the given function using the Quotient Rule

$\displaystyle f(x) = \frac{\tan x - 1}{\sec x}$

Solving for $f'(x)$

$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{\sec x \displaystyle \frac{d}{dx} (\tan x - 1) - \left[ (\tan x - 1) \frac{d}{dx} (\sec x) \right]}{(\sec x)^2}
&&
\\
\\
f'(x) =& \frac{(\sec x)(\sec^2 x) - [(\tan x - 1) (\sec x \tan x)]}{\sec ^2 x}
&& \text{Simplify the equation}
\\
\\
f'(x) =& \frac{\sec ^3 x - \sec x \tan ^2 x + \sec x \tan x}{\sec ^2 x }
&& \text{Factor out $\sec x$}
\\
\\
f'(x) =& \frac{\cancel{\sec x} (\sec ^2 x - \tan^2 x + \tan x)}{\sec^{\cancel{2}} x}
&& \text{Applying Pythagorean identity $\sec^2 x - \tan^2 x = 1$}
\\
\\
f'(x) =& \frac{1 \tan x}{\sec x}

\end{aligned}
\end{equation}
$


b.) Write the expression for $f(x)$ in terms of $\sin x + \cos x$ and simplify it then find $f'(x)$


$
\begin{equation}
\begin{aligned}

f(x) =& \frac{\tan x - 1}{\sec x}
&& \text{Applying the reciprocal identities}
\\
\\
f(x) =& \left( \frac{\sin x}{\cos x} - 1 \right) (\cos x)
&& \text{Simplify the equation}
\\
\\
f(x) =& \left( \frac{\sin x - \cos x}{\cancel{\cos x}} \right) (\cancel{\cos} x)
&& \text{Cancel like terms}
\\
\\
f(x) =& \sin x \cos x
&& \text{Simplified expression of $f(x)$}
\\
\\


\end{aligned}
\end{equation}
$


Solving for $f'(x)$


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{d}{dx} (\sin x) - \frac{d}{dx} (\cos x)
&& \text{Derive directly}
\\
\\
f'(x) =& \cos x - (- \sin x)
&& \text{Simplify the equation}
\\
\\
f'(x) =& \cos x + \sin x
&& \text{}
\\
\\


\end{aligned}
\end{equation}
$



c.) Prove that the answers to part (a) and (b) are equivalent.

We can express the answer in part (a) in terms of $\sin x$ and $\cos x$. We get


$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{1 + \tan x}{\sec x}
&& \text{Group the equation}
\\
\\
f'(x) =& \frac{1}{\sec x} + \frac{\tan x}{\sec x}
&& \text{Apply the reciprocal identity}
\\
\\
f'(x) =& \cos x + \left( \frac{\sin x}{\cancel{\cos x}} \right) (\cancel{\cos x})
&& \text{Simplify the equation}
\\
\\
f'(x) =& \cos x + \sin x
&&

\end{aligned}
\end{equation}
$


The simplified expression in part (a) is equal to the answer in part (b)