Single Variable Calculus, Chapter 3, 3.4, Section 3.4, Problem 30

Assume that $f(x) = \sec x$, find $\displaystyle f''\left( \frac{\pi}{4} \right)$

Solving for $f'(x)$



$
\begin{equation}
\begin{aligned}

f'(x) =& \frac{d}{dx} (\sec x)
\\
\\
f'(x) =& \sec x \tan x


\end{aligned}
\end{equation}
$



Solving for $f''(x)$


$
\begin{equation}
\begin{aligned}

f''(x) =& \sec x \frac{d}{dx} (\tan x) + \tan x \frac{d}{dx} (\sec x)
\\
\\
f''(x) =& (\sec x)(\sec^2 x) + (\tan x) (\sec x \tan x)
\\
\\
f''(x) =& \sec^3 x + \sec x \tan^2 x


\end{aligned}
\end{equation}
$



Substitute $\displaystyle f''\left( \frac{\pi}{4} \right) $


$
\begin{equation}
\begin{aligned}

f'' \left( \frac{\pi}{4} \right) =& \sec^3 x + \sec x \tan^2 x
&&
\\
\\
f'' \left( \frac{\pi}{4} \right) =& \sec^3 \left( \frac{\pi}{4} \right) + \sec \left( \frac{\pi}{4} \right) \tan^2 \left( \frac{\pi}{4} \right)
&& \text{Simplify the equation}
\\
\\
f'' \left( \frac{\pi}{4} \right) =& \sqrt[2]{2} + \sqrt{2}
&& \text{Combine like terms}
\\
\\
f'' \left( \frac{\pi}{4} \right) =& \sqrt[3]{2}

\end{aligned}
\end{equation}
$