Find \(f'' \left( \dfrac{\pi }{4}\right) \) if \(f( x) =\sec x\).

If the trigonometric function begins with the letter \(c\), that is, cosine, cotangent, or cosecant, then its derivative has a minus sign.

Solution If \(f( x) =\sec x\), then \(f^\prime (x) =\sec x\tan x\) and \[ \begin{eqnarray*} f'' ( x) &=&\dfrac{d}{dx}( \sec x\tan x) \underset{\underset{\color{#0066A7}{\hbox{Use the Product Rule.}}}{{{\color{#0066A7}\uparrow }}}}= \sec x\!\left( \dfrac{d}{dx}\tan x\right) +\left( \dfrac{d}{dx}\sec x\right) \tan x \\ \\ &=&\sec x\cdot \sec ^{2}x+( \sec x\tan x) \tan x=\sec ^{3}x+\sec x\tan ^{2}x \end{eqnarray*} \]

Now, \[ \begin{eqnarray*} f'' \left( \dfrac{\pi }{4}\right) =\sec ^{3}\!\left( \dfrac{\pi }{4} \right) +\sec\! \left( \dfrac{\pi }{4}\right) \tan ^{2}\!\left( \dfrac{\pi }{4} \right) \underset{\underset{\color{#0066A7}{\hbox{sec} \dfrac{\pi }{4}=\sqrt{2};\hbox{ tan}\dfrac{\pi }{4}=1}} {{{\color{#0066A7}\uparrow }}}}= ( \sqrt{2}) ^{3}+\sqrt{2}\cdot 1^{2}=2 \sqrt{2}+\sqrt{2}=3\sqrt{2} \\ \end{eqnarray*} \]