Answer
$f'(\frac{\pi}{4}) = 3~\sqrt{2}$
Work Step by Step
$f(x) = \lim\limits_{t \to x}~\frac{sec~t-sec~x}{t-x}$
Note that $f(x) = g'(x)$ where $g(x) = sec~x$
Then $f'(x) = g''(x)$
We can find $g'(x)$:
$g'(x) = sec~x~tan~x$
We can find $g''(x)$:
$g''(x) = sec~x~tan^2~x+sec^3~x$
We can find $g''(\frac{\pi}{4})$:
$g''(\frac{\pi}{4}) = sec~\frac{\pi}{4}~tan^2~\frac{\pi}{4}+sec^3~\frac{\pi}{4}$
$g''(\frac{\pi}{4}) = (\sqrt{2})~(1)^2+ (\sqrt{2})^3$
$g''(\frac{\pi}{4}) = \sqrt{2}+ 2~\sqrt{2}$
$g''(\frac{\pi}{4}) = 3~\sqrt{2}$
Therefore, $f'(\frac{\pi}{4}) = 3~\sqrt{2}$
