Answer
$3\sqrt{2}$
Work Step by Step
We know that $\lim\limits_{x \to x_{0}}\frac{f(x)-f(x_{0})}{x-x_{0}}=f'(x_{0})=\frac{d}{dx}f(x)|_{x=x_{0}}$
So this $f(x)= \lim\limits_{t \to x}\frac{\sec(t)-\sec(x)}{t-x}$ implies this $f(x)= \frac{d}{dx}(\sec(x))$
We know that $\frac{d}{dx}(\sec(x))=\sec(x) \cdot \tan(x)$ so:
$f(x)=\sec(x) \cdot \tan(x)$
Using the product rules for derivatives, the first derivative of $f$ with respect to $x$ is:
$f'(x)=(\sec(x))' \cdot \tan(x)+\sec(x) \cdot (\tan(x))'$
$f'(x)=(\sec(x) \cdot \tan(x)) \cdot \tan(x)+\sec(x) \cdot (\sec(x)))^{2}$
Substitute $x=\frac{\pi}{4}$ into the equation:
$f'(\frac{\pi}{4})=(\sec(\frac{\pi}{4}) \cdot \tan(\frac{\pi}{4})) \cdot \tan(\frac{\pi}{4})+\sec(\frac{\pi}{4}) \cdot (\sec(\frac{\pi}{4})))^{2}$
$f'(\frac{\pi}{4})=3\sqrt{2}$
