Answer
$\displaystyle \frac{d}{dx} [\displaystyle \sec^{-1}x]=\frac{1}{|x|\sqrt{x^{2}-1}}$
Work Step by Step
Let $f(x)=\sec x.$
Then,$\quad f'(x)=\sec x\tan x\quad$and
$f^{-1}(x)=\sec^{-1}x$
By Theorem 1,
$\displaystyle \frac{df^{-1}}{dx}=\frac{1}{f'\left[f^{-1}(x)\right]}$
$=\displaystyle \frac{1}{\sec[f^{-1}(x)]\tan[f^{-1}(b)]}$
$=\displaystyle \frac{1}{\sec[\sec^{-1}(x)]\tan[\sec^{-1}(x)]}$
We have the first term in the denominator, $\sec[\sec^{-1}(x)]=x.$
For the other term, sketch a right triangle with hypotenuse $x$ and one of the legs: $1$ (see below)
The other leg is $\sqrt{x^{2}-1}$.
The angle between the hypotenuse and $1$ is $y=\sec^{-1}x$
$\displaystyle \tan[\sec^{-1}(x)]=\frac{\sqrt{x^{2}-1}}{1}=\sqrt{x^{2}-1}$,
$\displaystyle \frac{df^{-1}}{dx}=\frac{1}{x\sqrt{x^{2}-1}}$ ... (which works for positive x)
Now, since $\sec^{-1}x$ is an increasing function, its derivative is positive.
We ensure that it is positive when x is negative by placing $|x|$ instead of $x$ in the denominator.
$\displaystyle \frac{d}{dx} [\displaystyle \sec^{-1}x]=\frac{1}{|x|\sqrt{x^{2}-1}}$
