We would like to find the derivative of $\operatorname{arcsec}(x) = y$. Rearranging this, we get $x = \sec(y)$. Taking the derivative of both sides, we get $1 = \sec(y)\tan(y)y^\prime$. Thus, $$y^\prime = \frac{1}{\sec(y)\tan(y)}.$$ Now, draw a right triangle with an acute angle $y$, hypotenuse $x$, and a side adjacent to angle $y$ with length $1$. By the Pythagorean theorem, the side opposite angle $y$ is equal to $\sqrt{x^2-1}$. Thus, we have $x = \sec(y)$, which we already knew and got from the triangle, and $$\tan(y) = \frac{1}{\sqrt{x^2-1}}$$ from the triangle we drew. We substitute these values in to our derivative expression to get $$y' = \frac{1}{x\sqrt{x^2-1}}.$$
However, $y^\prime$ should equal $$\frac1{|x|\sqrt{x^2-1}}$$ (note the absolute value). Since the difference occurred in the $|x|$ part, that means I must've done something wrong when I said $\sec(y) = x$, but I do not know why this assumption is wrong. Please correct my proof.
Best Answer
The sloppy reasoning occurs with the right triangle you wrote. You must be careful about the range (values) of the arcsec function. $y=\text{arcsec}(x)$ can lie either in $[0,\pi/2)$ or in $(\pi/2,\pi]$. (These correspond, respectively, to $x>0$ and $x<0$.)
$\tan(y)<0$ when $y\in (\pi/2,\pi]$, and so $\tan(y)=-\sqrt{x^2-1}$ in that event. Working with your formula for $y'$, we note that when $x<0$, $$\sec(y)\tan(y)= x(-\sqrt{x^2-1})= (-x)\sqrt{x^2-1} =|x|\sqrt{x^2-1},$$ and this explains the formula. To be honest, it's a bit sneaky to move the negative to the other term, but it allows us to write a single formula, rather than writing down cases. (There's no problem, of course, when $x>0$.)