Differentiating secant inverse I want to show that $\cfrac{d}{dx}[\operatorname{arcsec} x]=\cfrac{1}{\vert x \vert\sqrt{x^2-1}}$. Here is my attempt:
\begin{align*} y=\sec^{-1}x &\iff x=\sec y \\ \frac{d}{dx}[x]&=\frac{d}{dx}[\sec y] \\ 1 &=\frac{\cos^2y}{\sin y} \frac{dy}{dx} \\ \frac{dy}{dx} &= \frac{\sin y}{\cos^2y} \end{align*}
I tried using the Pythagorean identity for 1 on the lefthand side, but couldn't see what else that would bring. I'm starting to think implicit differentiation isn't going to let me derive the expression I'm looking for.
Any ideas?

You have
$$x = \sec y = \frac{1}{\cos y} \implies \cos y = \frac{1}{x} \tag{1}\label{eq1A}$$
Also, you have, assuming that $\sin y \ge 0$, that
$$\sin^2 y + \cos^2 y = 1 \implies \sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{|x|} \tag{2}\label{eq2A}$$
Note you have a mistake in your differentiation where you got the reciprocal of what you should have had. In particular,
$$\begin{equation}\begin{aligned}
\frac{d}{dx}[\sec y] & = \frac{d}{dx}\left(\frac{1}{\cos y}\right) \\
& = \left(-\frac{(-\sin y)}{\cos^2 y}\right)\frac{dy}{dx} \\
& = \left(\frac{\sin y}{\cos^2 y}\right)\frac{dy}{dx}
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This then gives
$$\frac{dy}{dx} = \frac{\cos^2 y}{\sin y} \tag{4}\label{eq4A}$$
Using \eqref{eq1A} and \eqref{eq2A} in \eqref{eq4A} then gives the expression you're trying to prove, i.e, you get
$$\begin{equation}\begin{aligned}
\frac{dy}{dx} & = \frac{\frac{1}{x^2}}{\frac{\sqrt{x^2 - 1}}{|x|}} \\
& = \frac{|x|}{x^2\sqrt{x^2 - 1}} \\
& = \frac{1}{|x|\sqrt{x^2 - 1}}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
