# 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?

• $\frac {dy}{dx} = \frac {1}{\sec y\tan y}$ apply the identity $\sec^2 y = 1 + \tan^2 y$ Commented Mar 9, 2020 at 3:42

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{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}\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{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}\tag{5}\label{eq5A}