Why the Arcsecant Derivative Needs Care
The inverse secant function is defined only for inputs that occur as values of secant on its chosen principal range. Its derivative also depends on which angle in that range corresponds to a given input. That choice determines the sign of the tangent that appears during implicit differentiation.
The resulting derivative is
The absolute value is not an optional adjustment: it records the sign information imposed by the principal range. We will derive the formula from the defining equation, identify its domain, and check the result at specific inputs.
Definition and Domain
Throughout this derivation, \(\operatorname{arcsec}x\) denotes the principal inverse of secant with angle values in \([0,\pi]\), excluding \(\pi/2\), where secant is undefined. In other words, if \(y=\operatorname{arcsec}x\), then \(\sec y=x\) and \(y\) lies in that principal range.
To see the domain, recall that \(\sec y=1/\cos y\) wherever \(\cos y\ne0\). Since \(-1\le\cos y\le1\), every defined value of secant is at most \(-1\) or at least \(1\). On the positive side of the principal range, \(0\le y<\pi/2\), secant takes values from \(1\) upward. On the negative side, \(\pi/2<y\le\pi\), it takes values from negative infinity up to \(-1\). The values \(1\) and \(-1\) occur at \(y=0\) and \(y=\pi\), respectively.
The function is defined at \(x=1\) and \(x=-1\), but those endpoints are not interior points of its domain. The derivative formula will apply on the two open intervals \((-\infty,-1)\) and \((1,\infty)\).
Deriving the Formula by Implicit Differentiation
Let \(y=\operatorname{arcsec}x\). By definition, \(\sec y=x\). For \(x\) in either open interval of the domain, \(y\) is not an endpoint of the principal range. Differentiate the defining equation with respect to \(x\), using the chain rule on \(\sec y\):
Because \(\sec y=x\), this gives
This intermediate expression is not yet the final formula. In particular, replacing \(\tan y\) by the positive square root of \(\tan^2 y\) would lose the sign on the negative-input branch. We need to use the principal range to determine that sign.
For \(x>1\), the corresponding angle satisfies \(0<y<\pi/2\), where \(\tan y>0\). For \(x<-1\), it satisfies \(\pi/2<y<\pi\), where \(\tan y<0\). Also, the identity \(1+\tan^2 y=\sec^2 y\) gives
Since \(\sqrt{x^2-1}\) is positive on both open intervals, the sign information gives
Equivalently, on both intervals, \(\tan y=\operatorname{sgn}(x)\sqrt{x^2-1}\), where \(\operatorname{sgn}(x)\) is \(1\) for positive \(x\) and \(-1\) for negative \(x\). Thus \(x\tan y=|x|\sqrt{x^2-1}\). Substituting into the differentiated equation proves the result.
Why the Absolute Value Appears
The square-root identity determines only the magnitude of \(\tan y\): it says \(|\tan y|=\sqrt{x^2-1}\). The principal range supplies the missing sign. When \(x>1\), both \(x\) and \(\tan y\) are positive. When \(x<-1\), both are negative. Consequently, their product \(x\tan y\) is positive on either branch and equals \(|x|\sqrt{x^2-1}\).
This is why the derivative is positive on both intervals, even though the angle \(y\) lies on different sides of \(\pi/2\). Omitting the absolute value would incorrectly give a negative derivative for negative inputs.
Worked Examples
Worked Example: Differentiate at \(x=2\)
Because \(2>1\), the derivative formula applies. First calculate the square-root factor:
Therefore,
Check directly in the differentiated defining relation. Since \(\operatorname{arcsec}2=\pi/3\), we have \(\sec(\pi/3)=2\) and \(\tan(\pi/3)=\sqrt{3}\). The equation \(x\tan y\,dy/dx=1\) then gives \(2\sqrt{3}\,dy/dx=1\), hence \(dy/dx=1/(2\sqrt{3})\), as found.
Worked Example: Differentiate at \(x=-2\)
The input \(-2\) lies in the negative part of the domain, so the derivative exists. Evaluate each factor:
Thus
To check, the principal angle is \(\operatorname{arcsec}(-2)=2\pi/3\). At that angle, \(\sec(2\pi/3)=-2\) and \(\tan(2\pi/3)=-\sqrt{3}\). Substitution into \(x\tan y\,dy/dx=1\) gives \((-2)(-\sqrt{3})\,dy/dx=1\), or \(2\sqrt{3}\,dy/dx=1\). This agrees with the formula.
Worked Example: Differentiate at \(x=\sqrt{5}\)
Since \(\sqrt{5}>1\), the derivative formula applies. The absolute value is \(|\sqrt{5}|=\sqrt{5}\), and the radicand is
It follows that
For a check, let \(y=\operatorname{arcsec}(\sqrt{5})\). The principal angle lies in \((0,\pi/2)\), so \(\tan y>0\). The identity \(\tan^2y=\sec^2y-1\) gives \(\tan^2 y=5-1=4\), hence \(\tan y=2\). Therefore \(x\tan y\,dy/dx=(\sqrt{5})(2)\,dy/dx=1\), so \(dy/dx=1/(2\sqrt{5})\).
A Common Sign Error
A frequent mistake is to use \(\tan y=\sqrt{x^2-1}\) for every \(x\). That statement is false when \(x<-1\): the principal angle is then in \((\pi/2,\pi)\), where tangent is negative. For example, at \(x=-2\), the incorrect substitution gives
But \(\operatorname{arcsec}(-2)=2\pi/3\), and \(\tan(2\pi/3)=-\sqrt{3}\), not \(+\sqrt{3}\). Correcting the sign gives
The safe procedure is to determine the sign from the principal angle before simplifying. Alternatively, combine the signs in \(x\tan y\) only after establishing which branch contains \(y\).
Domain Endpoints and Where the Derivative Exists
The function values at the endpoints are \(\operatorname{arcsec}(1)=0\) and \(\operatorname{arcsec}(-1)=\pi\). However, the derivative formula has a zero factor under the square root at both inputs, so its denominator vanishes there. The formula does not give a finite derivative at either endpoint.
There is also a domain reason to distinguish these points: the function’s domain has no open interval around \(1\) or \(-1\). Even if one considers the appropriate one-sided difference quotient at an endpoint, it does not approach a finite number. For instance, for \(h>0\), set \(y=\operatorname{arcsec}(1+h)\). Then \(y\to0\) as \(h\to0^+\), and \(h=\sec y-1\). Since \(\sec\) is differentiable at \(0\) and \(\sec'(0)=\sec(0)\tan(0)=0\), we have \((\sec y-1)/y\to0\). Thus \(y/h\to+\infty\), so there is no finite right derivative at \(1\). The endpoint \(-1\) has the corresponding one-sided behavior.
Derivation in Four Steps
Set \(y=\operatorname{arcsec}x\), so that \(\sec y=x\) on the stated principal range.
Obtain \(\sec y\tan y\,dy/dx=1\), and use \(\sec y=x\) to get \(dy/dx=1/(x\tan y)\).
From \(\tan^2 y=x^2-1\), the magnitude of \(\tan y\) is \(\sqrt{x^2-1}\); its sign is positive for \(x>1\) and negative for \(x<-1\).
On either branch, \(x\tan y=|x|\sqrt{x^2-1}\), giving the derivative formula for \(x\in(-\infty,-1)\cup(1,\infty)\).
Check Your Understanding
Use the principal range and the defining relation, not just the algebraic square-root identity, when answering.
- What is the domain of \(\operatorname{arcsec}x\), and on which open intervals does its derivative formula apply?
- Starting from \(\sec y=x\), what equation results after implicit differentiation?
- Why is \(\tan y\) negative when \(x<-1\) under the chosen principal range?
- Evaluate \(\left.\frac{d}{dx}\operatorname{arcsec}x\right|_{x=3}\).
- What incorrect derivative value results at \(x=-2\) if \(\tan y\) is mistakenly taken to be positive?
