Why the Absolute Value Matters
The derivatives of arcsecant and arccosecant look similar, and both involve a square root. The square root alone, however, does not determine the sign of the derivative. That sign also depends on which part of the inverse function’s domain is being used.
For arcsecant, a formula without an absolute value can give the wrong sign whenever the input is negative. For arccosecant, the same omission causes a sign error there as well. To see why, we will keep track of the signs of the relevant trigonometric functions as we derive each formula.
Domains and Principal Values
The real arcsecant and arccosecant functions are defined for \(|x|\geq 1\). Using the standard principal-value conventions, we can write \(\arcsec x=\arccos(1/x)\), with values in \([0,\pi]\) other than \(\pi/2\), and \(\arccsc x=\arcsin(1/x)\), with values in \([-\pi/2,\pi/2]\) other than \(0\).
The derivative formulas apply on the open parts of the domain, where \(|x|>1\). At \(x=1\) and \(x=-1\), the square root in the formulas is zero, and neither inverse function has a finite derivative there. The domain has separate positive and negative pieces, so we must track the sign on each piece rather than assume the input is positive.
Deriving the Arcsecant Formula
Let \(y=\arcsec x\), so \(\sec y=x\). Differentiating this identity with respect to \(x\) gives \(\sec y\tan y\,\frac{dy}{dx}=1\), and hence \(\frac{dy}{dx}=\frac{1}{x\tan y}\). This expression is not yet enough: we need the sign of \(\tan y\) for the chosen principal value.
If \(x>1\), then \(y\) lies between \(0\) and \(\pi/2\), so \(\tan y>0\). If \(x<-1\), then \(y\) lies between \(\pi/2\) and \(\pi\), so \(\tan y<0\). In both cases, \(\sec y=x\) gives \(\tan^2 y=\sec^2 y-1=x^2-1\). Thus the magnitude of \(\tan y\) is \(\sqrt{x^2-1}\), but its sign matches the sign of \(x\).
Here, \(\operatorname{sign}(x)=1\) for \(x>0\) and \(\operatorname{sign}(x)=-1\) for \(x<0\). Substituting this signed value into the differentiated identity gives
The last equality follows because \(x\,\operatorname{sign}(x)=|x|\). Therefore the derivative is positive on both pieces of the domain, even though \(\tan y\) changes sign between them.
Deriving the Arccosecant Formula
Now let \(y=\arccsc x\), so \(\csc y=x\). Differentiating gives \(-\csc y\cot y\,\frac{dy}{dx}=1\), so \(\frac{dy}{dx}=-\frac{1}{x\cot y}\). Again, the sign of the trigonometric factor matters.
On the principal range for arccosecant, \(y\) is positive when \(x>1\) and negative when \(x<-1\). In both cases \(\cot y\) has the same sign as \(x\): for positive \(y\) in this range, \(\cot y>0\); for negative \(y\), \(\cot y<0\). Also, \(\cot^2 y=\csc^2 y-1=x^2-1\). Therefore \(\cot y=\operatorname{sign}(x)\sqrt{x^2-1}\), and substitution yields
The negative sign is present on both pieces of the domain. In particular, the derivative of arccosecant is negative even when \(x< -1\).
Compare the Signs on Each Piece
The square root \(\sqrt{x^2-1}\) is positive for both \(x>1\) and \(x<-1\). The absolute value \(|x|\) is also positive on both pieces. It follows directly from the formulas that arcsecant is increasing and arccosecant is decreasing on each of their domain intervals.
| Input | \(\arcsec x\) derivative | \(\arccsc x\) derivative |
|---|---|---|
| \(x>1\) | Positive | Negative |
| \(x<-1\) | Positive | Negative |
A tempting but incorrect step is to replace \(|x|\) with \(x\), as though every input were positive. That replacement changes the sign of the denominator when \(x<0\), even though the square root stays positive. It therefore changes the derivative’s sign on the negative part of the domain.
Worked Examples
Worked Example: Arcsecant at a Negative Input
Find \(\frac{d}{dx}\arcsec x\) at \(x=-2\). Since \(|-2|=2\) and \((-2)^2-1=3\), the derivative is
The result is positive, as required by the formula on the negative part of the domain. If we incorrectly omitted the absolute value, we would get \(\frac{1}{(-2)\sqrt{3}}=-\frac{1}{2\sqrt{3}}\), the wrong sign. Substituting into the correct derivative formula confirms the value: \(1/(2\sqrt{3})\).
Worked Example: Arccosecant at a Negative Input
Find \(\frac{d}{dx}\arccsc x\) at \(x=-\sqrt{5}\). First, \(|-\sqrt{5}|=\sqrt{5}\) and \((-\sqrt{5})^2-1=5-1=4\), so \(\sqrt{4}=2\). Therefore
The derivative is negative. Omitting the absolute value would instead give \(-1/((-\sqrt{5})\cdot 2)=1/(2\sqrt{5})\), which is incorrect. Using the correct formula at \(x=-\sqrt{5}\) gives \(-1/(2\sqrt{5})\), confirming the computed value.
Worked Example: Arcsecant of a Function with Negative Input
Let \(f(t)=\arcsec(1-2t)\). Find \(f'(2)\). The inner input at \(t=2\) is \(1-2(2)=-3\), which satisfies the arcsecant domain condition. By the chain rule,
At \(t=2\), the denominator is \(|-3|\sqrt{(-3)^2-1}=3\sqrt{8}=6\sqrt{2}\). Thus
The derivative is negative because the inner function has derivative \(-2\), while arcsecant has a positive derivative at the input \(-3\). Substituting \(t=2\) into the displayed chain-rule formula gives the same value, \(-2/(6\sqrt{2})=-1/(3\sqrt{2})\).
A Useful Sign Check
When differentiating either function, separate the magnitude calculation from the sign calculation. The magnitude of the denominator is \(|x|\sqrt{x^2-1}\). The arcsecant derivative has a positive numerator; the arccosecant derivative has a negative numerator. This separation makes it harder to accidentally treat a negative input as positive or to let the sign of \(x\) alter the result incorrectly.
Verify that the input satisfies \(|x|>1\) at the point where a derivative is required.
Use \(|x|\sqrt{x^2-1}\) in the denominator for either derivative.
Arcsecant has a positive derivative; arccosecant has a negative derivative.
For a composed function, multiply by the derivative of its inner input.
Common Mistake: Treating the Input as Positive
For \(x=-3\), the incorrect arcsecant calculation \(\frac{1}{x\sqrt{x^2-1}}=\frac{1}{(-3)\sqrt{8}}<0\) predicts a negative derivative. But the principal-value derivation shows that \(\tan y<0\) as well as \(x<0\), so their product \(x\tan y\) is positive. The correct derivative is \(\frac{1}{|x|\sqrt{x^2-1}}=\frac{1}{3\sqrt{8}}>0\).
The error is not in evaluating the square root: \(\sqrt{x^2-1}\) is positive. It is in dropping the absolute value, which fails to account for the sign of the trigonometric factor on the negative branch. For arccosecant, omitting the absolute value creates the corresponding wrong sign on negative inputs.
Check Your Understanding
Use the domain conditions and the signed derivative formulas to answer these questions.
- On which real intervals are the arcsecant and arccosecant derivative formulas valid?
- What is the sign of \(\frac{d}{dx}\arcsec x\) when \(x<-1\), and why?
- Find \(\frac{d}{dx}\arccsc x\) at \(x=2\).
- What incorrect sign results from using \(1/(x\sqrt{x^2-1})\) for arcsecant when \(x< -1\)?
- For \(f(t)=\arcsec(1-2t)\), why does the derivative at \(t=2\) have a negative sign even though arcsecant itself has a positive derivative on its domain?
