Why Endpoint Slopes Need Special Care
Inverse trigonometric functions can have well-defined values at the edges of their domains even when their slopes become arbitrarily steep nearby. The key distinction is between a function value at an endpoint and a derivative there: the former may exist while the latter, defined by a finite difference-quotient limit, does not.
Endpoint questions also require attention to the domain. For arcsine and arccosine, the domain ends at \(-1\) and \(1\). For arcsecant and arccosecant, the domain consists of the regions at or outside those two values. Thus the permissible approach to an endpoint may be from the left or the right, and the side must be stated correctly.
Domains, Endpoint Values, and Interior Derivatives
We use the principal inverse trigonometric functions. The functions \(\arcsin x\) and \(\arccos x\) have domain \([-1,1]\), with values in \([-\pi/2,\pi/2]\) and \([0,\pi]\), respectively. The functions \(\arcsec x\) and \(\arccsc x\) have domain \((-\infty,-1]\cup[1,\infty)\), with their standard principal ranges. In particular, \(\arcsec x=\arccos(1/x)\) and \(\arccsc x=\arcsin(1/x)\) on this domain.
The endpoint function values are finite:
For arcsine and arccosine, values just inside the domain approach the endpoints from the right at \(-1\) and from the left at \(1\). For arcsecant and arccosecant, values in the domain approach \(-1\) from the left and \(1\) from the right. This difference matters when forming endpoint difference quotients.
Why the Interior Slopes Blow Up
The denominator of each derivative formula explains the basic behavior. For \(-1<x<1\), the identity \(1-x^2=(1-x)(1+x)\) shows that the denominator for arcsine and arccosine tends to zero at either endpoint. For \(|x|>1\), write \(x^2-1=(|x|-1)(|x|+1)\). In the arcsecant and arccosecant formulas, the denominator tends to zero as \(|x|\) approaches \(1\), while the factor \(|x|\) tends to \(1\).
More precisely, the interior derivatives have the following leading behavior. The symbol \(\sim\) means that the ratio of the two sides tends to \(1\) in the indicated limit.
For example, as \(x\to1^-\), the factor \(1+x\) tends to \(2\), so \(\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}\sim\sqrt{2(1-x)}\). The other arcsine and arccosine cases follow by the same factorization at the other endpoint. For \(|x|\to1^+\), the factor \(|x|+1\) tends to \(2\) and \(|x|\) tends to \(1\), giving the corresponding arcsecant and arccosecant behavior:
Thus the magnitudes of all four interior derivatives grow without bound near either endpoint. The signs differ: arcsine and arcsecant have positive interior derivatives, while arccosine and arccosecant have negative interior derivatives.
Endpoint Difference Quotients: The Side Is Part of the Claim
A derivative at a point is defined by a limit of difference quotients. At a boundary point of a function’s domain, only the side lying in the domain is available. A finite one-sided derivative requires that the corresponding difference quotient approach a finite real number. If its magnitude instead grows without bound, there is no finite one-sided derivative, even though it is common informally to describe the graph as having a vertical tangent or an unbounded slope.
Consider \(\arcsin x\) at \(x=1\). Let \(t>0\) tend to \(0\), and set \(x=\cos t\). Then \(x<1\), so these inputs approach \(1\) from the left, as required by the domain \([-1,1]\). For small positive \(t\), \(\arcsin(\cos t)=\pi/2-t\), and \(1-\cos t=2\sin^2(t/2)\).
Here we used the standard limit \(\sin u/u\to1\) as \(u\to0\). The quotient diverges positively, so there is no finite left-hand derivative of \(\arcsin x\) at \(x=1\). The side is essential: this calculation uses inputs \(x<1\), not \(x>1\).
The other endpoint cases have the same domain-side discipline. The derivative sign or the endpoint quotient sign can be read from the function’s local change and the sign of the input change.
| Function and endpoint | Permitted approach | Interior derivative behavior |
|---|---|---|
| \(\arcsin x\), \(x=1\) | From the left | Positive, unbounded |
| \(\arcsin x\), \(x=-1\) | From the right | Positive, unbounded |
| \(\arccos x\), \(x=1\) | From the left | Negative, unbounded in magnitude |
| \(\arccos x\), \(x=-1\) | From the right | Negative, unbounded in magnitude |
| \(\arcsec x\), \(x=1\) | From the right | Positive, unbounded |
| \(\arcsec x\), \(x=-1\) | From the left | Positive, unbounded |
| \(\arccsc x\), \(x=1\) | From the right | Negative, unbounded in magnitude |
| \(\arccsc x\), \(x=-1\) | From the left | Negative, unbounded in magnitude |
Worked Examples
Worked Example: Arcsine Slopes Close to \(1\)
Estimate how the derivative of \(\arcsin x\) changes between \(x=0.99\) and \(x=0.9999\), and state the direction of approach to the endpoint.
Both inputs are less than \(1\), so they are in the domain and approach the endpoint \(1\) from the left. At \(x=0.99=99/100\),
Therefore,
At \(x=0.9999=9999/10000\),
so
The derivative is positive at both inputs and much larger at the input closer to \(1\). The approximations are rounded to two decimal places; the exact expressions are shown above. Substituting each input into \(1/\sqrt{1-x^2}\) gives the respective exact values, checking both calculations.
Worked Example: Arccosine at the Left Endpoint
Determine the sign and finiteness of the one-sided endpoint difference quotient for \(\arccos x\) at \(x=-1\).
The domain is \([-1,1]\), so inputs approach \(-1\) from the right. Set \(x=-\cos t\) for small \(t>0\); then \(-1<x<1\), \(x\to-1^+\), and \(\arccos(-\cos t)=\pi-t\). The endpoint value is \(\arccos(-1)=\pi\). Thus
The permitted right-hand difference quotient diverges to negative infinity, so no finite right-hand derivative exists at \(-1\). This conclusion matches the negative interior derivative, whose magnitude grows without bound. The identity \(1-\cos t=2\sin^2(t/2)\) and the limit \(\sin u/u\to1\) verify the limiting calculation.
Worked Example: Arcsecant and Arccosecant Near \(1\)
Compare the derivative values of \(\arcsec x\) and \(\arccsc x\) at \(x=1.01=101/100\), and compare their endpoint difference quotients at \(x=1\).
First, \(x>1\), so it is an interior point of the arcsecant and arccosecant domains. We have
Also, \(|x|=101/100\). Therefore,
At the endpoint, \(\arcsec 1=0\) and \(\arccsc 1=\pi/2\). To see the permitted approach, write \(x=\sec t\) for small \(t>0\). Then \(x>1\), \(\arcsec(\sec t)=t\), and \(\arccsc(\sec t)=\pi/2-t\). The endpoint quotients are
Indeed, \(\sec t-1=(1-\cos t)/\cos t\sim t^2/2\) as \(t\to0^+\), so the first quotient is asymptotic to \(2/t\) and the second to \(-2/t\). These signs agree with the interior derivative formulas. The displayed approximate derivative values are rounded to two decimal places; substituting \(x=101/100\) into the formulas gives the exact values shown.
A Chain-Rule Caution at an Endpoint
The chain rule requires differentiability of the functions involved at the point where it is applied. An interior formula for an inverse trigonometric function cannot automatically be used when its input reaches a domain endpoint. Even if a formal expression resembles a product of zero and an unbounded quantity, that expression is not an ordinary product of finite real numbers and does not determine a derivative.
For example, \(F(t)=\arcsin(1-t^2)\) is defined for \(-\sqrt{2}\le t\le\sqrt{2}\). At \(t=0\), its value is \(\pi/2\). For nonzero \(t\) near zero, the chain rule applies because \(1-t^2<1\), and gives
For \(t>0\), this tends to \(-\sqrt{2}\) as \(t\to0^+\); for \(t<0\), it tends to \(+\sqrt{2}\) as \(t\to0^-\). These unequal limits are consistent with the local expression \(F(t)=\pi/2-\arccos(1-t^2)\), and show why simply substituting \(t=0\) into the chain-rule expression would be invalid. The derivative at \(t=0\) does not exist, even though \(F(0)\) does.
What to Remember
Endpoint analysis has three separate parts: check the function’s domain, identify the permitted direction of approach, and distinguish the finite endpoint value from the derivative limit. At all four inverse-trigonometric endpoints considered here, the interior derivative magnitude becomes unbounded. The endpoint difference quotients also have unbounded magnitude, with signs determined by the function’s local change and the permitted side.
Check Your Understanding
Use domain sides, derivative signs, and endpoint values to answer these questions.
- From which side can inputs in the domain of \(\arcsin x\) approach \(x=1\)? What happens to its interior derivative there?
- At \(x=-1\), does the domain of \(\arccos x\) permit a left-hand or a right-hand approach? What is the sign of the endpoint difference quotient near that point?
- Which sign does the interior derivative of \(\arcsec x\) have near \(x=1\), and from which side does its domain approach that endpoint?
- Why does the existence of \(\arccsc(1)\) not imply that \(\arccsc x\) has a finite right-hand derivative at \(x=1\)?
- For \(F(t)=\arcsin(1-t^2)\), why is direct substitution of \(t=0\) into the interior chain-rule expression invalid?
