Endpoint Behavior of Arcsine and Arccosine Derivatives

Endpoint Values and Endpoint Derivatives

The inverse sine and inverse cosine functions are defined at both ends of their domains: \(x=-1\) and \(x=1\). Their function values are finite there. Their derivatives, however, are not finite at either endpoint. Distinguishing these facts matters whenever a derivative formula contains a denominator that becomes zero at an endpoint.

Write the inverse sine function as \(\arcsin x\), and the inverse cosine function as \(\arccos x\). Their domains are both the closed interval \([-1,1]\). The derivative formulas derived for interior inputs are

The condition \(-1<x<1\) is essential. At \(x=-1\) or \(x=1\), the denominator is zero, so these expressions do not give a real, finite derivative. We will show more than that the formulas cannot be evaluated there: the relevant one-sided difference quotients become unbounded.

Endpoint Function Values

The inverse sine function returns angles in \([-\pi/2,\pi/2]\), and the inverse cosine function returns angles in \([0,\pi]\). Using the angles whose sine or cosine is \(1\) or \(-1\), their endpoint values are

These are ordinary function values. For example, \(\arcsin(1)\) is the angle in the inverse sine range whose sine is \(1\). Knowing that value does not tell us that the function has a finite slope at \(x=1\).

At an endpoint of a function's domain, the derivative is assessed using a difference quotient from the side that remains in the domain. If that one-sided limit is a finite real number, it can be called a finite one-sided derivative. A quotient that tends to \(+\infty\) or \(-\infty\) is unbounded, not a finite derivative.

Why the Interior Derivative Formulas Stop at the Endpoints

For \(-1<x<1\), the formulas contain \(\sqrt{1-x^2}\), which is positive. At either endpoint,

Thus direct substitution would require division by zero. This tells us that the displayed formulas cannot be evaluated at the endpoints; by itself, it does not yet explain what happens to the derivative. To decide whether a finite one-sided derivative exists, we examine the difference quotients.

Deriving the Endpoint Behavior for Arcsine

Consider first the endpoint \(x=1\). Approach it from within the domain by writing \(x=1-h\), where \(h>0\). Let

As \(h\) approaches \(0\) from above, \(1-h\) approaches \(1\), so \(\theta\) approaches \(\pi/2\) and \(\delta\) approaches \(0\) from above. Since \(\sin\theta=1-h\), and \(\theta=\pi/2-\delta\), the cofunction identity gives

Therefore \(h=1-\cos\delta=2\sin^2(\delta/2)\). The endpoint difference quotient is

The standard limit \(\lim_{u\to0}\sin u/u=1\) implies that the squared factor approaches \(1\). Meanwhile \(1/(\delta/2)\) grows without bound as \(\delta\to0^+\). Hence this one-sided difference quotient tends to \(+\infty\), not to a finite real number.

At \(x=-1\), approach from the right and write \(x=-1+h\), with \(h>0\). Set \(\theta=\arcsin(-1+h)\) and \(\delta=\theta+\pi/2\). As \(h\to0^+\), \(\delta\to0^+\). Because \(\theta=-\pi/2+\delta\),

The difference quotient from the domain side is therefore

So neither endpoint has a finite one-sided derivative for \(\arcsin x\). At both ends, the function's secant slopes become arbitrarily large and positive as the input approaches the endpoint from inside the domain.

Arccosine Endpoint Behavior

For every \(x\in[-1,1]\), the inverse trigonometric values satisfy

This follows because the angles in the stated ranges are complementary when their sine and cosine correspond to the same input. The identity also relates the endpoint difference quotients: the constant \(\pi/2\) cancels, and the varying term has the opposite sign. Thus at either endpoint the one-sided difference quotient for \(\arccos x\) is the negative of the corresponding quotient for \(\arcsin x\). Since the latter tends to \(+\infty\), the former tends to \(-\infty\).

Consequently, \(\arccos x\) has no finite one-sided derivative at either endpoint. Its secant slopes become unbounded in the negative direction as the input approaches either endpoint from within the domain.

Worked Example: Comparing Values and Derivatives at the Endpoints

Worked Example: Checking an Interior Input

Worked Example: Endpoint Quotients and Their Signs

Common Pitfall: Calling an Infinite Slope a Derivative

A common error is to substitute \(x=1\) into \(1/\sqrt{1-x^2}\), obtain division by zero, and write that \(\arcsin'(1)=\infty\). This is not correct. The formula is only established for \(-1<x<1\), and an infinite limit is not a finite derivative.

For example, at \(x=1\),

is undefined, not equal to a real number called infinity. The correct conclusion comes from the endpoint difference quotient: it tends to \(+\infty\), so there is no finite one-sided derivative. Similarly, the arccosine quotient tends to \(-\infty\), not to a finite derivative.

The opposite mistake is to assume that a function cannot have an endpoint value unless its derivative exists there. The values \(\arcsin(1)=\pi/2\) and \(\arccos(1)=0\) are well-defined even though neither function has a finite derivative at \(1\).

When This Distinction Matters

The endpoint behavior tells you where the derivative formulas may be used: only for inputs strictly between \(-1\) and \(1\). It also helps you interpret the graph. Near the endpoints, the arcsine graph has increasingly steep positive secant slopes, while the arccosine graph has increasingly steep negative secant slopes. The functions remain defined at the endpoints even as these slopes become unbounded.

Before applying either derivative formula, check that the input lies in \((-1,1)\). If it equals an endpoint, do not substitute into the formula or report an infinite derivative. Instead, distinguish the endpoint function value from the derivative question and, when needed, examine the one-sided difference quotient.