Deriving The Arcsine Derivative

Why the Principal Range Matters

The sine function is not one-to-one on all real numbers, so it does not have an inverse function on its entire domain. The inverse sine function, written \(\arcsin(x)\), is defined using the restriction of sine to the interval \([-\pi/2,\pi/2]\). This choice makes the inverse single-valued, and it will also determine the sign needed to find the derivative.

The goal is to derive the slope of the graph of \(y=\arcsin(x)\). The key idea is to use the equivalent equation \(\sin(y)=x\), differentiate implicitly, and then recover \(\cos(y)\) from the identity \(\sin^2(y)+\cos^2(y)=1\). That identity gives two possible signs for cosine; the principal range tells us which one is correct.

The Inverse-Derivative Principle

The relevant inverse-function principle is this: if a differentiable function \(g\) is one-to-one near \(y_0\), if \(g'(y_0)\ne 0\), and if its inverse is defined near \(x_0=g(y_0)\), then the inverse is differentiable at \(x_0\) and

For sine restricted to \((-\pi/2,\pi/2)\), the function is one-to-one, and its derivative is \(\cos(y)\), which is positive throughout that open interval. Therefore, at each interior angle, the inverse-derivative principle applies. We can also see the derivative directly by differentiating the defining relation, as follows.

Deriving the Arcsine Derivative

Let \(x\in(-1,1)\), and set \(y=\arcsin(x)\). By definition, \(\sin(y)=x\), and the principal range gives \(y\in(-\pi/2,\pi/2)\). The chain rule applied to both sides of the defining relation gives

Because \(y\) lies strictly between \(-\pi/2\) and \(\pi/2\), \(\cos(y)>0\). We may divide by \(\cos(y)\), obtaining \(dy/dx=1/\cos(y)\). To express this in terms of \(x\), use \(\sin(y)=x\) and the Pythagorean identity:

Taking a square root gives \(\cos(y)=\sqrt{1-x^2}\), with the positive sign because of the principal range. Therefore:

Why the Derivation Works

The implicit differentiation step is not just a convenient manipulation. It expresses how small changes in the input and output are related by the defining equation. At an interior point, the inverse-function principle justifies differentiating the inverse relation: since the derivative of sine at \(y\) is \(\cos(y)\ne 0\), the inverse slope is the reciprocal slope, \(1/\cos(y)\). The identity \(\sin^2(y)+\cos^2(y)=1\) then converts that slope into a formula involving the input \(x\).

The restriction to interior inputs is important. At \(x=1\) and \(x=-1\), the corresponding angles are \(\pi/2\) and \(-\pi/2\), where \(\cos(y)=0\). The implicit equation would become \(0\cdot dy/dx=1\), so dividing by \(\cos(y)\) is not possible. The inverse-function principle also requires a nonzero derivative of the original function, and that condition fails at both endpoint angles.

In fact, the slopes become unbounded as the input approaches either endpoint from inside the domain. For example, if \(x=\sin(y)\) with \(y<\pi/2\) and \(y\to\pi/2\), the mean value theorem gives a number \(c\) between \(y\) and \(\pi/2\) such that

As \(y\to\pi/2\), \(c\to\pi/2\), so \(\cos(c)\to0^+\) and this quotient grows without bound. The corresponding one-sided slope at \(x=1\) is not finite; the same reasoning applies at \(x=-1\). Thus the derivative formula correctly has the open interval \((-1,1)\) as its domain.

Worked Examples

A Common Sign Error

From \(\cos^2(y)=1-x^2\), taking square roots gives \(\cos(y)=\pm\sqrt{1-x^2}\). A common mistake is to choose the negative square root without checking which angles the inverse function returns. For instance, at \(x=0\), that choice would give \(\cos(y)=-1\) and would lead to the claimed slope \(-1\). But \(\arcsin(0)=0\), and \(\cos(0)=1\), so the implicit slope equation is \(1\cdot dy/dx=1\), giving slope \(1\). The principal range makes cosine positive at every interior angle, so the negative sign is wrong.

Another important boundary check is not to substitute \(x=1\) or \(x=-1\) into the formula and treat the resulting division by zero as an ordinary derivative. The formula was derived under the condition \(\cos(y)\ne0\), which excludes the endpoint angles. At those inputs, the slope is not finite.

Why This Derivative Is Useful

The formula describes how quickly the principal angle changes as its sine value changes. Near \(x=0\), the derivative is \(1\); as \(x\) approaches either endpoint, the denominator \(\sqrt{1-x^2}\) approaches zero and the slope grows. This behavior is consistent with the inverse relationship: sine changes very little near its maximum and minimum, so reversing that relationship requires a large change in angle for a small change in input.

The derivation also illustrates a general method for inverse functions. Begin with the equation defining the inverse, differentiate the relation while accounting for the dependence of the inverse output on the input, and then use the defining range to resolve any ambiguity. Here, the principal range is not merely a convention for choosing an angle; it is essential information in the derivative calculation.