Introduction
The inverse cosine function reverses the cosine function on a chosen interval. Its derivative has a negative sign, and that sign is not an arbitrary convention: it follows from the way cosine changes on the interval used to define the inverse.
We will derive the formula by writing the inverse relation as an equation involving cosine, differentiating that equation, and then using the principal range of arccosine to determine the sign of a square root. The range matters: without it, the equation for the sine would allow two signs.
Definition and Principal Range
The function \(\arccos\) assigns to each \(x\) in \([-1,1]\) the unique angle \(y\) in \([0,\pi]\) whose cosine is \(x\). Equivalently, if \(y=\arccos x\), then \(\cos y=x\) and \(0\le y\le\pi\).
The interval \([0,\pi]\) is the principal range: cosine is one-to-one on it, so each \(x\in[-1,1]\) corresponds to exactly one angle. For interior inputs \(-1<x<1\), the angle lies in \((0,\pi)\). On that open interval, \(\sin y>0\). This fact will select the sign needed in the derivative formula.
Deriving the Formula
Let \(y=\arccos x\), so that \(\cos y=x\). For \(-1<x<1\), we have \(0<y<\pi\), and therefore \(\sin y\ne 0\). The inverse function derivative rule applies locally when a differentiable function has nonzero derivative. Since the derivative of cosine at \(y\) is \(-\sin y\), which is nonzero here, \(y\) is differentiable as a function of \(x\) at every interior input.
Now differentiate the defining equation with respect to \(x\). The left side requires the chain rule because \(y\) depends on \(x\); the right side differentiates to \(1\).
Because \(\sin y\ne0\), we can solve for the derivative:
To express this in terms of \(x\), use the identity \(\sin^2 y+\cos^2 y=1\). Since \(\cos y=x\),
Taking square roots initially gives \(\sin y=\pm\sqrt{1-x^2}\). But \(y\in(0,\pi)\), and sine is positive throughout this interval. Thus \(\sin y=\sqrt{1-x^2}\), with the positive square root. Substitution gives the result.
Why the Principal Range Determines the Sign
The sign choice is essential. The equation \(\sin^2 y=1-x^2\) alone determines only the square of \(\sin y\), not its sign. The range restriction \(0\le y\le\pi\) resolves the ambiguity: for interior \(x\), \(y\) is strictly between \(0\) and \(\pi\), where sine is positive.
The negative sign in the derivative then comes from differentiating cosine: \(\frac{d}{dy}\cos y=-\sin y\). Since the sine in the denominator is positive, the derivative of arccosine is negative throughout its interior domain. This agrees with the fact that cosine decreases as its angle moves from \(0\) to \(\pi\); its inverse therefore decreases as well.
Worked Examples
Worked Example: Find the Derivative at \(x=0\)
First, \(0\) is in the interval \((-1,1)\), so the derivative formula applies. Substitute \(x=0\):
Check using the defining relation. At \(x=0\), \(y=\arccos 0=\frac{\pi}{2}\), so \(\sin y=1\). The differentiated equation gives \(-\sin y\,\frac{dy}{dx}=1\), hence \(-1\cdot\frac{dy}{dx}=1\), or \(\frac{dy}{dx}=-1\). This agrees with the formula.
Worked Example: Find the Derivative at \(x=\frac{3}{5}\)
Since \(-1<\frac35<1\), the derivative exists. Substitute and simplify:
Check through the implicit relation. The angle \(y=\arccos(\frac35)\) lies in \((0,\frac{\pi}{2})\). Its cosine is \(\frac35\), and its sine is positive. The identity \(\sin^2y=1-\cos^2y\) gives \(\sin y=\sqrt{1-\frac9{25}}=\frac45\). Therefore the differentiated equation yields \(\frac{dy}{dx}=-1/(\frac45)=-\frac54\), as required.
Worked Example: Find the Derivative at \(x=-\frac{3}{5}\)
This input is also strictly inside the domain. The formula gives:
Check the sign using the principal range. Here \(y=\arccos(-\frac35)\) lies between \(\frac{\pi}{2}\) and \(\pi\), where sine remains positive. The identity gives \(\sin y=\sqrt{1-(-\frac35)^2}=\frac45\). Thus \(-\sin y\,\frac{dy}{dx}=1\) becomes \(-\frac45\frac{dy}{dx}=1\), so \(\frac{dy}{dx}=-\frac54\). In particular, the negative input does not make the derivative positive.
Where the Derivative Exists
The function \(\arccos x\) is defined at \(x=-1\) and \(x=1\), but being defined does not guarantee that it has a finite derivative there. The derivation requires division by \(\sin y\). At \(x=1\), the corresponding angle is \(y=0\), and at \(x=-1\), it is \(y=\pi\). In both cases \(\sin y=0\), so the implicit derivative equation cannot be solved for a finite \(\frac{dy}{dx}\).
The displayed derivative formula also signals this limitation: at either endpoint, \(1-x^2=0\), so its denominator is zero. More precisely, the endpoint behavior rules out a finite one-sided derivative. At \(x=1\), let \(y=\arccos x\), so \(y\to0^+\) as \(x\to1^-\). The one-sided difference quotient at \(1\) is
Using \(\sin u/u\to1\) as \(u\to0\), the last expression is asymptotic to \(-2/y\), which tends to \(-\infty\). At the other endpoint, put \(t=\pi-y\), so \(t\to0^+\) as \(x\to-1^+\). Since \(x=\cos y=-\cos t\), the one-sided quotient is
Thus there is no finite derivative at either endpoint, even when one-sided difference quotients are considered. The derivative exists precisely for \(-1<x<1\).
Common Sign Error
A common mistake is to take the positive square root after using the identity for sine, but then lose the negative sign from differentiating cosine. For example, a wrong calculation might read
The first implication is incorrect: dividing both sides by \(-\sin y\) gives a negative reciprocal, not a positive one. The correction is
A quick check is \(x=0\). The wrong result gives slope \(+1\), whereas the differentiated defining equation at \(y=\frac{\pi}{2}\) gives slope \(-1\). Keeping the negative sign from \(\frac{d}{dy}\cos y=-\sin y\) prevents this error.
Why the Derivation Is Useful
The method shows how the inverse cosine derivative follows from three ingredients: the defining equation, implicit differentiation, and the chosen range. Each contributes something necessary. Differentiating the equation produces the reciprocal of the cosine slope; the trigonometric identity rewrites that slope in terms of \(x\); and the principal range determines that the square root is positive before the overall negative sign is applied.
It also explains the domain restriction rather than merely attaching it to a memorized formula. At interior inputs, the cosine slope \(-\sin y\) is nonzero, so the local inverse has a finite derivative. At the endpoints, that slope vanishes, and the inverse develops an unbounded slope. The formula and its domain therefore express the same geometric and algebraic limitation.
Check Your Understanding
Use the defining relation and the principal range to reason through these questions.
- Why does \(y=\arccos x\) lie in \((0,\pi)\) when \(-1<x<1\)?
- After differentiating \(\cos y=x\), what equation relates \(\sin y\) and \(\frac{dy}{dx}\)?
- Why is \(\sin y=\sqrt{1-x^2}\), rather than its negative, for interior inputs?
- Compute the derivative of \(\arccos x\) at \(x=\frac45\).
- Why does the derivative formula not establish a finite derivative at \(x=1\) or \(x=-1\)?
