Why the Negative Sign Matters
The derivative of an arccosine composition combines two ideas: the derivative of the inverse cosine function and the chain rule. The sign is important. Unlike the derivative of arcsine, the derivative of arccosine is negative on its interior domain.
The basic rule is not automatically valid at every input where the expression is defined. The inner function must be differentiable, and its value must lie strictly between \(-1\) and \(1\). At an endpoint, the formula has a zero denominator, so the composition must be examined separately.
The Rule and Its Conditions
The real-valued function \(\arccos(u)\) is defined for \(-1\leq u\leq 1\), with output in \([0,\pi]\). Its derivative at an input \(u\) exists for \(-1<u<1\) and is \[ \frac{d}{du}\arccos(u)=-\frac{1}{\sqrt{1-u^2}}. \] For a composition, the chain rule multiplies this outer derivative by the derivative of the inner function.
The square root in the denominator is positive at interior inputs. The minus sign comes from differentiating the outer function \(\arccos\), not from the inner function. If \(g'(x)\) is itself negative, the two negative signs can make the resulting derivative positive.
Why the Formula Works
Here is a derivation at an interior point. Let \(y(x)=\arccos(g(x))\). Since \(g(x)\) is strictly between \(-1\) and \(1\), the value \(y(x)\) lies strictly between \(0\) and \(\pi\), and the defining relation for arccosine gives \(\cos(y(x))=g(x)\). Differentiate both sides with respect to \(x\):
Because \(0<y(x)<\pi\), its sine is positive. From \(\cos(y(x))=g(x)\) and \(\sin^2(y)+\cos^2(y)=1\), we obtain
This quantity is nonzero at an interior input, so we can divide by it. Solving for \(y'(x)\) gives
This also explains the negative sign: differentiating \(\cos(y(x))\) produces \(-\sin(y(x))y'(x)\). The derivation relies on an interior input; at an endpoint, \(\sin(y)=0\), so division in this argument is not valid.
Worked Examples
Worked Example: A Linear Inner Function
Differentiate \(y=\arccos(2x-1)\), and find the derivative at \(x=\frac14\).
Set \(g(x)=2x-1\), so \(g'(x)=2\). The expression is defined when \(-1\leq 2x-1\leq 1\), or \(0\leq x\leq 1\). At interior points \(0<x<1\), the chain rule gives
At \(x=\frac14\), the inner input is \(2(\frac14)-1=-\frac12\), which is strictly between \(-1\) and \(1\). Therefore,
Check using the defining relation. At this point \(y=\arccos(-\frac12)=\frac{2\pi}{3}\), so \(\sin y=\frac{\sqrt3}{2}\). Differentiating \(\cos y=2x-1\) gives \(-\sin(y)y'=2\). Substituting the result gives \(-(\frac{\sqrt3}{2})(-\frac4{\sqrt3})=2\), as required.
Worked Example: A Squared Inner Function
Differentiate \(y=\arccos(x^2)\), and find the derivative at \(x=\frac12\).
Here \(g(x)=x^2\) and \(g'(x)=2x\). The real-valued expression requires \(-1\leq x^2\leq1\), so \(-1\leq x\leq1\). At points with \(-1<x<1\), \(x^2<1\), and the derivative rule gives
At \(x=\frac12\), the inner input is \((\frac12)^2=\frac14\), an interior value. Thus
For a check, \(\cos y=\frac14\) at this point, and \(y\in[0,\pi]\), so \(\sin y=\sqrt{1-\frac1{16}}=\frac{\sqrt{15}}4\). The differentiated relation is \(-\sin(y)y'=g'(x)=1\). Substitution gives \(-(\frac{\sqrt{15}}4)(-\frac4{\sqrt{15}})=1\), confirming the derivative.
Worked Example: A Rational Inner Function
Differentiate \(y=\arccos\left(\frac{x-1}{x+1}\right)\), and find the derivative at \(x=2\).
The inner function is \(g(x)=\frac{x-1}{x+1}\), with \(x\ne-1\). By the quotient rule,
Where \(-1<g(x)<1\), the composition derivative is
At \(x=2\), \(g(2)=\frac13\), so the input is interior. Also \(g'(2)=\frac{2}{9}\). Therefore,
To check, \(\cos y=\frac13\) implies \(\sin y=\sqrt{1-\frac19}=\frac{2\sqrt2}{3}\). Differentiating \(\cos y=g(x)\) at \(x=2\) requires \(-\sin(y)y'=g'(2)=\frac29\). The computed derivative gives \(-(\frac{2\sqrt2}{3})(-\frac{1}{3\sqrt2})=\frac29\), as required.
Endpoint Inputs Need Separate Analysis
If \(g(x)=1\) or \(g(x)=-1\), the denominator in the chain-rule expression is zero. Do not substitute an endpoint into the formula and treat the resulting division by zero as a derivative. The formula simply does not apply there. Depending on how \(g\) approaches the endpoint, the composition may have no finite derivative, or it may have a derivative that needs to be found by a separate argument.
Worked Example: An Endpoint with No Finite Derivative
Consider \(y=\arccos(1+2x)\). Its real domain is \(-1\leq x\leq0\). At \(x=0\), the inner input is \(1\), so the chain-rule formula is not valid. We can check the derivative from within the domain using a difference quotient. For \(h<0\), the identity \(\arccos(1-u)=2\arcsin(\sqrt{u/2})\), valid for \(0\leq u\leq2\), gives
Since \(\arcsin(z)/z\to1\) as \(z\to0\), this quotient behaves as
Thus there is no finite derivative at \(x=0\); in particular, the value is not \(-2\). At the interior point \(x=-\frac12\), however, the input is \(1+2(-\frac12)=0\) and the derivative rule applies:
As a check at this interior point, \(y=\arccos(0)=\frac\pi2\), so \(-\sin(y)y'=2\) becomes \(-(1)(-2)=2\). The equation holds.
A Common Sign Error
A frequent mistake is to omit the negative sign from the derivative of arccosine. For example, for \(y=\arccos(3x)\), an incorrect calculation is
This has the wrong sign. The outer derivative is negative, and the inner derivative is \(3\), so at interior points where \(|3x|<1\), the correct calculation is
A quick check confirms the correction: at \(x=0\), the function is decreasing as its input \(3x\) increases, and the derivative is \(-3\), not \(3\).
Organizing the Calculation
For a new composition, separate the domain check from the differentiation. The following sequence helps prevent both endpoint errors and sign errors.
Write \(g(x)\) and calculate \(g'(x)\).
The expression is real-valued where \(-1\leq g(x)\leq1\); the derivative rule requires the strict inequalities \(-1<g(x)<1\).
Use \(-g'(x)/\sqrt{1-[g(x)]^2}\), retaining the negative sign.
Substitute into the differentiated relation \(-\sin(y)y'=g'(x)\), or evaluate carefully at the requested point.
Why This Rule Is Useful
Arccosine compositions occur whenever an angle is defined by a changing quantity, such as an angle whose cosine depends on position or time. The derivative describes how quickly that angle changes. Its sign reflects both the decreasing behavior of arccosine and the direction in which the inner input changes.
The main practical distinction is between an interior input and an endpoint input. At interior inputs, the chain-rule formula is direct and reliable. At endpoints, check the domain and use the definition of derivative or another suitable argument; a formula with a zero denominator cannot settle the question.
Check Your Understanding
For each question, first check whether the relevant input is interior before using the derivative rule.
- Differentiate \(y=\arccos(5x-2)\) at points where the derivative rule applies.
- Find the derivative of \(y=\arccos(x^3)\) at \(x=\frac12\).
- For \(y=\arccos(1-x)\), identify the \(x\)-values where the chain-rule formula applies.
- Explain why differentiating \(\cos(y)=g(x)\) produces a negative sign in the arccosine composition rule.
- What additional analysis is needed when \(g(x)=1\) or \(g(x)=-1\)?
