Why the Arccosine Derivative Is Negative
The inverse cosine function, written \(\arccos(x)\), returns an angle whose cosine is \(x\). Its derivative is negative wherever the derivative exists in the interior of its domain. This sign is not an arbitrary feature of the formula: as an angle increases from \(0\) to \(\pi\), its cosine decreases.
To derive the formula, let \(y=\arccos(x)\). Then \(\cos(y)=x\). Implicit differentiation relates the rate of change of \(y\) to the rate of change of \(x\). The principal range of arccosine then determines which square root to use when rewriting \(\sin(y)\).
What Arccosine Means
Cosine is not one-to-one on all real numbers, so an inverse function requires a restricted interval on which cosine is one-to-one. For arccosine, the chosen interval is \([0,\pi]\). The input to arccosine must be a cosine value from this interval, so it lies in \([-1,1]\).
On the interior \(0<y<\pi\), \(\sin(y)>0\), while \(\cos(y)\) decreases as \(y\) increases. In fact, the derivative of cosine is \(-\sin(y)\), which is negative on this interval. By the Mean Value Theorem, cosine is strictly decreasing on \([0,\pi]\): for any \(a,b\) with \(0\leq a<b\leq\pi\), there is a \(c\in(a,b)\) such that
Therefore, when \(x\) increases, the angle whose cosine is \(x\) must decrease. This explains the expected negative sign before any algebra is done.
Deriving the Formula by Implicit Differentiation
Set \(y=\arccos(x)\). The definition gives the identity \(\cos(y)=x\). Here \(y\) depends on \(x\), so differentiating \(\cos(y)\) requires the chain rule: its derivative with respect to \(x\) is \(-\sin(y)\frac{dy}{dx}\). Differentiating both sides gives
For \(-1<x<1\), the corresponding angle satisfies \(0<y<\pi\), so \(\sin(y)>0\). In particular, \(\sin(y)\neq 0\), and we can solve for the derivative:
To express this answer using \(x\), use the identity \(\sin^2(y)+\cos^2(y)=1\) and \(\cos(y)=x\). These give \(\sin^2(y)=1-x^2\). Since \(0<y<\pi\) implies \(\sin(y)>0\), the correct square root is \(\sin(y)=\sqrt{1-x^2}\), not its negative. Substitution yields the derivative formula.
The derivative formula is established on the open interval \((-1,1)\). At \(x=-1\) and \(x=1\), the denominator in the formula is zero, so this derivation does not give a finite derivative there.
Worked Examples
Worked Example: Derive the Derivative from the Defining Identity
Let \(y=\arccos(x)\), where \(-1<x<1\). Find \(\frac{dy}{dx}\) by implicit differentiation and check the result in the differentiated identity.
By the definition of arccosine, \(\cos(y)=x\). Differentiate each side with respect to \(x\). The chain rule on the left gives
Thus
Solving for \(\frac{dy}{dx}\) gives \(\frac{dy}{dx}=-\frac{1}{\sin(y)}\). Since \(y\in(0,\pi)\), sine is positive. Also, \(\cos(y)=x\), so
Therefore
To check, substitute this expression for \(\frac{dy}{dx}\) into the differentiated identity. Since \(\sin(y)=\sqrt{1-x^2}\),
This matches the derivative of the right side, \(x\), so the result satisfies the differentiated relation.
Worked Example: Find the Derivative at \(x=\frac{3}{5}\)
Find \(\left.\frac{d}{dx}\arccos(x)\right|_{x=3/5}\), then check it using the corresponding angle.
Since \(-1<\frac{3}{5}<1\), the derivative formula applies. Substitute \(x=\frac{3}{5}\):
For a check, let \(y=\arccos(\frac{3}{5})\). Then \(\cos(y)=\frac{3}{5}\), and \(y\in(0,\pi)\). Its sine is positive, so
The implicit derivative equation is \(-\sin(y)\frac{dy}{dx}=1\). Substituting \(\sin(y)=\frac{4}{5}\) and \(\frac{dy}{dx}=-\frac{5}{4}\) gives
The check succeeds. The derivative is \(-\frac{5}{4}\).
Worked Example: Find the Derivative at \(x=-\frac{\sqrt{3}}{2}\)
Find \(\left.\frac{d}{dx}\arccos(x)\right|_{x=-\sqrt{3}/2}\), and verify the sign using implicit differentiation.
The input is in \((-1,1)\), so substitute it into the formula:
The corresponding angle is \(y=\arccos(-\frac{\sqrt{3}}{2})=\frac{5\pi}{6}\), which is in the principal range \([0,\pi]\). Thus \(\sin(y)=\frac{1}{2}\). Substituting the proposed derivative into the differentiated identity gives
This agrees with the derivative of \(x\). The negative value also matches the fact that arccosine decreases as its input increases.
Why the Principal Range Matters
The equation \(\sin^2(y)=1-x^2\) determines the square of \(\sin(y)\), not its sign. Taking a square root without considering the range of \(y\) can therefore lead to an incorrect derivative. For arccosine, \(y\in[0,\pi]\), and sine is nonnegative throughout that interval and positive in its interior. That fact fixes the sign of \(\sin(y)\) for the derivative calculation.
The negative sign in the final result has two connected explanations. Algebraically, differentiating \(\cos(y)=x\) produces the factor \(-\sin(y)\), and \(\sin(y)\) is positive in the interior of the principal range. Geometrically, cosine decreases on \([0,\pi]\), so its inverse decreases as well.
Common Pitfall: Losing the Negative Sign
A frequent error is to differentiate \(\cos(y)=x\) as if \(y\) were independent of \(x\), or to omit the negative sign in the derivative of cosine. That can produce the incorrect equation \(\sin(y)\frac{dy}{dx}=1\), and hence a positive derivative.
For example, at \(x=\frac{3}{5}\), that incorrect equation would give \(\frac{dy}{dx}=\frac{1}{\sin(y)}=\frac{5}{4}\), because \(\sin(y)=\frac{4}{5}\). But substituting this proposed value into the correctly differentiated relation gives
not \(1\). The proposed derivative fails the check. The correct chain-rule equation is \(-\sin(y)\frac{dy}{dx}=1\), which gives \(\frac{dy}{dx}=-\frac{5}{4}\) at this input.
Another version of the same mistake is to write \(\sin(y)=\pm\sqrt{1-x^2}\) and leave the sign undecided. The principal range resolves it: for \(y\in(0,\pi)\), \(\sin(y)>0\). The derivative still has a negative sign because the minus sign comes from differentiating cosine.
Using the Result Correctly
When evaluating the derivative of arccosine at a particular input, first make sure the input lies in the open interval \((-1,1)\), where the formula applies. Then substitute the input into \(1-x^2\), simplify the square root, and keep the leading negative sign.
For the derivative formula, require \(-1<x<1\).
Evaluate \(-\frac{1}{\sqrt{1-x^2}}\), keeping the minus sign.
The result must be negative because arccosine is decreasing on its domain.
This formula describes the rate at which the output angle changes when the input cosine value changes. The magnitude \(\frac{1}{\sqrt{1-x^2}}\) depends on the input, while the negative sign records the decreasing relationship. The derivation relies on implicit differentiation, the chain rule, the identity \(\sin^2(y)+\cos^2(y)=1\), and the principal range of arccosine.
Check Your Understanding
Answer these questions using the defining equation \(\cos(y)=x\) and the derivative formula.
- What interval contains the values of \(y=\arccos(x)\)?
- Why is \(\sin(y)=\sqrt{1-x^2}\), rather than the negative square root, when \(-1<x<1\)?
- Differentiate \(\cos(y)=x\) implicitly and solve for \(\frac{dy}{dx}\) before rewriting the result in terms of \(x\).
- Find \(\left.\frac{d}{dx}\arccos(x)\right|_{x=0}\), and check your answer in the differentiated identity.
- Explain why the derivative must be negative on \((-1,1)\).
