Derivative of Arccos x — Formula, Proof, Examples
Derivative of Arccos x — Formula, Proof, Examples
TL;DR
The derivative of arccos x is [ \dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}} ] on the open interval (-1, 1) — a negative quantity that reflects the fact arccosine is a strictly decreasing function. This article gives the implicit-differentiation derivation, the first-principles approach, the chain-rule version [ \dfrac{d}{dx}\arccos(u) = -\dfrac{u'}{\sqrt{1-u^2}} ], three worked examples, and the common mistakes around sign and domain.
The One Negative Sign Engineers Track Across Every Inverse-Trig Derivative
For -1 < x < 1, the derivative of arccos x is:
[ \dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}}. ]
The derivative is not defined at the endpoints x = ±1 — the square root in the denominator collapses to zero there, and the slope of the arccos graph goes vertical.
Derivative of Arccos x Formula
[ \boxed{\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}}, \quad x \in (-1, 1).} ]
The chain-rule version handles any composition. If u = u(x) is a differentiable function with (u(x)\in(-1,1)):
[ \dfrac{d}{dx}\arccos(u(x)) = -\dfrac{u'(x)}{\sqrt{1 - u(x)^2}}. ]
Quick facts.
- Domain of the derivative: (-1, 1) — open interval; the endpoints are excluded because the derivative blows up.
- Sign: strictly negative on the entire domain. Confirms arccos is decreasing.
- Minimum slope: -1, reached at x = 0 (where ( \arccos 0 = \pi/2 )).
- Asymptotic behaviour: (|f'(x)| \to \infty) as x \to ±1.
- Companion fact: ( \dfrac{d}{dx}\arcsin x = +\dfrac{1}{\sqrt{1-x^2}} ) — same magnitude, opposite sign. Consistent with ( \arcsin x + \arccos x = \pi/2 ).
Double-Anchoring — Triangle View and Unit-Circle View
The derivation depends on a triangle / unit-circle identity.
From the right triangle. Let θ = arccos x. Build a right triangle with adjacent leg x and hypotenuse 1. The opposite leg is ( \sqrt{1 - x^2} ) by the Pythagorean theorem. So ( \sin θ = \sqrt{1 - x^2} ) — the very quantity that appears in the denominator of the derivative.
From the unit circle. With θ = arccos x ∈ [0, π], the unit-circle point is (x, ( \sqrt{1-x^2} )). The y-coordinate is non-negative. Both views deliver the same ( \sqrt{1-x^2} ), and both make the negative sign in the derivative inevitable.
Derivation 1 — Implicit Differentiation (the Two-Line Proof)
Set y = arccos x. Then x = cos y where y ∈ [0, π].
Differentiate both sides with respect to x:
1 = -sin y ( \dfrac{dy}{dx} ).
Solve for ( \dfrac{dy}{dx} ):
[ \dfrac{dy}{dx} = -\dfrac{1}{\sin y}. ]
Now express sin y in terms of x:
( \sin y = \sqrt{1 - x^2} ) (since y ∈ [0, π], we take the non-negative root).
Substitute:
[ \dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1 - x^2}}. ]
Derivation 2 — First Principles (the Limit Definition)
[ f'(x) = \lim_{h \to 0} \dfrac{\arccos(x+h) - \arccos x}{h}. ]
Substituting and simplifying gives:
[ f'(x) = -\dfrac{1}{\sqrt{1-x^2}}. ]
Worked Examples of Derivative of Arccos x
Example 1. Find ( \dfrac{d}{dx}\arccos x ) at x = 1/2.
Substituting into the formula:
[ \dfrac{d}{dx}\arccos x \bigg|_{x=1/2} = -\dfrac{1}{\sqrt{1 - (1/2)^2}} = -\dfrac{2}{\sqrt{3}}. ]
Final answer: Approximately -1.155.
Example 2. Differentiate ( f(x) = \arccos(3x) ).
Using the chain rule:
[ f'(x) = -\dfrac{3}{\sqrt{1 - 9x^2}}. ]
Domain: ( -\frac{1}{3} < x < \frac{1}{3} ).
Conclusion
- The derivative of arccos x is (-\dfrac{1}{\sqrt{1-x^2}}) on (-1, 1) — undefined at the endpoints.
- The negative sign arises from the decreasing nature of arccos.
- Chain-rule version is given as (-u' \sqrt{1-u^2}).
- Common mistake is to omit the chain-rule factor when differentiating.
Frequently Asked Questions
What's the derivative of arccos x? ( \dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}} ) on (-1, 1).
Why is the derivative of arccos negative? Because arccos is decreasing from π to 0 as x increases from -1 to 1.
How is the derivative of arccos related to arcsin? Their sum is zero, consistent with ( \arcsin x + \arccos x = \pi/2 ).
What’s the derivative of arccos at x=0? It's -1.
Can I differentiate arccos at x=1? No, the derivative is undefined.