The Slope That Runs Away At The Edges
Ask a computer to find the angle between two directions and it reaches for arccosine - but push the two directions almost into line and the answer becomes wildly sensitive, a hair's change in input swinging the angle by a lot. That sensitivity is exactly what the derivative of cos inverse measures, and its formula, $-\dfrac{1}{\sqrt{1 - x^{2}}}$, blows up as $x$ nears $\pm 1$. The edges are where arccosine is most delicate.
That is why the derivative is not a formula to memorise and forget: it tells engineers and graphics programmers precisely where an angle computed from a ratio can be trusted and where a tiny measurement error explodes into a large angular one.
What Is The Derivative Of Cos Inverse?
The derivative of cos inverse $x$ - the inverse cosine, also written $\arccos x$ - is:
$$\dfrac{d}{dx}\cos^{-1}x = -\dfrac{1}{\sqrt{1 - x^{2}}}, \quad \text{for } -1 < x < 1$$
Two features of this formula carry the whole meaning.
The sign is negative. Inverse cosine is a decreasing function: as its input $x$ rises from $-1$ to $1$, the output angle falls from $\pi$ to $0$. A decreasing function has a negative slope everywhere, which is where the minus sign comes from.
The domain excludes the endpoints. At $x = \pm 1$ the square root $\sqrt{1 - x^{2}}$ becomes $0$, so the fraction is undefined; the derivative only exists on the open interval $(-1, 1)$.
This is the mirror image of the derivative of inverse sine, $\dfrac{d}{dx}\sin^{-1}x = +\dfrac{1}{\sqrt{1 - x^{2}}}$ - same size, opposite sign. Inverse cosine belongs to the family of inverse trigonometric functions, and it undoes the cosine function: where cosine takes an angle to a ratio, $\arccos$ takes a ratio back to an angle.
Cos Inverse And Arccos - The Same Function, Two Names
Before the proof, one clarification that saves confusion. $\cos^{-1}x$ and $\arccos x$ are the same function - two notations for the inverse cosine, so they have the same derivative. This article leads with the "$\cos^{-1}$" notation you meet first in school; the companion derivative of arccos article leads with the "arccos" naming used in higher courses and reference tables. Same result, $-\dfrac{1}{\sqrt{1 - x^{2}}}$, reached the same way - pick whichever notation your course uses.
One caution: $\cos^{-1}x$ (the inverse function) is not $\dfrac{1}{\cos x}$ (the reciprocal, which is secant). The "$-1$" is an inverse-function label, not an exponent.
How Do You Prove The Derivative Of Cos Inverse?
The cleanest proof uses implicit differentiation, and it needs one setup term first. The principal branch of arccosine is the slice of output angles that makes it a genuine function: $\arccos x$ returns an angle $y$ in $[0, \pi]$. That range matters, because it decides a sign later.
Start by naming the angle.
$$\text{Let } y = \cos^{-1}x, \quad \text{so } \cos y = x, \quad \text{with } y \in [0, \pi]$$
Differentiate both sides with respect to $x$, remembering the chain rule on the left:
$$-\sin y \times \dfrac{dy}{dx} = 1$$
$$\dfrac{dy}{dx} = -\dfrac{1}{\sin y}$$
Now replace $\sin y$ with something in $x$. From the Pythagorean identity $\sin^{2}y + \cos^{2}y = 1$:
$$\sin y = \sqrt{1 - \cos^{2}y} = \sqrt{1 - x^{2}}$$
The positive root is the correct choice because $y$ lies in $[0, \pi]$, the principal branch, where sine is never negative. Substitute:
$$\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^{2}}}$$
The principal branch is not a footnote - it is the reason the root is positive, which is the reason the derivative is cleanly negative rather than ambiguous in sign.
How Does The Derivative Of Cos Inverse Relate To Sin Inverse?
There is a one-line confirmation using the identity $\cos^{-1}x + \sin^{-1}x = \dfrac{\pi}{2}$. Differentiating both sides, the constant $\dfrac{\pi}{2}$ has derivative $0$:
$$\dfrac{d}{dx}\cos^{-1}x + \dfrac{d}{dx}\sin^{-1}x = 0$$
$$\dfrac{d}{dx}\cos^{-1}x = -\dfrac{d}{dx}\sin^{-1}x = -\dfrac{1}{\sqrt{1 - x^{2}}}$$
The two inverse derivatives are exact negatives of each other, which is a memorable way to recover the sign if it ever slips.
Examples Of The Derivative Of Cos Inverse
Example 1
Differentiate $y = \cos^{-1}x$ and state its value at $x = 0$.
The derivative is $-\dfrac{1}{\sqrt{1 - x^{2}}}$. At $x = 0$:
$$-\dfrac{1}{\sqrt{1 - 0}} = -\dfrac{1}{1} = -1$$
Final answer: the slope at $x = 0$ is $-1$.
Example 2
Differentiate $y = \cos^{-1}x$. First instinct, then the correct route.
The tempting move is to recall that inverse-trig derivatives look like $\dfrac{1}{\sqrt{1 - x^{2}}}$ and write it with a plus sign.
Check it against the shape of the function. Arccosine decreases - its graph falls from $\pi$ to $0$ - so its slope must be negative everywhere. A positive derivative would describe a rising function, which contradicts the graph. So the plus sign is wrong.
The rescue is the sign that the proof forces: $-\sin y$ appears when cosine is differentiated, carrying a minus into the final result.
$$\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^{2}}}$$
Final answer: $\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - x^{2}}}$.
Example 3
Differentiate $y = \cos^{-1}(2x)$ using the chain rule.
The outer derivative is $-\dfrac{1}{\sqrt{1 - u^{2}}}$ with $u = 2x$, and the inner derivative is $2$:
$$\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - (2x)^{2}}} \times 2 = -\dfrac{2}{\sqrt{1 - 4x^{2}}}$$
Final answer: $-\dfrac{2}{\sqrt{1 - 4x^{2}}}$.
Example 4
Find $\dfrac{d}{dx}\left(\cos^{-1}x + \sin^{-1}x\right)$.
The two derivatives are exact opposites, so they cancel:
$$-\dfrac{1}{\sqrt{1 - x^{2}}} + \dfrac{1}{\sqrt{1 - x^{2}}} = 0$$
Final answer: $0$ - which confirms the identity $\cos^{-1}x + \sin^{-1}x = \dfrac{\pi}{2}$ is constant.
Example 5
Differentiate $y = \cos^{-1}(x^{2})$.
Inner function $u = x^{2}$, inner derivative $2x$:
$$\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{1 - (x^{2})^{2}}} \times 2x = -\dfrac{2x}{\sqrt{1 - x^{4}}}$$
Final answer: $-\dfrac{2x}{\sqrt{1 - x^{4}}}$.
Example 6
Evaluate the derivative of $\cos^{-1}x$ at $x = \dfrac{1}{2}$.
Substitute $x = \dfrac{1}{2}$ into $-\dfrac{1}{\sqrt{1 - x^{2}}}$:
$$-\dfrac{1}{\sqrt{1 - \frac{1}{4}}} = -\dfrac{1}{\sqrt{\frac{3}{4}}} = -\dfrac{1}{\frac{\sqrt{3}}{2}} = -\dfrac{2}{\sqrt{3}}$$
Final answer: $-\dfrac{2}{\sqrt{3}}$.
The first-instinct error students reach for across these is copying the inverse-sine derivative and keeping its plus sign - the $\sqrt{1 - x^{2}}$ denominator is remembered correctly, but the minus that marks arccosine as a decreasing function is dropped.
Why The Derivative Of Cos Inverse Matters - "How sensitive an angle is to its ratio"
A derivative measures sensitivity, and the derivative of cos inverse answers a precise question: if you recover an angle from a cosine ratio, how much does the angle move when the ratio moves? That is its fundamental utility.
Angle from a ratio. Whenever a quantity is known as a cosine value - the alignment of two vectors, a phase, a direction - arccosine recovers the angle, and its derivative says how trustworthy that recovery is.
The endpoints are fragile. Because the derivative runs to $-\infty$ as $x \to \pm 1$, a small error in a ratio near $\pm 1$ produces a large error in the angle. This is a real precision limit in numerical computation, not an abstract curiosity.
The middle is stable. Near $x = 0$ the derivative is only $-1$, so the angle responds gently to changes in the ratio - the region where arccosine is safest to use.
What competitor explainers usually skip is why the derivative and the graph agree. Arccosine's graph is steepest at its ends and gentlest in the middle, and the derivative's size - small near $x = 0$, huge near $x = \pm 1$ - is exactly that steepness written as a formula. The blow-up at the edges is not a defect; it is the graph standing vertical.
Common Mistakes With The Derivative Of Cos Inverse
Mistake 1: Keeping the plus sign
Where it slips in: Any time the inverse-sine and inverse-cosine derivatives are recalled together.
Don't do this: Writing $\dfrac{d}{dx}\cos^{-1}x = +\dfrac{1}{\sqrt{1 - x^{2}}}$.
The correct way: Arccosine decreases, so its derivative is negative: $-\dfrac{1}{\sqrt{1 - x^{2}}}$. The plus version is the inverse sine derivative.
The memorizer who stores both formulas as "one over root one minus x squared" loses track of which one carries the minus.
Mistake 2: Reading $\cos^{-1}x$ as $\dfrac{1}{\cos x}$
Where it slips in: Interpreting the "$-1$" superscript as a power.
Don't do this: Treating $\cos^{-1}x$ as $\sec x$ and differentiating that instead.
The correct way: $\cos^{-1}x$ is the inverse function (the angle whose cosine is $x$), not the reciprocal. Its derivative is $-\dfrac{1}{\sqrt{1 - x^{2}}}$, nothing to do with secant.
Mistake 3: Forgetting the chain rule on a composite
Where it slips in: Differentiating $\cos^{-1}$ of something more complicated than plain $x$.
Don't do this: Writing $\dfrac{d}{dx}\cos^{-1}(2x) = -\dfrac{1}{\sqrt{1 - 4x^{2}}}$ without the inner factor.
The correct way: Multiply by the inner derivative. Since $\dfrac{d}{dx}(2x) = 2$, the answer is $-\dfrac{2}{\sqrt{1 - 4x^{2}}}$.
Key Takeaways
The derivative of $\cos^{-1}x$ is $-\dfrac{1}{\sqrt{1 - x^{2}}}$, defined only on the open interval $(-1, 1)$.
The negative sign comes from arccosine being a decreasing function, confirmed by implicit differentiation.
The principal branch $[0, \pi]$ is why $\sin y = +\sqrt{1 - x^{2}}$ — the positive root is not optional.
$\cos^{-1}x$ is the inverse, not the reciprocal; it is not $\sec x$.
Its derivative is the exact negative of the inverse-sine derivative.
To master inverse-trig differentiation with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for calculus, or live math classes online with peers from 20+ countries.
A Practical Next Step
Practice these to solidify your understanding: differentiate $\cos^{-1}(3x)$, then $\cos^{-1}(x^{3})$, then evaluate the derivative of $\cos^{-1}x$ at $x = \dfrac{\sqrt{3}}{2}$, checking the sign is negative each time. If the sign or the inner factor slips, come back to the proof above. Want a live Bhanzu trainer to check your inverse-trig work? Book a free demo class.
Read More
Arccosine — the inverse cosine function this derivative acts on.
Inverse Trigonometric Ratios — how the inverse functions are defined and bounded.
Differentiation of Trigonometric Functions — the full derivative rule set.
Cofunction Identities — the complementary relationship behind the $\dfrac{\pi}{2}$ identity.
Sin Cos Tan — the primary ratios the inverses reverse.
Was this article helpful?
Your feedback helps us write better content