Derivative of arccos x: Formula, Proof and Examples

The derivative of \arccos x is -\dfrac{1}{\sqrt{1 - x^2}}. It looks just like the derivative of arcsin x with a minus sign in front, and there is a neat reason for that.

Graph of arccos x and its derivative

Proof of the derivative of arccos x

Let y = \arccos x, so \cos y = x with 0 \le y \le \pi. Use implicit differentiation:

On [0, \pi] sine is never negative, so \sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}. That gives

A faster proof using arcsin

For every x in [-1, 1],

Picture a right triangle: if one acute angle has sine x, the other has cosine x, and the two angles add to \frac{\pi}{2}. Differentiate both sides. The right side is a constant, so its derivative is 0:

So the arccos derivative is the arcsin derivative with the sign flipped. This is a good way to remember it.

Chain rule version: derivative of arccos(u)

Worked examples

Example 1: product rule. For f(x) = x^2\arccos x, the product rule gives

Example 2: slope at a point. At x = \frac{\sqrt{3}}{2}, the slope of \arccos x is

Example 3: tangent line. At x = 0 the curve passes through \left(0, \frac{\pi}{2}\right) with slope -1, so the tangent line is y = \frac{\pi}{2} - x.

Common mistakes

1. Dropping the minus sign. It is the only difference from arcsin, and the easiest thing to lose.
2. Thinking \cos^{-1} x = \sec x. The reciprocal of cosine is secant. \cos^{-1} is the inverse function.
3. Leaving out u'. \frac{d}{dx}\arccos(5x) = -\frac{5}{\sqrt{1 - 25x^2}}, not -\frac{1}{\sqrt{1 - 25x^2}}.
4. Using it at x = \pm 1. The tangent is vertical there, so there is no derivative.

You will rarely see -\frac{1}{\sqrt{1 - x^2}} integrated as arccos, since -\arcsin x + C is the same family of answers. See the arccosine function page for the graph and range, and practise with inverse trig derivative problems.