What is the Derivative of arccos(x)?

Derivative of arccos(x)

The derivative of

\arccos(x)

-\frac{1}{\sqrt{1 - x^2}}

.

Solution. Let

y = \cos^{-1}(x)

. Then

x = \cos(y)

and

0 \leq y \leq \pi

. We want to differentiate with respect to

x

\begin{align*} \frac{d}{dx} x = \frac{d}{dx} \cos(y) &\iff 1 = \frac{d(\cos(y))}{dy}\frac{dy}{dx} \\ &\iff 1 = -\sin(y) \frac{dy}{dx} \\ &\iff \frac{dy}{dx} = -\frac{1}{\sin(y)}, \end{align*}

where we have seen here that

\frac{d(\cos(y))}{dy} = -\sin(y)

. We have that

0 \leq y \leq \pi

and therefore

\sin(y) \geq 1

. So we can apply the next identity:

\begin{align*} \sin^2(y) + \cos^2(y) = 1 &\iff \sin^2(y) = 1 - \cos^2(y) \\ &\iff \sin(y) = \sqrt{1 - \cos^2(y)} \\ &\iff \sin(y) = \sqrt{1 - x^2} \end{align*}

where

\cos^2(y) = x^2

since

\cos(y) = x

. Combining everything, we get that the derivative of

\arccos(x)

is:

\begin{align*} \frac{d}{dx} \arccos(x) = \frac{d}{dx} \cos^{-1}(x) = \frac{dy}{dx} = -\frac{1}{\sin(y)} = -\frac{1}{\sqrt{1-x^2}}, \quad x \in (-1,1). \end{align*}

So, the derivative of

\arccos(x)

-\frac{1}{\sqrt{1-x^2}}