Derivative of arccos(x)
The derivative of the inverse cosine is $\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}}$ for $-1 < x < 1$ – exactly the derivative of arcsin with a minus sign.
See the derivative of arccos(x) step by step in the calculator →
📈 Graph of arccos(x) and its derivative
🔍 Why is the derivative of arccos(x) equal to $-\frac{1}{\sqrt{1-x^2}}$?
Short proof: for all $x$ in $[-1,1]$, $\arccos x = \frac{\pi}{2} - \arcsin x$. The constant $\frac{\pi}{2}$ has derivative $0$, so $(\arccos x)' = -(\arcsin x)' = -\frac{1}{\sqrt{1-x^2}}$.
Direct proof: with $y = \arccos x$ we have $\cos y = x$, $0 \le y \le \pi$. Then $-\sin y\cdot y' = 1$ and
where $\sin y \ge 0$ because $y$ lies between $0$ and $\pi$.
🔗 Derivative of arccos(x) with an inner function
If the argument is a function $u(x)$, the chain rule adds the inner derivative:
✏️ Worked examples
| Function | Derivative | Note | |
|---|---|---|---|
| $\arccos(2x)$ | $-\frac{2}{\sqrt{1-4x^2}}$ | Valid for $|x| < \frac12$. | steps → |
| $\arccos(x^2)$ | $-\frac{2x}{\sqrt{1-x^4}}$ | $u = x^2$, $u^2 = x^4$. | steps → |
| $\arccos\left(\frac{1}{x}\right)$ | $\frac{1}{|x|\sqrt{x^2-1}}$ | For $|x| > 1$; this is the derivative of $\operatorname{arcsec} x$, because $\operatorname{arcsec}x = \arccos\frac1x$. | steps → |
| $\arccos x + \arcsin x$ | $0$ | The sum is the constant $\frac{\pi}{2}$. | steps → |
⚠️ Common mistakes
Dropping the minus sign – arccos is a decreasing function.
Using the formula at $x = \pm 1$, where the derivative does not exist.
❓ Frequently asked questions
What is the derivative of cos^-1(x)?
$\cos^{-1}x$ is $\arccos x$, so the derivative is $-\frac{1}{\sqrt{1-x^2}}$.
Why do arcsin and arccos have opposite derivatives?
Because $\arcsin x + \arccos x = \frac{\pi}{2}$ is constant, the two derivatives must add up to $0$.
What is the derivative of arccos(2x)?
$-\frac{2}{\sqrt{1-4x^2}}$.