The derivative of $\cos^{-1}(x)$ (also written as $\arccos(x)$) with respect to $x$ is:

$\frac{d}{dx} \left(\cos^{-1}(x)\right) = -\frac{1}{\sqrt{1 - x^2}}$

Explanation:

The derivative is derived using the chain rule and the identity that relates the derivative of the inverse trigonometric functions. Specifically:

1. Start with $y = \cos^{-1}(x)$, which implies $x = \cos(y)$.
2. Differentiate both sides with respect to $x$: $\frac{d}{dx}(x) = \frac{d}{dx}(\cos(y))$ $1 = -\sin(y) \cdot \frac{dy}{dx}$
3. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{1}{\sin(y)}$
4. Since $\sin(y) = \sqrt{1 - \cos^2(y)} = \sqrt{1 - x^2}$, we have: $\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}$

Thus, the derivative of $\cos^{-1}(x)$ is $-\frac{1}{\sqrt{1 - x^2}}$.

Would you like more details or have any questions?

Here are 5 related questions to explore further:

1. What is the derivative of $\sin^{-1}(x)$?
2. How do you find the derivative of $\tan^{-1}(x)$?
3. What is the second derivative of $\cos^{-1}(x)$?
4. How is the chain rule applied in finding derivatives of inverse trigonometric functions?
5. How does the domain of $\cos^{-1}(x)$ affect its derivative?

Tip: Remember that the domain of $\cos^{-1}(x)$ is $-1 \leq x \leq 1$, which means the derivative is only defined within this interval.