1. The problem is to find the derivative of the function $y = x^{\cos^{-1} x}$. 2. We use the formula for the derivative of $y = f(x)^{g(x)}$: $$\frac{dy}{dx} = y \left(g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)}\right)$$ where $f(x) = x$ and $g(x) = \cos^{-1} x$. 3. Calculate the derivatives: - $f'(x) = 1$ - $g'(x) = \frac{d}{dx} \cos^{-1} x = -\frac{1}{\sqrt{1-x^2}}$ 4. Substitute into the formula: $$\frac{dy}{dx} = x^{\cos^{-1} x} \left(-\frac{1}{\sqrt{1-x^2}} \ln x + \cos^{-1} x \frac{1}{x}\right)$$ 5. This is the derivative of the given function. Final answer: $$\boxed{\frac{dy}{dx} = x^{\cos^{-1} x} \left(\frac{\cos^{-1} x}{x} - \frac{\ln x}{\sqrt{1-x^2}}\right)}$$