1. **Problem statement:** Find the derivative of the function $$y = \arccos\left(\sin\left(e^x\right)\right)$$. 2. **Recall the chain rule:** The derivative of $$\arccos u$$ with respect to $$x$$ is $$\frac{dy}{dx} = -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}$$. 3. **Identify the inner function:** Here, $$u = \sin\left(e^x\right)$$. 4. **Find the derivative of $$u$$:** $$\frac{du}{dx} = \cos\left(e^x\right) \cdot \frac{d}{dx} e^x = \cos\left(e^x\right) \cdot e^x$$. 5. **Apply the chain rule:** $$\frac{dy}{dx} = -\frac{1}{\sqrt{1-\sin^2\left(e^x\right)}} \cdot \cos\left(e^x\right) e^x$$. 6. **Simplify inside the square root:** Recall $$1-\sin^2(\theta) = \cos^2(\theta)$$, so $$\sqrt{1-\sin^2\left(e^x\right)} = \sqrt{\cos^2\left(e^x\right)} = |\cos\left(e^x\right)|$$. 7. **Final simplified derivative:** $$\frac{dy}{dx} = -\frac{\cos\left(e^x\right) e^x}{|\cos\left(e^x\right)|}$$. This derivative can be written as $$\boxed{\frac{dy}{dx} = -e^x \cdot \mathrm{sgn}\left(\cos\left(e^x\right)\right)}$$, where $$\mathrm{sgn}$$ denotes the sign function capturing the absolute value division.