1. The problem is to find the derivative of the function $$y = \cos(\sin x)$$ with respect to $$x$$. 2. We use the chain rule for differentiation, which states that if $$y = f(g(x))$$, then $$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$$. 3. Here, $$f(u) = \cos u$$ and $$g(x) = \sin x$$. 4. The derivative of $$f(u) = \cos u$$ is $$f'(u) = -\sin u$$. 5. The derivative of $$g(x) = \sin x$$ is $$g'(x) = \cos x$$. 6. Applying the chain rule: $$\frac{dy}{dx} = -\sin(\sin x) \cdot \cos x$$ 7. Therefore, the derivative is $$-\cos x \sin(\sin x)$$. This matches the first option given.