1. **State the problem:** Find the limit $$\lim_{x \to \pi} \frac{f'(x) - f'(\pi)}{\pi - x}$$ where $$f(x) = e^x \cos x + \sin x$$. 2. **Recall the formula:** This limit resembles the definition of the derivative of $$f'(x)$$ at $$x=\pi$$, but with a negative sign in the denominator. Specifically, $$\lim_{x \to a} \frac{g(x) - g(a)}{a - x} = -g'(a)$$ for a function $$g$$. Here, $$g(x) = f'(x)$$ and $$a = \pi$$. 3. **Find $$f'(x)$$:** Using the product rule and derivatives, $$f(x) = e^x \cos x + \sin x$$ $$f'(x) = \frac{d}{dx}(e^x \cos x) + \frac{d}{dx}(\sin x)$$ $$= e^x \cos x - e^x \sin x + \cos x$$ 4. **Find $$f'(\pi)$$:** Evaluate at $$x=\pi$$: $$f'(\pi) = e^{\pi} \cos \pi - e^{\pi} \sin \pi + \cos \pi$$ Since $$\cos \pi = -1$$ and $$\sin \pi = 0$$, $$f'(\pi) = e^{\pi}(-1) - e^{\pi}(0) + (-1) = -e^{\pi} - 1$$ 5. **Define $$g(x) = f'(x)$$ and find $$g'(x)$$:** $$g(x) = e^x \cos x - e^x \sin x + \cos x$$ Differentiate term by term: $$g'(x) = \frac{d}{dx}(e^x \cos x) - \frac{d}{dx}(e^x \sin x) - \sin x$$ Calculate each derivative: $$\frac{d}{dx}(e^x \cos x) = e^x \cos x - e^x \sin x$$ $$\frac{d}{dx}(e^x \sin x) = e^x \sin x + e^x \cos x$$ So, $$g'(x) = (e^x \cos x - e^x \sin x) - (e^x \sin x + e^x \cos x) - \sin x$$ Simplify: $$g'(x) = e^x \cos x - e^x \sin x - e^x \sin x - e^x \cos x - \sin x = -2 e^x \sin x - \sin x$$ 6. **Evaluate $$g'(\pi)$$:** $$g'(\pi) = -2 e^{\pi} \sin \pi - \sin \pi = 0$$ since $$\sin \pi = 0$$. 7. **Calculate the limit:** Using the limit definition, $$\lim_{x \to \pi} \frac{f'(x) - f'(\pi)}{\pi - x} = \lim_{x \to \pi} \frac{g(x) - g(\pi)}{\pi - x} = -g'(\pi) = -0 = 0$$ **Final answer:** $$\boxed{0}$$