1. **State the problem:** Find the derivative of $A(x)$ at $x=\pi$, where $$A(x) = \int_2^x \frac{\cos t}{1+t} \, dt.$$ 2. **Recall the Fundamental Theorem of Calculus:** If $$F(x) = \int_a^x f(t) \, dt,$$ then $$F'(x) = f(x).$$ This means the derivative of an integral with a variable upper limit is just the integrand evaluated at that upper limit. 3. **Apply the theorem:** Here, $$A'(x) = \frac{\cos x}{1+x}.$$ 4. **Evaluate at $x=\pi$:** $$A'(\pi) = \frac{\cos \pi}{1 + \pi}.$$ 5. **Simplify:** Since $\cos \pi = -1$, $$A'(\pi) = \frac{-1}{1 + \pi}.$$ **Final answer:** $$\boxed{A'(\pi) = \frac{-1}{1 + \pi}}.$$