## Calculus (3rd Edition)

$-\frac{1}{1+\pi}$
By making use of Theorem 1 (FTC II), we have $$A'(x)=\frac{d}{d x} \int_{2}^{x} \frac{\cos t}{1+t} d t=\frac{\cos x}{1+x}$$ Hence $$A'(\pi)= \frac{\cos \pi}{1+\pi} =-\frac{1}{1+\pi}$$