$12.39 f(x)=\frac{\cos x}{1+\sin x} ; f^{\prime}\left(\frac{\pi}{2}\right)=$ ?
12.39 We have

\[
\begin{aligned}
& f^{\prime}(x)=\frac{-\sin x(1+\sin x)-\cos x \cdot \cos x}{(1+\sin x)^{2}}= \\
& =\frac{-\sin x-\sin ^{2} x-\cos ^{2} x}{(1+\sin x)^{2}}=-\frac{1+\sin x}{(1+\sin x)^{2}}=-\frac{1}{1+\sin x} \\
& f^{\prime}\left(\frac{\pi}{2}\right)=-\frac{1}{2}
\end{aligned}
\]

Answer: $-0.5$.