12.7 $y=\frac{1-\cos 2 x}{1+\cos 2 x}$.

Calculate the values of the derivatives of the given functions at the specified values of the independent variable (12.8-12.9):
12.7 Let's use the quotient rule of differentiation (12.4) and the table of derivatives:

$$
\begin{aligned}
& y^{\prime}=\left(\frac{1-\cos 2 x}{1+\cos 2 x}\right)^{\prime}= \\
& =\frac{(1-\cos 2 x)^{\prime}(1+\cos 2 x)-(1-\cos 2 x)(1+\cos 2 x)^{\prime}}{(1+\cos 2 x)^{2}}= \\
& =\frac{-(-2 \sin 2 x)(1+\cos 2 x)-(1-\cos 2 x)(-2 \sin 2 x)}{(1+\cos 2 x)^{2}}=
\end{aligned}
$$

$$
=\frac{2 \sin 2 x(1+\cos 2 x+1-\cos 2 x)}{(1+\cos 2 x)^{2}}=\frac{4 \sin 2 x}{(1+\cos 2 x)^{2}}
$$