\[\boxed{\mathbf{55}\mathbf{.}}\]
\[\textbf{а)}\ f(x) = \log_{4}{12x} - \log_{2}x;\ \ \]
\[x > 0\]
\[f^{'}(x) = \frac{12}{12x \cdot \ln 4} - \frac{1}{x \cdot \ln 2} =\]
\[= \frac{1}{x \cdot \ln 4} - \frac{1}{x \cdot \ln 2} =\]
\[= \frac{1}{2x \cdot \ln 2} - \frac{1}{x \cdot \ln 2} =\]
\[= \frac{1 - 2}{2x \cdot \ln 2} = - \frac{1}{2x \cdot \ln 2}.\]
\[\textbf{б)}\ f(x) = \log_{4}( - x) +\]
\[+ \log_{2}{( - x)};\ \ x < 0\]
\[f^{'}(x) = \frac{- 1}{- x \cdot \ln 4} + \frac{- 1}{- x \cdot \ln 2} =\]
\[= \frac{1}{x \cdot \ln 4} + \frac{1}{x \cdot \ln 2} =\]
\[= \frac{1}{2x \cdot \ln 2} + \frac{1}{x \cdot \ln 2} =\]
\[= \frac{1 + 2}{2x \cdot \ln 2} = \frac{3}{2x \cdot \ln 2}.\]
\[\textbf{в)}\ f(x) = \ln{2x};\ \ x > 0\]
\[f^{'}(x) = \frac{2}{2x} = \frac{1}{x}.\]
\[\textbf{г)}\ f(x) = \ln(5x - 10);\ \ x > 2;\]
\[f^{'}(x) = \frac{1}{5x - 10} \cdot (5x - 10)^{'} =\]
\[= \frac{5}{5x - 10} = \frac{1}{x - 2}\text{.\ }\]