## Thomas' Calculus 13th Edition

$$\frac{{dy}}{{dx}} = \frac{1}{{\ln 25}}\left( {1 - \frac{1}{x}} \right)$$
\eqalign{ & y = {\log _{25}}{e^x} - {\log _5}\sqrt x \cr & {\text{write as}} \cr & y = {\log _{25}}{e^x} - {\log _5}{x^{1/2}} \cr & {\text{use the logarithmic property }}{\log _a}{b^n} = n{\log _a}b \cr & y = x{\log _{25}}e - \frac{1}{2}{\log _5}x \cr & {\text{Find the derivative of }}y{\text{ with respect to }}x \cr & \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x{{\log }_{25}}e} \right] - \frac{d}{{dx}}\left[ {\frac{1}{2}{{\log }_5}x} \right] \cr & \frac{{dy}}{{dx}} = {\log _{25}}e\frac{d}{{dx}}\left[ x \right] - \frac{1}{2}\frac{d}{{dx}}\left[ {{{\log }_5}x} \right] \cr & {\text{use the rule }}\frac{d}{{dx}}\left[ {{{\log }_a}x} \right] = \frac{1}{{\left( {\ln a} \right)x}}.{\text{ Then}} \cr & \frac{{dy}}{{dx}} = {\log _{25}}e\left( 1 \right) - \frac{1}{2}\left( {\frac{1}{{\left( {\ln 5} \right)x}}} \right) \cr & {\text{simplifying}} \cr & \frac{{dy}}{{dx}} = {\log _{25}}e - \frac{1}{{2\left( {\ln 5} \right)x}} \cr & \frac{{dy}}{{dx}} = {\log _{25}}e - \frac{1}{{\left( {\ln 25} \right)x}} \cr & {\text{Use the logarithmic property }}{\log _a}x = \frac{{\ln x}}{{\ln a}} \cr & \frac{{dy}}{{dx}} = \frac{{\ln e}}{{\ln 25}} - \frac{1}{{\left( {\ln 25} \right)x}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{\ln 25}} - \frac{1}{{\left( {\ln 25} \right)x}} \cr & \frac{{dy}}{{dx}} = \frac{1}{{\ln 25}}\left( {1 - \frac{1}{x}} \right) \cr}