Answer
$$y'=\frac{x-1}{2x\ln5}$$
Work Step by Step
$$y=\log_{25}e^x-\log_5\sqrt x$$
Recall Theorem 7: $$\frac{d}{dx}\log_au=\frac{1}{u\ln a}\frac{du}{dx}$$
That means: $$(\log_{25}e^x)'=\frac{1}{e^x\ln25}(e^x)'=\frac{e^x}{e^x\ln25}=\frac{1}{\ln25}=\frac{1}{\ln5^2}=\frac{1}{2\ln5}$$ $$(\log_5\sqrt x)'=\frac{1}{\sqrt x\ln5}(\sqrt x)'=\frac{1}{\sqrt x\ln5}\frac{1}{2\sqrt x}=\frac{1}{2x\ln5}$$
Apply here to find $y'$: $$y'=(\log_{25}e^x)'-(\log_5\sqrt x)'=\frac{1}{2\ln5}-\frac{1}{2x\ln5}$$
$$y'=\frac{x-1}{2x\ln5}$$
