Answer
$y'= \frac{ \ln x+1}{x \ln x \ln 2} $
Work Step by Step
The derivative of $y$ is:
$$y'=\frac{(\ln(x \log_{5}(x)))'}{\ln 2} $$
Using the chain rule it follows:
$$y'=\frac{(x \log_{5}(x))'\cdot\frac{1}{x \log_{5}(x)}}{\ln 2} $$
$$y'=\frac{((x)' \log_{5}(x)+x(\log_{5}(x))')\cdot\frac{1}{x \log_{5}(x)}}{\ln 2} $$
$$y'=\frac{( \log_{5}(x)+x\frac{1}{x\ln 5})\cdot\frac{1}{x \log_{5}(x)}}{\ln 2} $$
$$y'=\frac{( \log_{5}(x)+\frac{1}{\ln 5})\cdot\frac{1}{x \log_{5}(x)}}{\ln 2} $$
$$y'=\frac{ \frac{1}{x}+\frac{1}{\ln 5\cdot x \log_{5}(x)}}{\ln 2} $$
$$y'=\frac{ \frac{1}{x}+\frac{1}{\ln 5\cdot x \frac{\ln x}{\ln 5}}}{\ln 2} $$
$$y'=\frac{ \frac{1}{x}+\frac{1}{x \ln x}}{\ln 2} $$
$$y'= \frac{ \ln x+1}{x \ln x \ln 2} $$
