Answer
$y'=\frac{1+lnx}{{x(lnx)(ln2)}}$
Work Step by Step
If we first simplify the given function using the properties of logarithms $\frac{d}{dx}(log_{e}(x))=\frac{1}{xlne}$
Then the differentiation becomes easier:
$y'=\frac{1}{(xlog_{5}x)ln2}[x\frac{d}{dx}(log_{5}x)+(log_{5}x)\frac{d}{dx}(x)$
$=\frac{1}{(xlog_{5}x)ln2}[\frac{1}{ln5}+(log_{5}x)]$
Re-arrange the above expression.
$y'=\frac{ln5}{xlnxln2}[\frac{1+lnx}{ln5}]$
Hence, $y'=\frac{1+lnx}{{x(lnx)(ln2)}}$
