Answer
$y'=\frac{ln5(log_{5}x)+1}{(xlog_{5}x)ln2ln5}$
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:
Apply product rule.
$y'=\frac{1}{(xlog_{5}x)ln2}[x\frac{d}{dx}(log_{5}x)+(log_{5}x)\frac{d}{dx}(x)$
$=\frac{1}{(xlog_{5}x)ln2}[x\times\frac{1}{xln5}+(log_{5}x)(1)]$
$=\frac{1}{(xlog_{5}x)ln2}[\frac{1}{ln5}+(log_{5}x)]$
$y'=\frac{ln5(log_{5}x)+1}{(xlog_{5}x)ln2ln5}$
Hence, $y'=\frac{ln5(log_{5}x)+1}{(xlog_{5}x)ln2ln5}$
