## Calculus 8th Edition

$y'=\frac{1+lnx}{{x(lnx)(ln2)}}$
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, $y'=\frac{1+lnx}{{x(lnx)(ln2)}}$