## Calculus: Early Transcendentals 8th Edition

$y'=\frac{ln5(log_{5}x)+1}{(xlog_{5}x)ln2ln5}$
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, $y'=\frac{ln5(log_{5}x)+1}{(xlog_{5}x)ln2ln5}$