## Calculus: Early Transcendentals 8th Edition

$y'=\frac{ln(x)+1}{xln(x)}$
Start with the function: $y=ln(xln(x))$. Let $u = xln(x)$. Use the product rule to find u': $u'=1*ln(x)+x*\frac{1}{x} = ln(x)+1$. Substitute u into the equation: $y=ln(u)$. Use chain rule to differentiate: $y=\frac{u'}{u}$. Substitute expressions for u and u' back into the equation: $y'=\frac{ln(x)+1}{xln(x)}$.