## University Calculus: Early Transcendentals (3rd Edition)

$(\ln t)^2+2\ln t$
$y=t(\ln t)^2$ We differentiate both sides to obtain: $y^{\prime}=t(\ln t)^2{\prime}+(\ln t)^2(t)^{\prime}$ $y^{\prime}=t(2\ln t)(\ln t)^{\prime}+(\ln t)^2(1)$ $y^{\prime}=t(2\ln t)(\frac{1}{t})+(\ln t)^2$ $y^{\prime}=(\ln t)^2+2\ln t$