Answer
$\displaystyle \frac{7t^{2}+3}{t(t^{2}+3)}$
Work Step by Step
By theorem 5.3/2: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$
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Here,$ u(t)=t(t^{2}+3)^{3}$
$\displaystyle \frac{du}{dt}$=...product rule...$=1\displaystyle \cdot(t^{2}+3)^{3}+t\cdot\frac{d}{dt}[(t^{2}+3)^{3}]$
$\displaystyle \frac{d}{dt}[(t^{2}+3)^{3}]$=...chain rule...=$ 3(t^{2}+3)^{2}\cdot 2t=6t(t^{2}+3)^{2}$
so,
$\displaystyle \frac{du}{dt}=(t^{2}+3)^{3}+t\cdot 6t(t^{2}+3)^{2}=(t^{2}+3)^{2}(t^{2}+3+6t^{2})=(t^{2}+3)^{2}(7t^{2}+3)$
So,$\ \ \ \displaystyle \frac{d}{dt}[\ln t(t^{2}+3)^{3}]=\frac{1}{t(t^{2}+3)^{3}}\cdot(t^{2}+3)^{2}(7t^{2}+3)=\frac{7t^{2}+3}{t(t^{2}+3)}$
