Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.1 Exercises - Page 325: 47

$\displaystyle \frac{2}{(t+1)}$

By theorem 5.3/2: $\displaystyle \frac{d}{dx}[\ln u]=\frac{1}{u}\cdot\frac{du}{dx}=\frac{u^{\prime}}{u},\ \ \ (u>0)$ Here, $u(t)=(t+1)^{2},\ \ \ $ $\displaystyle \frac{du}{dt}$=...chain rule...=$2(t+1)\cdot 1=2(t+1)$ $\displaystyle \frac{d}{dt}[\ln(t+1)^{2}]=\frac{1}{(t+1)^{2}}\cdot 2(t+1)=\frac{2}{(t+1)}$