-------- Original Message --------

Subject: Question

Date: Wed, 20 Jul 2011 21:37:34 -0500

From: Eric Post

To: Sungwook LeeCan you help me solve problem 11 on page 166 of 3.3 of the Essential Calculus book. I think I'm missing a step somewhere but I don't know where.

Solution: Eric, the question is to differentiate the function $F(t)=\ln\frac{(2t+1)^3}{(3t-1)^4}$. First step is to simplify the given logarithmic function using laws of logarithms:

\begin{eqnarray*}F(t)&=&\ln\frac{(2t+1)^3}{(3t-1)^4}\\&=&\ln(2t+1)^3-\ln(3t-1)^4\\&=&3\ln(2t+1)-4\ln(3t-1).\end{eqnarray*} Using the chain rule we find that $\frac{d}{dt}\ln(2t+1)=\frac{2}{2t+1}$ and $\frac{d}{dt}\ln(3t-1)=\frac{3}{3t-1}$. Hence,

\begin{eqnarray*}\frac{d}{dt}F(t)&=&3\frac{d}{dt}\ln(2t+1)-4\frac{d}{dt}\ln(3t-1)\\&=&\frac{6}{2t+1}-\frac{12}{3t-1}\\&=&-\frac{6(t+3)}{(2t+1)(3t-1)}.\end{eqnarray*}