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Calculus Question

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  • Started 2 years ago by lee

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  1. lee
    Key Master

    -------- Original Message --------
    Subject: Question
    Date: Wed, 20 Jul 2011 21:37:34 -0500
    From: Eric Post
    To: Sungwook Lee

    Can 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*}

    Posted 2 years ago #

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