# Chapter 4 - Calculating the Derivative - 4.2 Derivatives of Products and Quotients - 4.2 Exercises - Page 216: 13

${y^,}\, = \frac{{ - 17}}{{\,{{\left( {4 + t} \right)}^2}}}$

#### Work Step by Step

$\begin{gathered} y = \frac{{5 - 3t}}{{4 + t}} \hfill \\ Use\,\,the\,\,quotient\,\,rule\,\,to\,\,find\,\,{y^,}\, \hfill \\ {y^,} = \frac{{\,\left( {4 + t} \right)\,{{\left( {5 - 3t} \right)}^,} - \,\left( {5 - 3t} \right)\,{{\left( {4 + t} \right)}^,}}}{{\,{{\left( {4 + t} \right)}^2}}} \hfill \\ Then \hfill \\ {y^,}\, = \frac{{\left( {4 + t} \right)\,\left( { - 3} \right) - \,\left( {5 - 3t} \right)\,\left( 1 \right)}}{{\,{{\left( {4 + t} \right)}^2}}} \hfill \\ Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\ {y^,}\, = \frac{{ - 12 - 3t - 5 + 3t}}{{\,{{\left( {4 + t} \right)}^2}}} \hfill \\ {y^,}\, = \frac{{ - 17}}{{\,{{\left( {4 + t} \right)}^2}}} \hfill \\ \hfill \\ \end{gathered}$

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