## Calculus with Applications (10th Edition)

${h^,}\,\left( 3 \right) = \frac{{13}}{{16}}$
$\begin{gathered} Let\,\,h\,\left( x \right) = \frac{{f\,\left( x \right)}}{{g\,\left( x \right)}} \hfill \\ Find\,\,{h^,}\,\left( x \right)\,\,using\,\,the\,\,quotient\,\,rule \hfill \\ \,\,{\left[ {\frac{u}{v}} \right]^,} = \frac{{v{u^,} - u{v^,}}}{{{v^2}}} \hfill \\ Then \hfill \\ {h^,}\,\left( x \right) = \frac{{g\,\left( x \right){f^,}\,\left( x \right) - f\,\left( x \right){g^,}\,\left( x \right)}}{{\,\,{{\left[ {g\,\left( x \right)} \right]}^2}}} \hfill \\ Then \hfill \\ {h^,}\,\left( 3 \right) = \frac{{g\,\left( 3 \right){f^,}\,\left( 3 \right) - f\,\left( 3 \right){g^,}\,\left( 3 \right)}}{{\,\,{{\left[ {g\,\left( 3 \right)} \right]}^2}}} \hfill \\ Substitute\,\,the\,\,given\,\,values\,\,and\,\,simplify \hfill \\ {h^,}\,\left( 3 \right) = \frac{{\,\left( 4 \right)\,\left( 8 \right) - \,\left( 9 \right)\,\left( 5 \right)}}{{\,{{\left( 4 \right)}^2}}} \hfill \\ {h^,}\,\left( 3 \right) = \frac{{13}}{{16}} \hfill \\ \end{gathered}$