Answer
\[{k^,}\,\left( x \right) = \frac{{21}}{{{{\left( {4x + 7} \right)}^2}}}\]
Work Step by Step
\[\begin{gathered}
k\,\left( x \right) = \frac{{3x}}{{4x + 7}} \hfill \\
Find\,\,the\,\,derivative\,\,of\,\,the\,\,function \hfill \\
{k^,}\,\left( x \right) = \frac{d}{{dx}}\,\,\left[ {\frac{{3x}}{{4x + 7}}} \right] \hfill \\
Use\,\,the\,\,quotient\,\,rule \hfill \\
{k^,}\,\left( x \right) = \frac{{\,\left( {4x + 7} \right)\,{{\left( {3x} \right)}^,} - 3x\,{{\left( {4x + 7} \right)}^,}}}{{\,{{\left( {4x + 7} \right)}^2}}} \hfill \\
Then\, \hfill \\
{k^,}\,\left( x \right) = \frac{{\left( {4x + 7} \right)\,\left( 3 \right) - 3x\,\left( 4 \right)}}{{{{\left( {4x + 7} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms\, \hfill \\
{k^,}\,\left( x \right) = \frac{{12x + 21 - 12x}}{{{{\left( {4x + 7} \right)}^2}}} \hfill \\
{k^,}\,\left( x \right) = \frac{{21}}{{{{\left( {4x + 7} \right)}^2}}} \hfill \\
\hfill \\
\end{gathered} \]
