Answer
\[{y^,} = 2{x^{ - 1/2}} + \frac{3}{2}{x^{ - 3/2}}\]
Work Step by Step
\[\begin{gathered}
y = \frac{{4x - 3}}{{\sqrt x }} \hfill \\
\sqrt x \,\,can\,\,be\,\,written\,\,as\,\,{x^{1/2}} \hfill \\
y = \frac{{4x - 3}}{{{x^{1/2}}}} \hfill \\
Use\,\,the\,\,product\,\,rule\,\,to\,\,find\,\,{y^,} \hfill \\
{y^,} = \frac{{{x^{1/2}}\,{{\left( {4x - 3} \right)}^,} - \left( {4x - 3} \right)\,{{\left( {{x^{1/2}}\,} \right)}^,}}}{{\,{{\left( {{x^{1/2}}\,} \right)}^2}}} \hfill \\
Then \hfill \\
{y^,} = \frac{{{x^{1/2}}\,\left( 4 \right) - \,\left( {4x - 3} \right)\,\left( {\frac{1}{2}{x^{ - 1/2}}} \right)}}{{\,{{\left( {{x^{1/2}}} \right)}^2}}} \hfill \\
Simplify\,\,by\,\,multiplying\,\,and\,\,combining\,\,terms \hfill \\
{y^,} = \frac{{4{x^{1/2}} - 2{x^{1/2}} + \frac{3}{2}{x^{1/2}}}}{x} \hfill \\
{y^,} = \frac{{2{x^{1/2}} + \frac{3}{2}{x^{ - 1/2}}}}{x} \hfill \\
{y^,} = 2{x^{ - 1/2}} + \frac{3}{2}{x^{ - 3/2}} \hfill \\
\end{gathered} \]
