حد خواسته شده، مشتق تابع $f.g$ در $x=2$ است. لذا:
$\left( fg \right)\left( x \right)=\left( {{x}^{2}}-x \right)\sqrt{2x}$
${{\left( fg \right)}^{\prime }}\left( x \right)={f}'\left( x \right)g\left( x \right)+{g}'\left( x \right)f\left( x \right)=\left( 2x-1 \right)\sqrt{2x}+\left( {{x}^{2}}-x \right)\frac{2}{2\sqrt{2x}}$
بنابراین:
${{\left( fg \right)}^{\prime }}\left( 2 \right)=\left( 2\times 2-1 \right)\sqrt{2\times 2}+\left( {{2}^{2}}-2 \right)\frac{1}{\sqrt{2\times 2}}\Rightarrow {{\left( fg \right)}^{\prime }}\left( 2 \right)=6+\frac{2}{2}=7$