ابتدا از تابع داده شده مشتق می‌گیریم سپس به جای xها عدد 1 را قرار می‌دهیم. $f\left( x \right)=\frac{{{\left( 2x-1 \right)}^{2}}}{2{{x}^{2}}}$ ${f}'\left( x \right)=\frac{{{\left( {{\left( 2x-1 \right)}^{2}} \right)}^{\prime }}\left( 2{{x}^{2}} \right)-{{\left( 2x-1 \right)}^{2}}{{\left( 2{{x}^{2}} \right)}^{\prime }}}{{{\left( 2{{x}^{2}} \right)}^{2}}}=\frac{\left( 2\times 2\left( 2x-1 \right) \right)\left( 2{{x}^{2}} \right)-{{\left( 2x-1 \right)}^{2}}\left( 4x \right)}{{{\left( 2{{x}^{2}} \right)}^{2}}}=\frac{\left( 8x-4 \right)\left( 2{{x}^{2}} \right)-{{\left( 2x-1 \right)}^{2}}\left( 4x \right)}{4{{x}^{2}}}$ ${f}'\left( 1 \right)=\frac{\left( 8\times 1-4 \right)\left( 2{{\left( 1 \right)}^{2}} \right)-{{\left( 2\times 1-1 \right)}^{2}}\left( 4\times 1 \right)}{4{{\left( 1 \right)}^{4}}}=\frac{8-4}{4}=1$