طبق فرمول داده شده می‌توان گفت: $m=2$، $n=1$ و $u=3{{x}^{2}}-6x+7$ است. $u=3{{x}^{2}}-6x+7\Rightarrow {u}'=6x-6$ $f\left( x \right)=\sqrt[m]{{{u}^{n}}}\to {f}'\left( x \right)=\frac{n{u}'}{m\sqrt[m]{{{u}^{m-1}}}}$ ${f}'\left( x \right)=\frac{1\times \left( 6x-6 \right)}{2\sqrt[2]{{{\left( 3{{x}^{2}}-6x+7 \right)}^{2-1}}}}=\frac{6x-6}{2\sqrt{3{{x}^{2}}-6x+7}}=\frac{\cancel{6}\left( x-1 \right)}{\cancel{2}\sqrt{3{{x}^{2}}-6x+7}}=\frac{3\left( x-1 \right)}{\sqrt{3{{x}^{2}}-6x+7}}$