First we simplify

\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}+{3}{x}+{2}}}{{{x}+{1}}}}={\frac{{{\left({x}+{1}\right)}{\left({x}+{2}\right)}}}{{{x}+{1}}}}={x}+{2}\)

Then it's just

\(\displaystyle{f}'{\left({x}\right)}={1}\)

\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{x}^{{{2}}}+{3}{x}+{2}}}{{{x}+{1}}}}={\frac{{{\left({x}+{1}\right)}{\left({x}+{2}\right)}}}{{{x}+{1}}}}={x}+{2}\)

Then it's just

\(\displaystyle{f}'{\left({x}\right)}={1}\)