\(\displaystyle{f{{\left({x}\right)}}}={2}{x}{e}^{{{x}}}\)

Differentiate

\(\displaystyle{f}'{\left({x}\right)}={\frac{{{d}{\left({2}{x}{e}^{{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}\)

\(\displaystyle{f}'{\left({x}\right)}={2}{x}\times{\frac{{{d}{\left({e}^{{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}\times{\frac{{{d}{\left({2}{x}\right)}}}{{{\left.{d}{x}\right.}}}}\)

\(\displaystyle{2}{x}\times{\frac{{{d}{\left({e}^{{{x}}}\right)}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}\times{\frac{{{d}{\left({2}{x}\right)}}}{{{\left.{d}{x}\right.}}}}\) Using the product Rule

\(\displaystyle{f}'{\left({x}\right)}={2}{x}\times{\left({e}^{{{x}}}\right)}+{e}^{{{x}}}\times{\left({2}\right)}\)

\(\displaystyle{f}'{\left({x}\right)}={2}{x}\times{e}^{{{x}}}+{2}{e}^{{{x}}}\)

\(\displaystyle{f}'{\left({x}\right)}={e}^{{{x}}}{\left({2}{x}+{2}\right)}\)

Step 2

Slope of the tangent at \(\displaystyle{\left({0},\ {0}\right)}={f}'{\left({0}\right)}={e}^{{{0}}}\times{\left({0}+{2}\right)}={2}\)

Equation of the tangent is

\(\displaystyle{\frac{{{y}-{0}}}{{{x}-{0}}}}={2}\)

\(\displaystyle{y}={2}{x}\)

Step 3

Product of slopes of perpendicular lines is -1

Therefore slope of the normal line is \(\displaystyle-{\frac{{{1}}}{{{2}}}}\)

Equation of the normal line is

\(\displaystyle{\frac{{{y}-{0}}}{{{x}-{0}}}}=-{\frac{{{1}}}{{{2}}}}{x}\)