# Find equations of the tangent line and normal line to

Find equations of the tangent line and normal line to the given curve at the specified point. $$\displaystyle{y}={2}{x}{e}^{{{x}}},\ {\left({0},\ {0}\right)}$$

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Cleveland Walters
Step 1
$$\displaystyle{f{{\left({x}\right)}}}={2}{x}{e}^{{{x}}}$$
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
$$\displaystyle{\frac{{{y}-{0}}}{{{x}-{0}}}}=-{\frac{{{1}}}{{{2}}}}{x}$$
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Robert Pina
Step 1
The given equation is,
$$\displaystyle{y}={2}{x}{e}^{{{x}}}$$
Step 2
The slope is obtained as follows.
$$\displaystyle{y}={2}{x}{e}^{{{x}}}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={2}{\left({e}^{{{x}}}+{x}{e}^{{{x}}}\right)}$$
$$\displaystyle{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}_{{{x}={0}}}={2}{\left({e}^{{{0}}}+{0}\right)}={2}$$
$$\displaystyle{m}={2}$$
Step 3
The equation of tangent line is computed as follows.
$$\displaystyle{\left({y}-{y}_{{{1}}}\right)}={m}{\left({x}-{x}_{{{1}}}\right)}$$
$$\displaystyle{\left({y}-{0}\right)}={2}{\left({x}-{0}\right)}$$
$$\displaystyle{y}={2}{x}$$
Step 4
The equation of normal line is computed as follows.
$$\displaystyle{\left({y}-{y}_{{{1}}}\right)}=-{\frac{{{1}}}{{{m}}}}{\left({x}-{x}_{{{1}}}\right)}$$
$$\displaystyle{\left({y}-{0}\right)}=-{\frac{{{1}}}{{{2}}}}{\left({x}-{0}\right)}$$
$$\displaystyle{2}{y}=-{x}$$
nick1337

Step 1
To obtain slope, we will differentiate the given equation using product rule:
$$\frac{dy}{dx}=\frac{d}{dx}(2xe^{x})$$
$$\frac{dy}{dx}=2x\frac{d(e^{x})}{dx}+e^{x}\times\frac{d(2x)}{dx}$$
$$\frac{dy}{dx}=2xe^{x}+e^{x}\times2$$
$$\frac{dy}{dx}=2xe^{x}+2e^{x}$$
$$\frac{dy}{dx}=e^{x}(2x+2)$$
Now, slope of tangent at $$(0,\ 0)=e^{0}(2\times0+2)$$
NSK
So, m=2
Step 2
Now, the equation of tangent line
$$m=\frac{y-y_{1}}{x-x_{1}}$$
$$2=\frac{y-0}{x-0}$$
So, Equation of tangent line is y=2x
Step 3
We know that the slope of normal line is additive inverse of reciprocal of slope of tangent line.
Slope of normal line, $$m=\frac{-1}{2}$$
Now, the equation of normal line
$$m=\frac{y-y_{1}}{x-x_{1}}$$
$$\frac{-1}{2}=\frac{y-0}{x-0}$$
So, Equation of normal line is $$y=\frac{-1}{2}x$$