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Find the equation of the tangent line to the curve
y=2e^(x) at the point
(0,2)

y^(')=2e^(x)

Find the equation of the tangent line to the curve $y=2 e^{x}$ at the point $(0,2)$ $\newline$ $y^{\prime}=2 e^{x}$

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Find equations of tangent lines using limits

Full solution

Q. Find the equation of the tangent line to the curve

y=2 e^{x}

at the point

(0,2)

\newline

y^{\prime}=2 e^{x}

Find Derivative: First, we need to find the derivative of $y=2e^{x}$ to get the slope of the tangent line at any point $x$ . The derivative $y'$ is given as $y'=2e^{x}$ .
Calculate Slope at $(0,2)$ : Next, we substitute $x=0$ into the derivative to find the slope of the tangent line at the point $(0,2)$ . Plugging in, we get $y'(0)=2e^{(0)}$ .
Use Point-Slope Form: Since $e^{0}$ equals $1$ , the calculation simplifies to $y'(0)=2\cdot1=2$ . So, the slope of the tangent line at $(0,2)$ is $2$ .
Simplify Equation: Now, we use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is the point on the line. Substituting $m=2$ and $(x_1, y_1)=(0, 2)$ , we get $y - 2 = 2(x - 0)$ .
Simplify Equation: Now, we use the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is the point on the line. Substituting $m=2$ and $(x_1, y_1)=(0, 2)$ , we get $y - 2 = 2(x - 0)$ . Simplifying the equation, we find $y - 2 = 2x$ . Then, adding $2$ to both sides, we get $y = 2x + 2$ .

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