Single Variable Calculus

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Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 52

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Determine the equation of the tangent line to the curve $\displaystyle xe^y + ye^x = 1$ at the point $(0,1)$

Solving for the slope

$ \begin{equation} \begin{aligned} & \frac{d}{dx} (xe^y) + \frac{d}{dx} (ye^x) = \frac{d}{dx} (1) \\ \\ & \left[ (x) \frac{d}{dx} (e^y) + (e^y) \frac{d}{dx} (x) \right] + \left[ (y) \frac{d}{dx} (e^x) + (e^x) \frac{d}{dx} (y) \right] = 0 \\ \\ & x e^y \frac{dy}{dx} + e^y + ye^x + e^x \frac{dy}{dx} = 0 \\ \\ & xe^y y^1 + e^y + ye^x + e^x y^1 = 0 \\ \\ & xe^y y^1 + e^xy^1 = -e^y - ye^x \\ \\ & y^1 (xe^y + e^x) = -e^y - ye^x \\ \\ & y^1 = \frac{-e^y - ye^x}{xe^y + e^x} \\ \\ & y^1 = \frac{-e^1 - 1 (e^0)}{0 (e^1) + e^0} \\ \\ & y^1 = \frac{-e^1 - 1}{0 + 1} \\ \\ & y^1 = -e - 1 \end{aligned} \end{equation} $

Using Point Slope Form

$ \begin{equation} \begin{aligned} y - y_1 =& m(x - x_1) \\ \\ y - 1 =& (-e - 1) (x - 0) \\ \\ y =& x (-e - 1) + 1 \\ \\ y =& 1 - x (e + 1) \end{aligned} \end{equation} $

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