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Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 46

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Differentiate $\displaystyle y = \sqrt{1 + xe^{-2x}}$

$ \begin{equation} \begin{aligned} y' =& \frac{d}{dx} (\sqrt{1 + xe^{-2x}}) \\ \\ y' =& \frac{d}{dx} (1 + xe^{-2x})^{\frac{1}{2}} \\ \\ y' =& \frac{1}{2} (1 + xe^{-2x})^{\frac{-1}{2}} \frac{d}{dx} (1 + xe^{-2x}) \\ \\ y' =& \frac{1}{2 (1 + xe^{-2x})^{\frac{1}{2}}} \cdot \frac{d}{dx} \left( 1 + \frac{x}{e^{2x}} \right) \\ \\ y' =& \frac{1}{2 (1 + xe^{-2x})^{\frac{1}{2}}} \cdot \left[ 0 + \frac{\displaystyle (e^{2x}) \frac{d}{dx} (x) - (x) \frac{d}{dx} (e^{2x}) }{(e^{2x})^2} \right] \\ \\ y' =& \frac{1}{2 (1 + xe^{-2x})^{\frac{1}{2}}} \left[ \frac{\displaystyle e^{2x} - (x) (e^{2x} \frac{d}{dx} (2x) )}{(e^{4x})} \right] \\ \\ y' =& \frac{1}{2 (1 + xe^{-2x})^{\frac{1}{2}}} \left[ \frac{e^{2x} - 2xe^{2x}}{e^{4x}} \right] \\ \\ y' =& \frac{1}{2 (1 + xe^{-2x})^{\frac{1}{2}}} \left[ \frac{e^{2x } (1 - 2x)}{e^{4x}} \right] \\ \\ y' =& \frac{1}{2 (1 + xe^{-2x})^{\frac{1}{2}}} \left( \frac{1 - 2x}{e^{2x}} \right) \\ \\ y' =& \frac{1 - 2x}{2e^{2x} (1 + xe^{-2x})^{\frac{1}{2}}} \\ \\ & \text{ or } \\ \\ y' =& \frac{1 - 2x}{2e^{2x} \sqrt{1 + xe^{-2x}}} \end{aligned} \end{equation} $