## Calculus 8th Edition

$y'=\frac{e^{-2x}(1-2x)}{2\sqrt (1+xe^{-2x})}$
$y= \sqrt (1+xe^{-2x})$ Let $t=(1+xe^{-2x})$ This implies $\frac{dt}{dx}=e^{-2x}(1-2x)$ ...(1) Now, $y'=\frac{d}{dt}\sqrt (t)\frac{dt}{dx}$ $=\frac{1}{2\sqrt t}\frac{dt}{dx}$ From equation (1), we get Hence, $y'=\frac{e^{-2x}(1-2x)}{2\sqrt (1+xe^{-2x})}$