Answer
$y'=\frac{e^{-2x}(1-2x)}{2\sqrt (1+xe^{-2x})}$
Work Step by Step
$y= \sqrt (1+xe^{-2x})$
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, $y'=\frac{e^{-2x}(1-2x)}{2\sqrt (1+xe^{-2x})}$