iarc6io

2022-07-23

Differentiate the Function: $y=\sqrt{1+x{e}^{-2x}}$

gutsy8v

Expert

Solution:
$\phantom{\rule{0ex}{0ex}}y\mathrm{\prime }=\frac{1}{2}{\left(1+x{e}^{-2x}\right)}^{-\frac{1}{2}}\left(0+{e}^{-2x}-2x{e}^{-2x}\right)=\frac{\left(1-2x\right)}{2{e}^{2x}\sqrt{\left(1+x{e}^{-2x}\right)}}\phantom{\rule{0ex}{0ex}}$

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