Answer
$\dfrac{e^x|x+2|}{[1+(xe^x+e^x)^2]^{3/2}}$
Work Step by Step
As we are given that $y=xe^x$
Let us consider $f(x)=y=xe^x$
Write formula 11.
$\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$
$f'(x)=e^x+xe^x$ and $f''(x)=e^x(2+x)$
Thus, $\kappa(x)=\dfrac{|e^x(2+x)|}{[1+(1+x)e^x)^2]^{3/2}}$
or, $\kappa(x)=\dfrac{e^x|x+2|}{[1+(xe^x+e^x)^2]^{3/2}}$
