## Calculus: Early Transcendentals 8th Edition

$f'(x)=\frac{2x+2xe^x-x^2e^x}{(1+e^x)^2}$ $f''(x)=\frac{(1+e^x)(2+2e^x-x^2e^x)-2(2x+2xe^x-x^2e^x)(e^x)}{(1+e^x)^3}$
The quotient rule: $f(x)=\frac{u}{v}$, then $f'(x)=\frac{vu'-uv'}{v^2}$ In this case, $u=x^2$ and $v=1+e^x$ $f'(x)=\frac{(1+e^x)(2x)-(x^2)(e^x)}{(1+e^x)^2}=\frac{2x+2xe^x-x^2e^x}{(1+e^x)^2}$ For the second derivative, $u=2x+2xe^x-x^2e^x$ and $v=(1+e^x)^2$. Using the product rule ($(uv)'=u'v+v'u$) and the chain rule ($(u^n)'=nu^{n-1}\dot \thinspace u'$), we find: $u'=2+2e^x+2xe^x-2xe^x-x^2e^x=2+2e^x-x^2e^x$ $v'=2(1+e^x)(e^x)$ Using the quotient rule again: $f''(x)=\frac{(1+e^x)^2(2+2e^x-x^2e^x)-(2x+2xe^x-x^2e^x)(2(1+e^x)(e^x))}{(1+e^x)^4}$ $f''(x)=\frac{(1+e^x)(2+2e^x-x^2e^x)-2(2x+2xe^x-x^2e^x)(e^x)}{(1+e^x)^3}$