Differentiate:
$\displaystyle {\frac{1-xe^x}{x+e^x}}$
I'm new to this stuff and this one seems to be a bit tricky. If you could please break it down step by step that would be very helpful.
Use the Quotient Rule
$\displaystyle \frac{dy}{dx} = \frac{(x + e^x)\frac{d}{dx}(1 - xe^x) - (1 - xe^x)\frac{d}{dx}(x + e^x)}{(x + e^x)^2}$
$\displaystyle = \frac{(x + e^x)(-e^x - xe^x) - (1 - xe^x)(1 + e^x)}{(x + e^x)^2}$
$\displaystyle = \frac{-e^x(x + e^x)(1 + x) - (1 - xe^x)(1 + e^x)}{(x + e^x)^2}$