## Calculus (3rd Edition)

$$f'(x)= \frac{2 e^{x}-e^{x+1}-2 x e^{x}-1}{\left(2 e^{x}-1\right)^{2}}.$$
Recall that $(e^x)'=e^x$ Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since we have $$f(x)= \frac{e^{x+1}+x}{2e^x-1}$$ then the derivative $f'(x)$, using the chain and quotient rules, is given by \begin{align*}f'(x)&=\frac{(2e^x-1)(e^{x+1}+x)'-(e^{x+1}+x)(2e^x-1)'}{(2e^x-1)^2} \\ &=\frac{(2e^x-1)(e^{x+1}+1)-(e^{x+1}+x)(2e^x)}{(2e^x-1)^2} \\ &= \frac{2 e^{x}-e^{x+1}-2 x e^{x}-1}{\left(2 e^{x}-1\right)^{2}}. \end{align*}