Calculus (3rd Edition)

$$f'(x) =\frac{(3x-2)e^x}{(3x+1)^2}$$
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Recall that $(e^x)'=e^x$ Since we have $$f(x)= \frac{e^x}{3x+1}$$ then the derivative $f'(x)$, using the chain and quotient rules, is given by $$f'(x)=\frac{(3x+1)(e^x)'-e^x(3x+1)'}{(3x+1)^2} =\frac{(3x+1)e^x-3e^x}{(3x+1)^2}\\ =\frac{(3x-2)e^x}{(3x+1)^2}$$