## Calculus: Early Transcendentals 8th Edition

$$f'(v)=-\frac{2}{3v\sqrt[3]{v^2}}-2e^v$$
$$f(v)=\frac{\sqrt[3] v-2ve^v}{v}$$ $$f(v)=\frac{v^{1/3}-2ve^v}{v}$$ $$f(v)=v^{-2/3}-2e^v$$ Therefore, the derivative of $f(v)$ is $$f'(v)=\frac{d}{dv}(v^{-2/3})-2\frac{d}{dv}(e^v)$$ $$f'(v)=-\frac{2}{3}v^{-5/3}-2e^v$$ $$f'(v)=-\frac{2}{3v^{5/3}}-2e^v$$ $$f'(v)=-\frac{2}{3v\times v^{2/3}}-2e^v$$ $$f'(v)=-\frac{2}{3v\sqrt[3]{v^2}}-2e^v$$