Question

ate the function. $f(v)=\frac{sqrt3{v}-6ve^{v}}{v}$

Explanation:

Step1: Rewrite the function

Rewrite $f(v)=\frac{\sqrt[3]{v}-6ve^{v}}{v}$ as $f(v)=v^{-\frac{2}{3}} - 6e^{v}$. We use the rules of exponents $\frac{\sqrt[3]{v}}{v}=\frac{v^{\frac{1}{3}}}{v^{1}}=v^{\frac{1}{3}-1}=v^{-\frac{2}{3}}$.

Step2: Differentiate term - by - term

The derivative of $y = v^{-\frac{2}{3}}$ using the power rule $\frac{d}{dv}(v^{n})=nv^{n - 1}$ is $y^\prime=-\frac{2}{3}v^{-\frac{2}{3}-1}=-\frac{2}{3}v^{-\frac{5}{3}}$. The derivative of $y=-6e^{v}$ using the rule $\frac{d}{dv}(e^{v}) = e^{v}$ is $y^\prime=-6e^{v}$.

Step3: Combine the derivatives

The derivative of $f(v)$ is $f^\prime(v)=-\frac{2}{3}v^{-\frac{5}{3}}-6e^{v}$.

Answer:

$f^\prime(v)=-\frac{2}{3v^{\frac{5}{3}}}-6e^{v}$