Question

3-34:Differentiate the function.

32. \(f\left( v \right) = \frac{{\sqrt[3]{v} - 2v{e^v}}}{v}\)

Short answer

The derivative of given function is \(\frac{{df}}{{dv}} = - \frac{2}{3}{v^{ - 5/3}} + 2{e^v}\).

Step-by-step solution

Different rules of differentiation

The given function is \(f\left( v \right) = \frac{{\sqrt[3]{v} - 2v{e^v}}}{v}\).

The sum rule is:\(\frac{d}{{dx}}\left( {g\left( x \right) + f\left( x \right)} \right) = \frac{d}{{dx}}\left( {g\left( x \right)} \right) + \frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

And, the power rule is: \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)

Evaluating Derivative using all these rules

Differentiating with respect to\(v\)as:

\(\begin{align}f'\left( v \right) &= \frac{d}{{dv}}\left( {\frac{{\sqrt[3]{v} - 2v{e^v}}}{v}} \right)\\ &= \frac{d}{{dv}}\left( {\frac{{\sqrt[3]{v}}}{v} - \frac{{2v{e^v}}}{v}} \right)\\ &= \frac{d}{{dv}}\left( {{v^{ - \frac{2}{3}}} - 2{e^v}} \right)\\ &= - \frac{2}{3}{v^{ - \frac{5}{3}}} - 2{e^v}\\ &= - \frac{{2{v^{ - \frac{5}{3}}} + 6{e^v}}}{3}\\ &= - \frac{2}{3}{v^{ - 5/3}} + 2{e^v}\end{align}\)

Hence, the derivative of given function is \(\frac{{df}}{{dv}} = - \frac{2}{3}{v^{ - 5/3}} + 2{e^v}\).