Question

Differentiate the function.

16. \(p\left( v \right)=\frac{\ln v}{1-v}\)

Short answer

The derivative of the function is \(\frac{1}{{v\left( {1 - v} \right)}} + \frac{{\ln v}}{{{{\left( {1 - v} \right)}^2}}}\)

Step-by-step solution

Use the derivative of logarithmic function

Rule 2:The derivative of \(\ln x\) is,

\(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)

Evaluating the derivative of given function

\(p\left( v \right) = \frac{{\ln v}}{{1 - v}}\)

Differentiate \(p\left( v \right)\)with respect to\(v\),

\(\begin{aligned}{c}\frac{d}{{dv}}p\left( v \right)&= \frac{d}{{dv}}\left( {\frac{{\ln v}}{{1 - v}}} \right)\\&= \frac{1}{{\left( {1 - v} \right)}}\frac{{d\ln v}}{{dv}} + \ln v\frac{d}{{dv}}\left( {\frac{1}{{1 - v}}} \right)\\&= \frac{1}{{\left( {1 - v} \right)}} \times \frac{1}{v} + \ln v\frac{d}{{dv}}{\left( {1 - v} \right)^{ - 1}}\\&= \frac{1}{{v\left( {1 - v} \right)}} + \ln v\frac{{d{{\left( {1 - v} \right)}^{ - 1}}}}{{d\left( {1 - v} \right)}} \times \frac{{d\left( {1 - v} \right)}}{{dv}}\\&= \frac{1}{{v\left( {1 - v} \right)}} + \ln v\left( { - 1} \right){\left( {1 - v} \right)^{ - 1 - 1}} \times \left( {0 - 1} \right)\\&= \frac{1}{{v\left( {1 - v} \right)}} + \ln v{\left( {1 - v} \right)^{ - 2}}\\&= \frac{1}{{v\left( {1 - v} \right)}} + \frac{{\ln v}}{{{{\left( {1 - v} \right)}^2}}}\end{aligned}\)

Thus, the value of the derivative is \(\frac{1}{{v\left( {1 - v} \right)}} + \frac{{\ln v}}{{{{\left( {1 - v} \right)}^2}}}\).