Question

Find \(y'\) and \(y''\).

30. \(y = \frac{{\ln x}}{{1 + \ln x}}\)

Short answer

The values of \(y'\) and \(y''\) is \(y' = \frac{1}{{x{{\left( {1 + \ln x} \right)}^2}}}\) and \(y'' = - \frac{{3 + \ln x}}{{{x^2}{{\left( {1 + \ln x} \right)}^3}}}\).

Step-by-step solution

Derivative of logarithmic functions

The derivative of a logarithmicfunctionis shown below:

1. \(\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}\)
2. \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)
3. \(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\) or \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

 Step 2: Evaluate the values of  \(y'\) and \(y''\)

Use the quotient rule to evaluate the first derivative of the function as shown below:

\(\begin{aligned}{c}y'&= \frac{d}{{dx}}\left( {\frac{{\ln x}}{{1 + \ln x}}} \right)\\&= \frac{{\left( {1 + \ln x} \right)\frac{d}{{dx}}\left( {\ln x} \right) - \left( {\ln x} \right)\frac{d}{{dx}}\left( {1 + \ln x} \right)}}{{{{\left( {1 + \ln x} \right)}^2}}}\\&= \frac{{\left( {1 + \ln x} \right)\left( {\frac{1}{x}} \right) - \left( {\ln x} \right)\left( {\frac{1}{x}} \right)}}{{{{\left( {1 + \ln x} \right)}^2}}}\\&= \frac{{\frac{{1 + \ln x - \ln x}}{x}}}{{{{\left( {1 + \ln x} \right)}^2}}}\\&= \frac{1}{{x{{\left( {1 + \ln x} \right)}^2}}}\end{aligned}\)

The value of \(y'\) is \(y' = \frac{1}{{x{{\left( {1 + \ln x} \right)}^2}}}\).

Evaluate the second derivative of the function as shown below:

The value of \(y''\) is \(y'' = - \frac{{3 + \ln x}}{{{x^2}{{\left( {1 + \ln x} \right)}^3}}}\).

Thus, the values of \(y'\) and \(y''\) is \(y' = \frac{1}{{x{{\left( {1 + \ln x} \right)}^2}}}\) and \(y'' = - \frac{{3 + \ln x}}{{{x^2}{{\left( {1 + \ln x} \right)}^3}}}\).