Question

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

32. \(y = \ln \left( {1 + \ln x} \right)\)

Short answer

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

Step-by-step solution

Derivative of logarithmic functions

The derivative of a logarithmicfunction is 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 \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

Evaluate the values of  \(y'\) and \(y''\)

Evaluate the first derivative of the function as shown below:

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

Thus, the value of \(y'\) is \(y' = \frac{1}{{x\left( {1 + \ln x} \right)}}\).

Evaluate the second derivative of the function as shown below:

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

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

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