Question

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

29. \(y = \sqrt x \ln x\)

Short answer

The values are \(y' = \frac{{2 + \ln x}}{{2\sqrt x }}\) and \(y'' = - \frac{{\ln x}}{{4x\sqrt x }}\).

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

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

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

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

The value of \(y'\) is \(y' = \frac{{2 + \ln x}}{{2\sqrt x }}\).

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

\(\begin{aligned}{c}y''&= \frac{d}{{dx}}\left( {\frac{{2 + \ln x}}{{2\sqrt x }}} \right)\\&= \frac{{2\sqrt x \left( {\frac{1}{x}} \right) - \left( {2 + \ln x} \right)\left( {\frac{1}{{\sqrt x }}} \right)}}{{{{\left( {2\sqrt x } \right)}^2}}}\\&= \frac{{\left( {\frac{2}{{\sqrt x }}} \right) - \left( {\frac{{2 + \ln x}}{{\sqrt x }}} \right)}}{{4x}}\\&= \frac{{\frac{{2 - \left( {2 + \ln x} \right)}}{{\sqrt x }}}}{{4x}}\\&= \frac{{2 - 2 - \ln x}}{{\sqrt x \left( {4x} \right)}}\\&= - \frac{{\ln x}}{{4x\sqrt x }}\end{aligned}\)

The value of \(y''\) is \(y''= - \frac{{\ln x}}{{4x\sqrt x }}\).

Thus, the values of \(y'\) and \(y''\) is \(y' = \frac{{2 + \ln x}}{{2\sqrt x }}\) and \(y'' = - \frac{{\ln x}}{{4x\sqrt x }}\).