Question

Differentiate the function.

22.\(g\left( x \right) = {e^{{x^2}\ln x}}\)

Short answer

The derivative of the function is \(g'\left( x \right) = x{e^{{x^2}\ln x}}\left( {1 + 2\ln x} \right)\).

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

Differentiate the function

Use the product rule and the chain ruleto differentiate thefunction as shown below:

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

Thus, the derivative of the function is \(g'\left( x \right) = x{e^{{x^2}\ln x}}\left( {1 + 2\ln x} \right)\).